$n \in \mathbb{Z}$, hence $\Psi(b_s) = \Psi(a_s)$, $\Psi(b_t) = \Psi(a_t)$, $\Psi(b_n)+n\mu < \Psi(a_s)+s\mu$ whenever $n < s$ and $n > t$ and $\Psi(b_n)+n\mu \le \Psi(a_s)+s\mu$ whenever $n \in [s, t]$, hence finally, $\nu^+(h, \mu) = \nu^+(f, \mu)$ and $\nu^-(h, \mu) = \nu^-(f, \mu)$. $\square$

**Corollary 19.6.** Let $f(x) = \sum_{n=-\infty}^{+\infty} a_n x^n \in \mathcal{A}(\Gamma(0, r_1, r_2))$ (with $0 < r_1 < r_2$). The function $\mu \to \Psi(f, \mu)$ defined in $\log r_1, \log r_2$ is equal to $\sup_{n \in \mathbb{Z}}(\Psi(a_n)+n\mu)$. Next, we have $\Psi(f(x)) \le \Psi(f, \Psi(x))$ whenever $x \in \Gamma(0, r_1, r_2)$ and the equality holds in all of $\Gamma(0, r_1, r_2)$ except in finitely many classes of each circle $C(0, r)$ ($r_1 < r < r_2$). The right-side derivative (respectively the left-side derivative) of the function $\Psi(f, \cdot)$ at $\mu$ is equal to $\nu^+(f, \mu)$ (respectively to $\nu^-(f, \mu)$). Moreover, if the function in $\mu$ $\Psi(f, \mu)$ is not derivable at $\mu$, then $\mu$ lies in $\Psi(\mathbb{K})$.

Further, the function $\Psi(f, \cdot)$ is convex in $\log r_1, \log r_2$. Next, given another $g \in \mathcal{A}(\Gamma(0, r_1, r_2))$, the functions $\nu^+$ and $\nu^-$ satisfy

$$ \nu^+(fg, \mu) = \nu^+(f, \mu) + \nu^+(g, \mu), \quad \nu^-(fg, \mu) = \nu^-(f, \mu) + \nu^-(g, \mu). $$

Moreover, the function $\nu^+(f, \cdot)$ is continuous on the right and the function $\nu^-(f, \cdot)$ is continuous on the left at each point $\mu$. They are continuous at $\mu$ if and only if they are equal.

**Proof.** All statements hold in all annuli $\Gamma(0, r', r'')$ with $r_1 < r' < r'' < r_2$ because the restriction of $f$ to $\Gamma(0, r', r'')$ belongs to $H(\Gamma(0, r', r''))$. $\square$

**Proposition 19.7.** Let $\mu \in \mathbb{R}$ and let $f, g \in H(C(0, \theta^\mu))$. Then, $\nu^+(fg, \mu) = \nu^+(f, \mu) + \nu^+(g, \mu)$ and $\nu^-(fg, \mu) = \nu^-(f, \mu) + \nu^-(g, \mu)$.

**Proof.** By Proposition 19.2, the relations are obvious when $f$ and $g \in R(C(0, \theta^\mu))$ because there is an annulus $\Gamma(0, r_1, r_2) \supset C(0, \theta^\mu)$ such that $f, g \in R(\Gamma(0, r_1, r_2))$. Now, by Corollary 19.6, we may extend them to $H(C(0, \theta^\mu))$ by taking $h$ and $\ell \in R(C(0, \theta^\mu))$ such that $\Psi(f-h, \mu) < \Psi(f, \mu)$ and $\Psi(g-\ell, \mu) < \Psi(g, \mu)$. $\square$

**Proposition 19.8.** Let $r_1, r_2 \in \mathbb{R}$ and let $f, g \in \mathcal{A}(\Gamma(0, r_1, r_2))$ having no zero in $\Gamma(0, r_1, r_2)$ and satisfying $\nu(f, \mu) \ne \nu(g, \mu)$, $\forall \mu \in \log r_1, \log r_2$. Then, both $\nu^+(f+g, \mu)$ and $\nu^-(f+g, \mu)$ are equal either to $\nu(f, \mu)$ or to $\nu(g, \mu)$.
**Proof.** Let $\mu_j = \log(r_j)$, $j = 1, 2$. Since both $f, g$ have no zero in $\Gamma(0, r_1, r_2)$, $\nu(f, \mu)$ is a constant integer $s$ and $\nu(g, \mu)$ is a constant integer $t \neq s$. Consequently, $\Psi(f, \mu)$ is of the form $a + s\mu$ and $\Psi(g, \mu)$ is of the form $b + t\mu$, therefore the two functions in $\mu$ can coincide at most at one point in $[\mu_1, \mu_2]$. So, by Proposition 19.1, we have $\Psi(f + g, \mu) = \max(\Psi(f, \mu), \Psi(g, \mu))$ for all $\mu \in [\log(r_1), \log(r_2)]$ except maybe at all point. But then, by continuity, the equality holds in all $[\log(r_1), \log(r_2)]$.

Let us fix $\mu_0 \in ]\mu_1, \mu_2[$. Suppose $\Psi(f + g, \mu) = \Psi(f, \mu)$ in a neighborhood $]\mu_1, \mu_2[$ of $\mu_0$. Then, of course, $\nu(f + g, \mu_0) = \nu(f, \mu_0)$. Suppose now that $\Psi(f + g, \mu) = \Psi(f, \mu)$ in a left neighborhood $]\mu_1, \mu_0[$ of $\mu_0$ and $\Psi(f + g, \mu) = \Psi(g, \mu)$ in a right neighborhood $[\mu_0, \mu_2[$ of $\mu_0$, which implies $\Psi(f, \mu) > \Psi(g, \mu) \forall \mu \in ]\mu_1, \mu_0[$ and $\Psi(f, \mu) < \Psi(g, \mu) \forall \mu \in ]\mu_0, \mu_2[$. Then, we have $\nu(f + g, \mu) = \nu(f, \mu) \forall \mu \in ]\mu_1, \mu_0[$ and $\nu(f + g, \mu) = \nu(g, \mu) \forall \mu \in ]\mu_0, \mu_2[$. Consequently, since $\nu^+$ is continuous on the left and $\nu^-$ is continuous on the right, we can check that both $\nu^+(f + g, \mu_0)$ and $\nu^-(f + g, \mu_0)$ are equal either to $\nu(f\mu_0)$ or to $\nu(g, \mu_0)$. $\square$

**Theorem 19.9.** Let $f \in \mathcal{A}(\mathbb{K}\setminus d(0, R))$. There exists $q \in \mathbb{N}$ such that

$$ \lim_{r \to +\infty} |f|(r)r^q = +\infty. $$

**Proof.** Let $s \in ]R, +\infty[$ be such that $\nu^+(f, \log s) = \nu^-(f, \log s)$ and let $\tau = \nu^+(f, \log s)$. Thus, $\Psi(f, \mu)$ has a derivative at $\log s$ equal to $\tau$. Consequently, since by Proposition 19.1 $\Psi(f, \mu)$ is convex, we have $\Psi(f, \mu) - \Psi(f, \log s) \ge \tau(\mu - \log s)$. Therefore,

$$ \lim_{\mu \to +\infty} [\Psi(f, \mu) + (1 - \tau)\mu] = +\infty $$

i.e. $\lim_{r \to +\infty} |f|(r)r^{(1-\tau)} = +\infty$. Finally, we can take $q = \max(0, 1 - \tau)$. $\square$
# Chapter 20

## Vanishing Along a Monotonous Filter

Throughout this chapter, the set $D$ is supposed to be infraconnected.

From Chapter 7, we know that there exists a spherically complete algebraically closed extension $\tilde{K}$ of $K$ whose residue class field is not countable and whose valuation group is equal to $\mathbb{R}$. Given a subset $D$ of $K$, we denote by $\tilde{D}$ the subset

$$D \cup \left\{ \bigcup_{a \in \tilde{D}} \hat{d}(a, \delta(a, (K \setminus D))) \right\}.$$

The question of whether an analytic element can tend to zero along a monotonous filter is known to be one of the main problems which happen with $p$-adic analytic functions [35], [36], [44], [50]. Here, we do not describe $T$-filters. However, we describe sufficient conditions to prevent analytic elements to vanish along a monotonous filter, which is sufficient to study analytic and meromorphic functions inside disks or annuli.

We apply the results to characteristic functions and to the image of an infraconnected set.

**Definitions.** Let $f \in H(D)$ and let $\mathcal{F}$ be a monotonous filter on $D$. When $\mathcal{F}$ is decreasing (respectively increasing) of center $a$ and diameter $S$, $f$ is said to be *strictly vanishing* along $\mathcal{F}$ if $\lim_{\mathcal{F}} f(x) = 0$,
and if there exists $S' > S$ (respectively $S' < S$) such that for every $r \in ]S, S']$ (respectively $r \in [S', S]$), we have $_D\varphi_{a,r}(f) \neq 0$.

When $\mathcal{F}$ is decreasing with no center in $\mathbb{K}$, it admits a canonical basis $(D_n)_{n \in \mathbb{N}}$ with $D_n = d(a_n, r_n) \cap D$ and then $f$ is said to be strictly vanishing along $\mathcal{F}$ if $\lim_{\mathcal{F}} f(x) = 0$ and if there exists $S' > S$ such that $_D\varphi_{a_{n+1},r}(f) \neq 0$ whenever $r \in [r_n, S']$, whenever $n \in \mathbb{N}$. Actually, $\mathcal{F}$ admits a center $\alpha$ in $\hat{\mathbb{K}}$ and then the definition given for decreasing filters with a center in $\mathbb{K}$ also applies and is obviously equivalent.

Lemma 20.1 just translates these definitions by using the function $\Psi$.

**Lemma 20.1.** Let $f \in H(D)$ and let $\mathcal{F}$ be a decreasing (respectively an increasing) filter of center $a$ and diameter $S$ on $D$. Then, $f$ is strictly vanishing along $\mathcal{F}$ if and only if there exists $S' > S$ (respectively $S' \in ]0, S[$) such that $\Psi_a(f, \log S) = -\infty$, $\Psi_a(f, \mu) > -\infty$ whenever $\mu \in ]\log S, \log S[$] (respectively $[\log S', \log S[$).

Let $\mathcal{G}$ be a decreasing filter with no center, of diameter $S$, and canonical basis $(D_n)_{n \in \mathbb{N}}$, with $D_n = d(a_n, r_n) \cap D$. Then, $f$ is strictly vanishing along $\mathcal{G}$ if and only if there exists $S' > S$ such that $\lim_{n \to \infty} \Psi_{a_n}(f, \log r_n) = -\infty$ and $\Psi_{a_n}(f, \log r) > -\infty$ for $r_n \le r < S'$, $\forall n \in \mathbb{N}$.

**Lemma 20.2.** Let $f \in H(D)$ and let $\mathcal{F}$ be a monotonous filter such that $f$ is strictly vanishing along $\mathcal{F}$. Then, $f$ is properly vanishing along $\mathcal{F}$.

**Proof.** Let $S = \text{diam}(\mathcal{F})$. Let $(D_n)$ be a canonical basis of $\mathcal{F}$ and suppose that $f$ is not properly vanishing along $\mathcal{F}$. Since $f$ is vanishing along $\mathcal{F}$, there exists $q \in \mathbb{N}$ such that $f(x) = 0$ whenever $x \in D_q$. But then, we can check that $f$ does not satisfy the definition of an analytic element strictly vanishing along $\mathcal{F}$ because for every multiplicative semi-norm $_D\varphi_{a,r}$ whose circular filter is secant to $D_q$, we have $_D\varphi_{a,r}(f) = 0$. In particular, this applies to $_D\varphi_{a,r}$, for $r_q < r < S$ (respectively $S < r < r_q$), when $\mathcal{F}$ is increasing (respectively decreasing) of center $a$, and $D_q = D \cap \Gamma(a, r_q, S)$ (respectively $D_q = D \cap \Gamma(a, S, r_q)$), and this applies to $\varphi_{a_{q+1},r_q}$, when $\mathcal{F}$ has no center, whereas $D_q = D \cap d(a_{q+1}, r_q)$. $\square$
**Proposition 20.3.** Let $a \in \tilde{D}$ and $b \in D$ and let $f \in H(D)$ satisfy $f(b) \neq 0$ and $\phi_{a,r}(f) = 0$ for some $r \in ]0, |a-b|]$. If $\phi_{a,|a-b|}(f) = 0$, then $f$ is strictly vanishing along an increasing filter of center $b$ and diameter $S \le |a-b|$. If $\phi_{a,|a-b|}(f) \neq 0$, then $f$ is strictly vanishing along a decreasing filter of center $a$ and diameter $S \in [r, |a-b|]$.

**Proof.** First, suppose $\phi_{a,|a-b|}(f) = 0$, hence we have

$$ \Psi_b(f, \log |a-b|) = \Psi_a(f, \log |a-b|) = -\infty. $$

Since $f(b) \neq 0$, by Proposition 19.4, we know that $\lim_{\mu \to -\infty} \Psi_b(f, \mu) = \Psi(f(b))$, hence there exists a unique $\gamma \le \log |a-b|$ such that $\Psi_b(f, \gamma) = -\infty$ and $\Psi_b(f, \mu) > -\infty$ whenever $\mu \in ]\gamma, \log |a-b|[$. Therefore, $f$ is strictly vanishing along the increasing filter of center $b$ and diameter $S = \theta^\gamma$.

Now, suppose $\phi_{a,|a-b|}(f) \neq 0$. Since $\phi_{a,r}(f) = 0$, we have $\Psi_a(f, \log r) = -\infty$ and $\Psi_a(f, \log |a-b|) > -\infty$, hence there exists a unique $\gamma \in [\log r, \log |a-b|[$ such that $\Psi_a(f, \gamma) = -\infty$ and $\Psi_a(f, \mu) > -\infty$ whenever $\mu \in ]\gamma, \log |a-b|]$, so $f$ is strictly vanishing along the decreasing filter of center $a$ and diameter $S = \theta^\gamma$. $\square$

**Proposition 20.4.** Let $a \in \tilde{D}$ and $b \in D$ and let $f \in H(D)$ satisfy $f(b) \neq 0$ and $\phi_{a,r}(f) = 0$ for some $r \in \mathbb{R}_+$. Then $f$ is strictly vanishing along a monotonous filter with a center.

**Proof.** If $r \le |a-b|$, the statement comes directly from Proposition 20.3. If $r > |a-b|$, then we have $\Psi_b(f, \log r) = \Psi_a(f, \log r) = -\infty$, whereas $\lim_{\mu \to -\infty} \Psi_b(f, \mu) = \Psi(f(b))$, hence there exists a unique $\gamma \le \log r$ such that $\Psi_b(f, \gamma) = -\infty$ and $\Psi_b(f, \mu) > -\infty$ whenever $\mu < \gamma$. Thus, $f$ is strictly vanishing along the increasing filter of center $b$ and diameter $S = \theta^\gamma$. $\square$

**Proposition 20.5.** Let $\mathcal{F}$ be a monotonous filter on $D$ and let $f \in H(D)$ be strictly vanishing along $\mathcal{F}$. Then, $\mathcal{F}$ is pierced.

**Proof.** Suppose that $\mathcal{F}$ is increasing (respectively decreasing) of center $a$ and diameter $S$ and is not pierced. There exists an annulus $\Gamma(a, S, S')$ (respectively $\Gamma(a, S', S)$) included in $D$ such that $\Psi_a(f, \log S) = -\infty$ and $\Psi_a(f, \mu) > -\infty$ whenever $\mu \in [\log S', \log S[$ (respectively $\mu \in ]\log S, \log S']$). Hence, by Corollary 19.3, we know that $\Psi_a(f, \mu)$ is bounded in $] \log S', \log S[$ (respectively $] \log S, \log S[$], a contradiction to the hypothesis.
When $\mathcal{F}$ has no center, we consider a center $a$ of $\mathcal{F}$ in $\hat{\mathbb{K}}$ and we consider $f$ as an element of $\hat{H}(\hat{D})$. Then, the disks $\hat{d}(a_n, r_n)$ contain no hole of $\hat{D}$ when $n$ is big enough. Therefore, the restriction of $\hat{f}$ to $d(a_n, r_n)$ is a power series and therefore we have the same conclusion.

□

**Corollary 20.6.** Let $a, b \in D$ and let $f \in H(D)$ satisfy $f(b) \neq 0$ and $\mathcal{D}\varphi_{a,r}(f) = 0$ for some $r \in \mathbb{R}_+$. Then, $f$ is strictly vanishing along a pierced monotonous filter with a center.

**Proposition 20.7.** Let $a \in \tilde{D}$. Let $r, r' \in ]\log(\delta(a, D))$, $\log(\text{diam}(D))[$, with $r < r'$. Let $f \in H(D)$ be such that the function $\Psi_a(f, \mu)$ is neither bounded nor identically equal to $-\infty$ in $[\log r, \log r']$. Then there exists a monotonous filter $\mathcal{F}$ of center $a$ and diameter $s \in [r, r']$ such that $f$ is strictly vanishing along $\mathcal{F}$.

**Proof.** For convenience we assume $a = 0$. By compacity of $[\log r, \log r']$, there exists $\mu \in [\log r, \log r']$ such that $\Psi(f, \mu) = -\infty$. Since the function $\Psi(f, \mu)$ is not identically $-\infty$ in $[\log r, \log r']$, by continuity, either there exists $\xi, \zeta \in [\log r, \log r']$ with $\xi < \zeta$ such that $\Psi(f, \xi) = -\infty$, $\Psi(f, \mu) > -\infty$ whenever $\mu \in ]\xi, \zeta]$ and then $f$ is strictly vanishing along a decreasing filter of center 0 and diameter $\theta^\xi$, or there exist $\xi, \zeta \in [\log r, \log r']$ with $\xi < \zeta$ such that $\Psi(f, \zeta) = -\infty$, $\Psi(f, \mu) > -\infty$ whenever $\mu \in [\xi, \zeta[$ and then $f$ is strictly vanishing along an increasing filter of center 0 and diameter $\theta^\zeta$. This ends the proof.
□

**Proposition 20.8.** Let $f \in H(D)$ be vanishing along an increasing (respectively a decreasing) filter $\mathcal{F}$ of diameter $s$. Let $a$ be a center of $\mathcal{F}$ in $\hat{\mathbb{K}}$ and let $E = \hat{D} \cup (\hat{\mathbb{K}}\setminus\hat{d}(a, s^-))$ (respectively $E = \hat{D} \cup \hat{d}(a, s)$). Then, $f$ has continuation to an element $F$ of $\hat{H}(E)$ such that $F(x) = 0$ whenever $x \in \hat{\mathbb{K}}\setminus\hat{d}(a, s^-)$ (respectively $x \in \hat{d}(a, s)$).

**Proof.** We suppose $\mathcal{F}$ increasing. By Theorem 15.9, $f$ has an extension $\hat{f}$ to $\hat{D}$. For every $r > 0$, the set of classes of $C(a, r)$ which contain $\hat{f}$-holes is countable. Since the residue class field of $\hat{\mathbb{K}}$ is not countable, in $\hat{C}(a, r)$, there exist classes $\Lambda = d(b, r^-)$ which contain no $\hat{f}$-holes. Thereby, $\hat{f}$ has continuation to an infraconnected set $D'$ which contains $\hat{D}$ and satisfies $\tilde{D}' = \hat{\mathbb{K}}$ such that every hole is of the form $\hat{d}(\alpha, \rho^-)$, with $d(\alpha, \rho^-)$ a hole of $D$. In this set, we have
$$ (1) \quad \mathcal{D}' \varphi_{a,s}(\hat{f}) = \mathcal{D} \varphi_{\mathcal{F}}(f) = 0. $$

First, we suppose that $\mathcal{F}$ is increasing. Let $V = d(a, s^{-})$ and let $D'' = D' \setminus V$. Clearly, $V$ is a hole of $D''$. Then, as an element of $\hat{H}(D'')$, by Theorem 15.1, (1) implies

$$ (2) \quad (\hat{f})_V = 0. $$

Now, by Theorem 15.1, $\hat{f}$ has a decomposition of the form $g+h$ with $g \in H_0(D' \cup (\mathbb{K}\setminus\hat{d}(a, s^{-})))$ and $h \in H(D' \cup \hat{d}(a, s^{-}))$. By (2), it is seen that $\mathcal{D}' \varphi_{a,s}(h) = 0$, hence $h=0$ because $h$ belongs to $\hat{H}(\hat{d}(a, s))$. As a consequence, $\hat{f}$ belongs to $H_0(D' \cup (\mathbb{K}\setminus\hat{d}(a, s^{-})))$, and therefore, in $\mathbb{K}$, $f$ belongs to $H(D \cup (\mathbb{K}\setminus d(a, s^{-})))$. Moreover, by (2), we have $\hat{f}(x) = 0$ whenever $x \in D''$, hence $\hat{f}(x) = 0$ whenever $x \in \mathbb{K}\setminus d(a, s^{-})$.

If $\mathcal{F}$ is decreasing, we can easily perform a symmetric proof. $\square$

**Theorem 20.9.** Let $f \in H(D)$ be vanishing along an increasing (respectively a decreasing) filter $\mathcal{F}$ of center $a$ and diameter $s$. Let $E = D \cup (\mathbb{K}\setminus d(a, s^{-}))$ (respectively $E = D \cup d(a, s)$). Then, $f$ has continuation to an element of $H(E)$ such that $f(x) = 0$ whenever $x \in \mathbb{K}\setminus d(a, s^{-})$ (respectively whenever $x \in d(a, s)$).

**Proof.** By Proposition 20.8, $f$ has continuation to an element $\hat{f} \in \hat{H}(\hat{D} \cup (\mathbb{K}\setminus\hat{d}(a, s^{-})))$ (respectively to $H(\hat{D} \cup \hat{d}(a, s)))$. Therefore, in $\mathbb{K}$, $f$ belongs to $H(D \cup (\mathbb{K}\setminus d(a, s^{-})))$. Moreover, we have $\hat{f}(x) = 0$ whenever $x \in \mathbb{K}\setminus\hat{d}(a, S^{-})$ (respectively $x \in \hat{d}(a, s)$), hence $f(x) = 0$ whenever $x \in \mathbb{K}\setminus d(a, s^{-})$ (respectively $f(x) = 0$ whenever $x \in d(a, s)$). $\square$

**Definition.** Let $\mathcal{F}$ be a monotonous filter on $D$ and let $f \in H(D)$. Then, $f$ is said to be collapsing along $\mathcal{F}$ if there exists $b \in \mathbb{K}$ such that $f - b$ is vanishing along $\mathcal{F}$.

**Theorem 20.10.** Let $D$ be open and closed and let $\mathcal{F}$ be a monotonous filter on $D$. Let $f \in H(D)$ be such that $f' \in H(D)$. Then, $f$ is collapsing along $\mathcal{F}$ if and only if $f'$ is vanishing along $\mathcal{F}$.

**Proof.** Without loss of generality, we can suppose that $D$ is bounded because $\mathcal{F}$ is obviously secant with a bounded subset of $\mathbb{K}$.

Suppose first that $\mathcal{F}$ is decreasing, of center $a$ and diameter $R$. Without loss of generality, we can obviously assume $a = 0$. Suppose $f$ is collapsing along $\mathcal{F}$, of limit $\ell$. Since $\lim_{r \to R} \frac{|f - \ell|(r)}{r} = 0$,
by Theorem 18.25, we have $\lim_{r \to R} |f'(r)| = 0$, hence $f'$ is vanishing along $\mathcal{F}$.

Conversely, suppose that $f'$ is vanishing along $\mathcal{F}$. Let $(T_i)_{i \in I}$ be the family of holes of $D$ included in $K \setminus d(0, R)$ and let $(T_i)_{i \in J}$ be the family of holes of $D$ included in $d(0, R)$. Then, the pair $(I, J)$ makes a partition of the set of holes of $D$. Let $D_1 = D \cup d(0, R)$ and let $D_2 = D \cup (K \setminus d(0, R))$. By Corollary 15.2, we have $H(D) = H(D_1) \oplus H(D_2)$. For every $g \in H(D)$, we set $g = g_1 + g_2$ with $g_k \in H(D_k)$, $k = 1, 2$. Then, $\|g\|_D = \max(\|g_1\|_{D_1}, \|g_2\|_{D_2})$. By Theorem 18.19, we can check that the decomposition of $f'$ in the form $f' = (f')_1 + (f')_2$ is such that $(f')_k = (f_k)'$, $k = 1, 2$. When we take numbers $S' > R$ and $S'' < R$, we have $\lim_{S' \to R^+} \|f'\|_{D \cap d(0, S')} = 0 = \lim_{S'' \to R^-} \|f'\|_{D \setminus d(0, S'')}$, hence $\lim_{S' \to R^-} \|f'_1\|_{D \cap d(0, S')} = 0 = \lim_{S'' \to R^+} \|f'_1\|_{D \setminus d(0, S'')}$ and $\lim_{S' \to R^-} \|f'_2\|_{D \cap d(0, S')} = 0 = \lim_{S'' \to R^+} \|f'_2\|_{D \setminus d(0, S'')}$. Consequently, passing to the limit, we get $\varphi_{0,R}(f'_1) = \varphi_{0,R}(f'_2) = 0$. Therefore, by Corollary 20.6, $f'_2$, which belongs to $H(K \setminus d(0, R))$ and satisfies $\varphi_{0,R}(f'_2) = 0$, is identically zero in $K \setminus d(0, R)$. Consequently, $f'(x) = f'_1(x) \forall x \in D \cap (K \setminus d(0, R))$. On the other hand, since $f'_1$ belongs to $H(d(0, R))$ and satisfies $\varphi_{0,R}(f'_1) = 0$, it is identically zero in $d(0, R)$ and hence $f_1(x)$ is a constant $C$ in $d(0, R)$. Therefore, $\varphi(f_1 - C) = 0$, hence $f_1 - C$ is vanishing along $\mathcal{F}$. But since $f_2(x) = 0 \forall x \in D \setminus d(0, R)$, actually, $f - C$ is vanishing along $\mathcal{F}$. This finishes showing that $f$ is collapsing along $\mathcal{F}$ when $\mathcal{F}$ is a decreasing filter with a center.

Now, suppose that $\mathcal{F}$ has no center. We can place ourselves in an algebraically closed spherically complete extension of $K$ and prove the same property for the expansion of $f$, hence it holds for $f$. Finally, if $\mathcal{F}$ is increasing, we can make an inversion and prove the same. $\square$

Thanks to monotonous filters, we are now able to complete the study of the characteristic functions.

**Theorem 20.11.** Let $E$ be a subset of $K$ whose interior is not empty. Then, $E$ is not infraconnected if and only if there exists a proper subset $B$ of $E$ whose characteristic function belongs to $H(E)$.

**Proof.** If $E$ is not infraconnected, it admits an empty annulus $\Gamma(a, r', r'')$, and then by Proposition 11.15, the characteristic functions of $\mathcal{I}_E(\Gamma(a, r', r''))$ and $\mathcal{E}_E(\Gamma(a, r', r''))$ belong to $H(E)$.
Now, we suppose $E$ to be infraconnected and assume that there is a subset $B$ of $E$, $B \neq E$ and $B \neq \emptyset$, whose characteristic function $u$ belongs to $H(E)$. Since $u$, by definition, belongs to $H_b(E)$, it belongs to $H_b(\bar{E})$. And of course, $\bar{E}$ has an interior that is not empty. Hence, without loss of generality, we can assume that $E$ is closed. Let $A = E \setminus B$. Suppose $A$ and $B$ are different from $\emptyset$. Since $u$ is locally constant in all $E$, at least one of the two subsets $A$ and $B$ has an interior that is not empty. Without loss of generality, we can assume that the interior of $A$ is not empty and hence there exists $a \in A$ and $r > 0$ such that $d(a,r) \subset A$ and $u(x) = 0$ whenever $x \in d(a,r)$ and then we have $\partial_D \varphi_{a,r}(u) = 0$. By Proposition 20.3, there exists a monotonous filter $\mathcal{F}$ with center $\alpha \in E$ such that $u$ is strictly vanishing along $\mathcal{F}$, hence by Lemma 20.2, $f$ is properly vanishing along $\mathcal{F}$. But this contradicts the hypothesis "$f(x) = 0$ or $1$ for all $x \in E$". This finishes proving Theorem 20.11. $\square$

**Corollary 20.12.** Let $E \in \text{Alg.}$ The algebra $H(E)$ has non-trivial idempotents if and only if $E$ is not infraconnected.

**Theorem 20.13.** Let $f \in H(D)$. Then, $f(D)$ is infraconnected.

**Proof.** Let $D' = f(D)$ and let us suppose that $D'$ admits an empty annulus $\Gamma(a, r', r'')$. Let $D'' = \bar{D}'$. It is seen that $\Gamma(a, r', r'')$ also is an empty annulus of $D''$.

Let $A'' = \mathcal{I}_{D''}(\Gamma(a, r', r''))$ and let $B'' = \mathcal{E}_{D''}(\Gamma(a, r', r''))$. Let $u$ be the characteristic function of $A''$. By Proposition 11.15, we know that $u$ belongs to $H(D'')$. Since $D''$ is closed and contains $f(D)$, by Corollary 12.3, $u \circ f$ belongs to $H(D)$. Let $A = f^{-1}(A'')$ and $B = f^{-1}(B'')$. Obviously, we have $A \cap B = \emptyset$ and $A \cup B = D$. Now, $u \circ f(x) = 1 \ \forall x \in A$, $u \circ f(x) = 0 \ \forall x \in B$. But since $D$ is infraconnected, by Theorem 20.11, $H(D)$ contains no characteristic function of any proper subset. This ends the proof of Theorem 20.13. $\square$
# Chapter 21

## Quasi-Minorated Elements

Throughout this chapter, the set D is supposed to be infraconnected.

The main results given here were published in [35], [44], and [50]. According to the definition of quasi-minorated elements, Theorem 21.1 is easy.

**Theorem 21.1.** Let $f \in H(D)$. Then, $f$ is not quasi-minorated if and only if there exists a large circular filter $\mathcal{F}$ secant with $D$ such that $\text{D}\varphi_{\mathcal{F}}(f) = 0$.

**Proof.** By Lemmas 16.7 and 17.12, without loss of generality, we can assume that $D$ is bounded.

Suppose first that there exists a large circular filter $\mathcal{F}$ secant with $D$ such that $\text{D}\varphi_{\mathcal{F}}(f) = 0$. Let $(a_n)$ be a monotonous distances sequence thinner than $\mathcal{F}$ such that $\lim_{n \to +\infty} f(a_n) = 0$. Then, $f$ is not quasi-minorated.

Conversely, suppose that $f$ is not quasi-minorated. Then there exists a bounded sequence $(a_n)$ of $D$ such that $\lim_{n \to +\infty} f(a_n) = 0$ and such that one can't extract a sequence converging in $\mathbb{K}$. By Theorem 3.1, we can extract from the sequence $(a_n)$ either a monotonous distances sequence or a constant distances sequence. In both cases, there exists a circular filter $\mathcal{F}$ less thin than this subsequence, and hence, we have $\lim_{\mathcal{F}} f(x) = 0$. $\square$

**Theorem 21.2.** Let $f$ be a non-identically zero element of $H(D)$. Then, $f$ is not quasi-minorated if and only if there exists a pierced monotonous filter $\mathcal{F}$ on $D$ such that $f$ is strictly vanishing along $\mathcal{F}$.
**Proof.** By Theorem 21.1, if $f$ is vanishing along a monotonous filter; it is not quasi-minorated. Now suppose that $f$ is not quasi-minorated. Since $f$ is not identically zero, we can find a bounded sequence $(a_n)_{n \in \mathbb{N}}$ in $D$ such that $\lim_{n \to \infty} f(a_n) = 0$ such that no subsequence converges in $\mathbb{K}$.

Suppose first we can extract from the sequence $(a_n)_{n \in \mathbb{N}}$ a constant distances sequence $(a_{q(m)})_{m \in \mathbb{N}}$, let $c = a_{q(1)}$, let $r = |a_{q(1)} - a_{q(2)}|$, and let $\phi = \varphi_{b,r}$. Since $\lim_{m \to +\infty} f(a_{q(m)}) = 0$, we have $\varphi_{c,r}(f) = 0$. Since $f$ is not identically zero in $D$, there exists $b \in D$ such that $f(b) \neq 0$. If $b \in d(0, r)$, then by Proposition 20.3, $f$ is strictly vanishing along an increasing filter of center $c$, of diameter $\rho \in]0, r]$. If $b \notin d(c, r)$, by Proposition 20.3, if $\varphi_{c,|b-c|}(f) \neq 0$, then $f$ is strictly vanishing along a decreasing filter of center $c$, of diameter $\rho \in d(c, |b-c|)$. Finally, if $\varphi_{c,|b-c|}(f) = 0$, then $f$ is strictly vanishing along an increasing filter of center $b$, of diameter $\rho \in d(0, |b-c|)$, and by Proposition 20.5, that filter is pierced.

Suppose now we can't extract from the sequence $(a_n)_{n \in \mathbb{N}}$ a constant distances sequence. Then we can extract from the sequence $(a_n)_{n \in \mathbb{N}}$ a monotonous distances sequence $(a_{q(m)})_{m \in \mathbb{N}}$. There exists a unique monotonous filter $\mathcal{F}$ less thin than the subsequence $(a_{q(m)})_{m \in \mathbb{N}}$. Suppose first that $\mathcal{F}$ has a center $c$ and let $r = \text{diam}(\mathcal{F})$. Then, we have $\varphi_{c,r}(f) = 0$, and hence, the same reasoning shows that $f$ is strictly vanishing along a monotonous filter which by Proposition 20.5 is pierced. Finally, suppose $\mathcal{F}$ is decreasing, with no center. We can find a center in a spherically closed extension and make the same reasoning again. $\square$

**Corollary 21.3.** If $D$ has no monotonous pierced filter, every element of $H(D)$ different from zero is quasi-minorated and takes every value finitely many times.

**Definition.** A subset $D$ of $\mathbb{K}$ is said to be *analytic* if for every disk $d(a, r)$ included in $D$ and for every $f \in H(D)$, the property $f(x) = 0 \quad \forall x \in d(a, r)$ implies that $f$ is identically zero in $D$.

**Corollary 21.4.** If $D$ has no monotonous pierced filter, $D$ is analytic. Particularly, if $D$ has finitely many holes, $D$ is analytic. Particularly, if $D$ is infraconnected affinoid, it is analytic.

**Theorem 21.5.** Let $f_1, f_2$ be quasi-minorated elements of $H(D)$. If $f_1 f_2$ belongs to $H(D)$, then it is also quasi-minorated.
**Proof.** Indeed, suppose that $f_1f_2$ is not quasi-minorated. By Theorem 21.2, there exists a pierced monotonous filter $\mathcal{F}$ on $D$ such that $_D\varphi_{\mathcal{F}}(f_1f_2) = 0$. Hence, either $_D\varphi_{\mathcal{F}}(f_1) = 0$ or $_D\varphi_{\mathcal{F}}(f_2) = 0$. But, by Theorem 21.1, both options are impossible because both $f_1, f_2$ are quasi-minorated. Hence, so is $f_1f_2$. $\square$

**Theorem 21.6.** Let $f \in H(D)$. If $f$ is semi-invertible, then it is quasi-minorated.

**Proof.** Let $f$ be semi-invertible, of the form $P(x)g(x)$ with $g$ invertible in $H(D)$, and $P$ a polynomial whose zeros $a_1, \dots, a_q$ belong to $D$. We then suppose $f$ not quasi-minorated. By Theorem 21.5, $g$ is not quasi-minorated either. Hence, there exists a pierced monotonous filter $\mathcal{F}$ on $D$ such that $_D\varphi_{\mathcal{F}}(g) = 0$. But, by Lemma 11.3, that contradicts the hypothesis "$g$ invertible in $H(D)$". Hence, $f$ is quasi-minorated. $\square$

**Theorem 21.7.** Suppose that $\bar{D}$ is open. Then, an element of $H(D)$ is quasi-minorated if and only if it is quasi-invertible.

**Proof.** If $f$ is quasi-invertible, it is semi-invertible, and then, by Theorem 21.6, it is quasi-minorated. Now, assume $f$ to be quasi-minorated. We prove it to be quasi-invertible. As in Theorem 21.1, by Lemmas 16.7 and 17.12, without loss of generality, we can assume that $D$ is bounded. Let $S(D)$ be the set of polynomials whose zeros belong to $\bar{D} \setminus D$. If $D \in \text{Alg}$, then $H(D) = S(D)^{-1}H(\bar{D})$. By Corollary 11.13, there exists $Q \in S(D)$ and $h \in H(\bar{D})$ such that $f = \frac{h}{Q}$. If $h$ is not quasi-minorated, there exists a monotonous filter $\mathcal{F}$ on $D$ such that $_D\varphi_{\mathcal{F}}(h) = 0$ and then we have $_D\varphi_{\mathcal{F}}(f) = 0$: this contradicts the hypothesis "$f$ quasi-minorated" and therefore $h$ is quasi-minorated. Then, by Theorem 16.8, $h$ is quasi-invertible, in the form $Pg$ with $P$ a polynomial whose zeros are interior to $\bar{D}$ and $g$ an element invertible in $H(\bar{D})$. Since $\bar{D}$ is bounded and open, by Corollary 17.10, $D$ belongs to $\text{Alg}$, hence $\frac{g}{Q}$ is invertible in $H(D)$. Let $P = P_1P_2$ with $P_1$ (respectively $P_2$) the polynomial of the zeros of $P$ in $\frac{\circ}{\bar{D}} \cap D$ (respectively in $\bar{D} \setminus D$). Then, $\frac{P_2g}{Q}$ is invertible in $H(D)$, and then, $f$ is quasi-invertible. $\square$

**Theorem 21.8.** If $D$ belongs to $\text{Alg}$ and has no pierced filter, every element of $H(D)$ different from zero is quasi-invertible.
**Proof.** As in Theorem 21.1, by Lemmas 17.12 and 16.7, without loss of generality, we can assume that $D$ is bounded. Now, since $D$ has no pierced filter, by Corollary 21.3, every element of $H(D)$ is quasi-minorated. But since $D$ has no pierced filter, $\bar{D}$ is open, hence by Theorem 21.7, every element of $H(D)$ is quasi-invertible. $\square$

**Corollary 21.9.** Let $D$ be closed and let $\mathcal{T}$ be the set of holes of $D$. If $\{\tilde{T} | T \in \mathcal{T}\}$ is finite, then every element of $H(D)$ different from zero is quasi-invertible.

**Corollary 21.10.** If $D$ is closed and has finitely many holes, then every element of $H(D)$ different from zero is quasi-invertible.

**Corollary 21.11.** If $D$ is a disk $d(a,r)$ or $d(a,r^{-})$, or if $D$ is an annulus $\Gamma(a, r_1, r_2)$ (with $0 < r_1 < r_2$) or $\Delta(a, r_1, r_2)$ (with $0 < r_1 < r_2$) or a circle $C(a,r)$, then every element of $H(D)$ different from zero is quasi-invertible.

**Proof.** (Corollaries 21.9, 21.10, and 21.11) Indeed, $D$ has no pierced filter, hence the elements different from zero are quasi-minorated and are quasi-invertible because $D$ is open. $\square$

We see that when $D$ belongs to Alg, a quasi-minorated element that has no zero in $D$ actually is invertible in $H(D)$.

**Lemma 21.12.** Let $f \in H(\bar{D})$ be quasi-minorated in $H(D)$. Then, $f$ is quasi-minorated in $H(\bar{D})$.

**Proof.** Indeed, let $(a_n)_{n \in \mathbb{N}}$ be a bounded sequence in $\bar{D}$ such that $\lim_{n \to \infty} f(a_n) = 0$. There obviously exists a sequence $(b_n)_{n \in \mathbb{N}}$ in $D$ such that $\lim_{n \to \infty} a_n - b_n = 0$ and $\lim_{n \to \infty} f(b_n) = 0$. So, the sequence $(b_n)$ is bounded as the sequence $(a_n)$. Since $f$ is quasi-minorated in $H(D)$, then we can extract a Cauchy sequence $(b_{q(m)})_{m \in \mathbb{N}}$ from the sequence $(a_n)_{n \in \mathbb{N}}$ and then the sequence $(a_{q(m)})_{m \in \mathbb{N}}$ is a Cauchy subsequence of the sequence $(a_n)_{n \in \mathbb{N}}$. Thus, we have proven that $f$ is quasi-minorated in $H(\bar{D})$. $\square$

**Theorem 21.13.** Let $D \in \text{Alg}$. Let $f \in H(D)$ be quasi-minorated and have no zero in $D$. Then, $f$ is invertible in $H(D)$.
**Proof.** If $D$ is closed, the statement is given by Theorem 16.9. Consider now the general case. As in Theorem 21.1, by Lemmas 16.7 and 17.12, without loss of generality, we can assume that $D$ is bounded. Let $Q$ be the polynomial of the poles of $f$ in $\bar{D}\setminus D$ and let $h(x) = Q(x)f(x)$. By Theorem 21.5, $h$ is quasi-minorated in $H(D)$. But, by Corollary 21.11, $h$ belongs to $H(\bar{D})$, and therefore, by Lemma 21.12, it is quasi-minorated in $H(\bar{D})$. We prove that $h$ has finitely many zeros in $\bar{D}$. Indeed, we assume that $h$ admits infinitely many zeros in $\bar{D}$. So, we can find a sequence $(a_n)_{n\in\mathbb{N}}$ in $\bar{D}\setminus D$ such that $h(a_n) = 0$ whenever $n \neq m$, $n, m \in \mathbb{N}$. Since $h$ is quasi-minorated in $H(\bar{D})$ and since $D$ is bounded, we can extract a Cauchy subsequence from the sequence $(a_n)_{n\in\mathbb{N}}$. This Cauchy subsequence obviously converges to a point $a \in \bar{D}$ and therefore we have $h(a) = 0$. But, as $h$ has no zero in $D$, $a$ belongs to $\bar{D}\setminus D$. And now, since $D$ belongs to $\bar{D}$, then $a$ belongs to $\bar{D}$. But, by Corollary 16.3, a zero of $h$ which is interior to $\bar{D}$ is isolated in $\bar{D}$ and this contradicts the definition of $a$. Thus, we have proven that $h$ has finitely many zeros $(b_j)_{1\le j\le q}$ in $\bar{D}$, all of them in the set $\bar{D}\setminus D$ which is included in $\bar{D}$. Then, each zero $b_j$ has a multiplicity order $n_j$ $(1 \le j \le q)$. Let $P(x) = \prod_{j=1}^q (x-b_j)^{n_j}$ be the polynomial of the zeros of $h$ in $\bar{D}$. By Corollary 16.6, the function $g(x) = \frac{h(x)}{P(x)}$ belongs to $H(\bar{D})$ and obviously has no zero in $\bar{D}$. As we have already seen when $D$ is closed, $g$ is invertible in $H(\bar{D})$. Now, since both $P, Q$ have all their zeros in $\bar{D}\setminus D$, they are invertible in $R(D)$ and so is $\frac{P}{Q}g$. But then $f = \frac{P}{Q}g$ and hence $f$ is invertible in $H(D)$. $\square$

**Theorem 21.14.** Let $D$ be closed, bounded, having finitely many holes and let $f \in H(D)$. Then, $f(D)$ is an infraconnected closed and bounded subset of $\mathbb{K}$.

**Proof.** Since $D$ has finitely many holes, by Corollary 21.10, $f$ is quasi-minorated. By Theorem 20.13, $f(D)$ is infraconnected. Since $f$ is bounded on a closed bounded subset, $f(D)$ is bounded. Let $b$ belong to the closure of $f(D)$ and let $(a_n)_{n\in\mathbb{N}}$ be a sequence of $D$ such that $\lim_{n\to+\infty} f(a_n) = b$. Since the sequence $(a_n)$ is bounded, there exists a subsequence thinner than a circular filter $\mathcal{F}$ secant with $D$.
If $\mathcal{F}$ is large, then we have $D\varphi_{\mathcal{F}}(f) = 0$, and hence, by Proposition 21.1, $f$ is not quasi-minorated, a contradiction. Consequently, $\mathcal{F}$ converges to a point $a$. Since $D$ is closed, $a$ belongs to $D$. Consequently, $f(a) = b$ and hence $b$ belongs to $f(D)$. Therefore, $f(D)$ is closed. $\square$

Theorem 21.15 shows an example of very simple increasing $T$-filter, without describing the general theory of $T$-filters [38], [44], [53].

**Theorem 21.15.** Let $(r_n)_{n \in \mathbb{N}}$ be a sequence in $|\mathbb{K}|$ such that $0 < r_n < r_{n+1}$ and $\lim_{n \to +\infty} r_n = R$, and let $(q_n)_{n \in \mathbb{N}}$ be a sequence of $\mathbb{N}$ prime to the characteristic of $\mathbb{K}$ such that $q_n \le q_{n+1}$ and $\lim_{n \to +\infty} (\frac{r_n}{r_{n+1}})^{q_n} = 0$. Let $l \in ]0, R[$ and for each $n \in \mathbb{N}$, let $b_n \in C(0, (r_n)^{q_n})$, let $a_{n,1}, \dots, a_{n,q_n}$ be the $q_n$th roots of $b_n$, and let $E = d(0, R^-) \setminus (\cup_{n \in \mathbb{N}} (\cup_{j=1}^{q_n} d(a_{n,j}, l^-)))$. Set $f_n(x) = \prod_{k=1}^n \prod_{j=1}^{q_k} \left(1 - \left(\frac{x}{a_{k,j}}\right)\right)$. Then each $f_n$ belongs to $R(E)$ and the sequence $(f_n)_{n \in \mathbb{N}}$ converges in $H(E)$ to an element strictly vanishing along the pierced increasing filter of center 0 and diameter $R$.

**Proof.** Let us first show that the sequence $(f_n)_{n \in \mathbb{N}}$ converges in $H(E)$. We note that each pole $a_{k,j}$ of $f_n$ lies in a hole of diameter $l$ and is unique in that hole and in the class. Moreover, since each $q_n$ is prime to the residue characteristic of $\mathbb{K}$, each pole $a_{k,j}$ of $f_n$ is unique in the class $d(a_{k,j}, r_k^-)$. Consequently, we have $|f_n(x)| \le \frac{|x| |f_n(|x|)|}{l} \forall x \in E$ and hence

$$ (1) \qquad |f_n(x)| \le \frac{R|f_n(|x|)}{l} \quad \forall x \in E. $$

Let us now show that the sequence $(f_n)_{n \in \mathbb{N}}$ converges in $H(E)$. We first note that each factor $Q_k = \prod_{j=1}^{q_k} \left( \frac{1}{1 - \left( \frac{x}{a_{k,j}} \right)} \right)$ satisfies $|Q_k|(r) \le 1$ $\forall r < R$ because $|Q_k|(r) = 1 \forall r \le r_k$ and $|Q_k|(r) < r \forall r \in ]r_k, R[$. Consequently, we have

$$ (2) \qquad |f_n|(r) \le \prod_{k=0}^{n} \prod_{j=1}^{q_k} \left(\frac{r_k}{r}\right)^{q_k} \le \left(\frac{r_{n-1}}{r_n}\right)^{q_{n-1}} \quad \forall r \ge r_n. $$
On the other hand, when $r < r_n$, we have

$$|1 - Q_{n+1}|(r) = \left|1 - \frac{1}{1 - \left(\frac{x}{b_{n+1}}\right)^{q_{n+1}}}\right|(r) \le \left(\frac{r_n}{r_{n+1}}\right)^{q_{n+1}},$$

hence

$$(3) \qquad |f_{n+1} - f_n|(r) = |f_n|(r)|1 - Q_{n+1}|(r) \le |f_n|(r) \left(\frac{r_n}{r_{n+1}}\right)^{q_{n+1}} \quad \forall r < r_n.$$

By (2) and (3), we can see that $\lim_{n \to +\infty} (\sup_{r<R} |f_{n+1}(r) - f_n(r)|) = 0$, and hence by (1), we have $\lim_{n \to +\infty} \|f_{n+1} - f_n\|_E = 0$, hence the sequence $f_n$ converges to an element $f \in H(E)$. Moreover, by (1) and (2), we can see that $\lim_{|x| \to R^{-}} \lim_{x \in E} f(x) = 0$, so $f$ is vanishing along the increasing pierced filter $\mathcal{F}$ of center 0 and diameter $R$. Further, we may note that $|Q_n|(r) = 1 \forall r \le r_n$, hence, when $r \le r_n$, we have $|f|(r) = |f_s|(r) \forall s \ge n$. Consequently, $|f|(r) \ne 0 \forall r < R$ and hence $f$ is strictly vanishing along $\mathcal{F}$. □
# Chapter 22

## Zeros of Power Series

Most of classical results on zeros of polynomials are now extended to power series. In particular, power series converging inside a disk satisfy a Schwarz lemma that is even simpler than in $\mathbb{C}$.

Throughout this chapter, $r$ is a strictly positive real number and $r'$, $r''$ are strictly positive real numbers satisfying $r' < r''$.

**Theorem 22.1.** Let $f \in H(C(0, r))$. The number of zeros of $f$ in $C(0, r)$ is equal to $\nu^+(f, \log r) - \nu^-(f, \log r)$ (taking multiplicity into account).

**Proof.** This equality was given for a polynomial in Theorem 4.16. First, we prove it when $f$ is an element of $H(C(0, r))$ invertible in $H(C(0, r))$. By Proposition 19.7, we have $\nu^+(\frac{1}{f}, \log r) = -\nu^+(f, \log r)$ and $\nu^-(\frac{1}{f}, \log r) = -\nu^-(f, \log r)$. Since any $h \in H(C(0, r))$ satisfies $\nu^+(h, \log r) \ge \nu^-(h, \log r)$, we see that $\nu^+(f, \log r) = \nu^-(f, \log r)$.

We now consider the general case. By Corollary 21.11, $f$ has a factorization of the form $Pg$ with $P$ a polynomial whose zeros belong to $C(0, r)$ and $g$ invertible in $H(C(0, r))$. Then, $\nu^+(f, \log r) - \nu^-(f, \log r) = \nu^+(P, \log r) - \nu^-(P, \log r) = \deg(P)$, and this just ends the proof of Theorem 22.1. $\square$

**Corollary 22.2.** Let $a \in \mathbb{K}$ and $r > 0$. Let $f(x) \in H(C(a, r^{-}))$. Let $\hat{\mathbb{K}}$ be a complete algebraically closed extension of $\mathbb{K}$ and let $\hat{C}(a, r) = \{x \in \hat{\mathbb{K}} | |x - a| = r\}$. Then, the zeros of $f$ in $\hat{C}(a, r)$ are exactly
those of $f$ in $C(a, r)$ (taking multiplicity into account). Similarly, the zeros of $f$ in $\hat{d}(a, r) = \{x \in \mathbb{K} \mid |x - a| \le r\}$ (respectively in $\hat{d}(a, r^{-}) = \{x \in \mathbb{K} \mid |x - a| < r\}$) are exactly those of $f$ in $d(a, r)$ (respectively in $d(a, r^{-})$) (taking multiplicity into account).

**Corollary 22.3.** Let $f \in H(C(0, r))$ have $t$ zeros in $C(0, r)$.

$$ \text{Let } q = \nu^{+}(f, \log r) - \nu^{-}(f, \log r). \text{ Then, } r = \sqrt[q]{\left| \frac{a_t}{a_{q+t}} \right|}. $$

**Theorem 22.4.** Let $\Lambda = C(0, r)$ and let $f(x) = \sum_{-\infty}^{+\infty} a_n x^n$ be a convergent Laurent series in $\Lambda$, having no zero in $\Lambda$. Let $\mu = \log r$. Then, $\nu^{+}(f, \mu) = \nu^{-}(f, \mu) = q \in \mathbb{Z}$ and $|f(x)| = |a_q x^q|$ whenever $x \in \Lambda$. Moreover, if $q \neq 0$, then $f(\Lambda) = C(0, |a_q|r^q)$.

**Proof.** By Proposition 19.7, we see that $\nu^{+}(f, \mu) = \nu^{-}(f, \mu)$. Thus, we have $\Psi(f, \mu) = \Psi(a_q) + q\mu$ whenever $\mu \in I$. Consequently, $f(C(0, r)) \subset C(0, |a_q|r^q)$. Now, suppose $q \neq 0$ and let $s = |a_q|r^q$. Let $b \in C(0, s)$ and let $g = f - b$. So, by definition, $g(x) = \sum_{-\infty}^{-1} a_n x^n + (a_0 - b) + \sum_{1}^{+\infty} a_n x^n$, and by hypothesis, $|a_0| < |a_q|r^q$, hence $|a_0 - b| = |a_q|r^q$. Consequently, $\nu^{-}(g, \mu) < \nu^{+}(g, \mu)$, therefore, by Theorem 22.1, $g$ admits at least one zero in $C(0, r)$ and hence $b$ lies in $f(C(0, r))$. This proves that $f(C(0, r)) = C(0, |a_q|r^q)$. $\square$

**Corollary 22.5.** Let $f(x) = \sum_{-\infty}^{+\infty} a_n x^n$ be a convergent Laurent series in $\Lambda = \Gamma(a, R', R'')$ (respectively in $\Lambda = \Delta(a, R', R'')$), having no zero in $\Gamma(a, R', R'')$. Then we have $\nu^{+}(f, \mu) = \nu^{-}(f, \mu) = q \in \mathbb{Z}$ whenever $\mu \in [\log r', \log r'']$ (respectively $\mu \in [\log r', \log r'']$) and $|f(x)| = |a_q x^q|$ whenever $x \in \Lambda$. Moreover, if $q > 0$, putting $s' = |a_q|r'^q$, $s'' = |a_q|r''^q$, we have $f(\Lambda) = \Gamma(0, s', s'')$ (respectively $f(\Lambda) = \Delta(0, s', s'')$), and if $q < 0$, putting $s' = |a_q|r''^q$, $s'' = |a_q|r'^q$, we have $f(\Lambda) = \Gamma(0, s', s'')$ (respectively $f(\Lambda) = \Delta(0, s', s'')$).

**Corollary 22.6.** Let $\Lambda = \Gamma(0, r', r'')$ and let $f(x), g(x)$ be convergent Laurent series in $\Lambda$, having no zero in $\Lambda$ such that $|f|(r) = |g|(r) \forall r \in ]r', r'']$. Then, $\nu(f, \log r) = \nu(g, \log r) \forall r \in ]r', r'']$.

**Proof.** By hypothesis, we have $\Psi(f, \mu) = \Psi(g, \mu) \forall \mu \in ]\log r', \log r'']$. But $\nu(f, \log r) = \frac{d\Psi(f)}{d\mu}(\log r)$ and $\nu(g, \log r) = \frac{d\Psi(g)}{d\mu}(\log r)$, hence $\nu(f, \log r) = \nu(g, \log r)$. $\square$
**Theorem 22.7.** Let $f \in H(d(0, r))$. The number of zeros of $f$ in $d(0, r)$ is equal to $\nu^+(f, \log r)$ (taking multiplicity into account).

**Proof.** This equality was given for a polynomial in Corollary 4.17. First, we prove it when $f$ is an invertible element of $H(d(0, r))$. By Theorem 22.1, we have $\nu^+(f, \mu) = \nu^-(f, \mu) = 0 \quad \forall \mu \le \log(r)$. Consequently, $\nu^+(f, \log r) = 0$. Consider now the general case. Since, by Theorem 21.11, $f$ is quasi-invertible, it has a factorization of the form $P(x)h(x)$ with $P$ a polynomial whose zeros lie in $d(0, r)$ and $h$ is an invertible element of $H(d(0, r))$. Then, by Corollary 4.17, $\nu^+(P, \log(r))$ is the number of zeros of $P$ (hence of $f$), and by Proposition 19.7, we have $\nu^+(f, \log(r)) = \nu^+(P, \log(r)) + \nu^+(h, \log(r)) = \nu^+(P, \mu)$ which ends the proof. $\square$

**Corollary 22.8.** Let $\Lambda = \Gamma(0, r', r'')$ and let $f(x), g(x)$ be convergent Laurent series in $\Lambda$ such that $|f|(r) = |g|(r) \quad \forall r \in [r', r'']$. Then for each $r \in [r', r'']$, $f$ and $g$ have the same number of zeros in $C(0, r)$ (taking multiplicity into account).

**Theorem 22.9.** Let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in H(d(a, r))$ and let $s = \sup_{n \ge 1} |a_n| r^n$. Then, $f(d(a, r)) = d(a_0, s)$ and $\Psi_a(f-a_0, \log r) = \log s$.

**Proof.** Without loss of generality, we can suppose that $a=0$. Let $b \in d(a_0, s)$ and consider $f(x)-b=a_0-b+\sum_{n=1}^{+\infty} a_n x^n$. By hypothesis, $|a_0-b| \le s$, hence $\nu^+(f-b, \log(r)) = \nu^+(f-a_0, \log(r)) \ge 1$, and hence $f-b$ has at least one zero in $d(0, r)$. Consequently, $d(0, s) \subset f(d(0, r))$. Conversely, when $x \in d(0, r)$, we have $|f(x)-a_0| \le s$, hence $d(0, s) = f(d(0, r))$. As a consequence, $\Psi_a(f-a_0, \log r) = \log s$. $\square$

**Corollary 22.10.** Let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in \mathcal{A}_b(d(a, r^-))$ (not identically zero) and let $s = \sup_{n \ge 1} |a_n| r^n$. Then, $f(d(a, r^-)) = d(a_0, s^-)$.

**Proof.** On the one hand, $f(d(a, r^-))$ is obviously included in $d(a_0, s^-)$. On the other hand, given $b \in d(a_0, s^-)$ and $\rho \in [0, r]$ such that $\sup_{n \ge 1} |a_n| \rho^n \ge |b-a_0|$, by Theorem 22.9, $b$ belongs to $f(d(a, \rho))$ because $f \in H(d(a, \rho))$. $\square$

**Corollary 22.11.** Let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in H(d(a, r^-))$ (not identically zero) and let $s = \sup_{n \ge 1} |a_n| r^n$. Then, $f(d(a, r^-)) = d(a_0, s^-)$.
**Lemma 22.12.** Let $f \in H(d(0, r))$ satisfy $\nu^+(f, \log r) \ge 1$, let $b \in f(d(0, r))$, and let $g(x) = f(x) - b$. Then, we have $\nu^+(g, \log r) = \nu^+(f, \log r)$.

**Proof.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and let $t = \nu^+(f, \log r)$. By hypothesis, we have $|a_t|r^t \ge |a_n|r^n$ whenever $n < t$ and $|a_t|r^t > |a_n|r^n$ whenever $n > t$. Now let $g(x) = \sum_{n=0}^{\infty} b_n x^n$. Hence, $b_0 = a_0 - b$, $b_n = a_n$ whenever $n \ge 1$. By hypothesis, we have $|b_0| \le \sup_{n \ge 1} |a_n|r^n$, hence $|b_0| \le |b_t|r^t$, and finally, $\nu^+(g, \log r) = t$. $\square$

**Lemma 22.13.** Let $f \in H(d(a, r))$ have $t$ zeros in $d(a, r)$ with $t \ge 1$ (taking multiplicity into account) and let $b \in f(d(a, r))$. Then, $f-b$ also admits $t$ zeros in $d(a, r)$ (taking multiplicity into account).

**Proof.** We assume $a=0$. By Lemma 22.12, we know that $\nu^+(f, \log r) = t$. Hence, we have $\nu^+(f, \log r) = t$ and $\Psi(f, \log r) = \Psi(a_t) + t \log r$. Next, $\Psi(f(x)) \le \Psi(f, \log r)$ for all $x \in d(0, r)$, hence $\Psi(b) \le \Psi(f, \log r)$, and therefore, $\Psi(b) \le \Psi(a_t) + t \log r$. Hence, $\nu^+(f-b, \log r) = t$. That ends the proof. $\square$

**Theorem 22.14.** If an entire function $f \in \mathcal{A}(\mathbb{K})$ is bounded or has no zero, it is a constant.

**Proof.** Let $f(x) = \sum_{j=0}^{\infty} a_j x^j$. By Lemma 22.12, the number of zeros of $f$ in any disk $d(0, r)$ is equal to the biggest of the integers $l$ such that $|a_l|r^l = \sup_{j \in \mathbb{N}} |a_j|r^j$ and this is also equal to $\lim_{|x| \to r, |x| \ne r} |f(x)|$. Hence, if $f$ is bounded, or has no zero in $\mathbb{K}$, obviously we have $a_n = 0 \quad \forall n > 0$. $\square$

**Theorem 22.15.** Let $f(x) \in \mathcal{A}(\mathbb{K}) \setminus \mathbb{K}$. Then $f$ admits at least one zero in $\mathbb{K}$. Moreover, if $f$ is not a polynomial, then $f$ has infinitely many zeros in $\mathbb{K}$ and the zeros make a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$.

**Proof.** Suppose first that $f$ has finitely many zeros in $\mathbb{K}$. Then, by Theorem 14.16, there exists a polynomial $P$ such that $f$ factorizes in the form $Pg$ with $g \in \mathcal{A}(\mathbb{K})$ and $g(x) \ne 0 \forall x \in \mathbb{K}$. Hence, by Theorem 22.14, $g$ is a constant, hence $f$ is a polynomial. Next, by Theorem 22.7, $f$ has finitely many zeros in each disk, hence the sequence $(\alpha_n)_{n \in \mathbb{N}}$ tends to $+\infty$. $\square$
**Theorem 22.16.** Let $\Lambda$ be a disk of the form $d(a, r)$ (respectively $d(a, r^{-})$) and let $f \in H(\Lambda)$ have no zero in $\Lambda$. Then, $|f(x)|$ is equal to a constant in $\Lambda$ and $f$ is invertible in $H(\Lambda)$.

**Proof.** By Corollary 21.11 and Theorems 21.13, $f$ is invertible in $H(\Lambda)$. We may obviously assume $a = 0$. By Lemma 22.12, we have $\nu^{+}(f, \mu) = \nu^{-}(f, \mu) = 0$ whenever $\mu \le \log r$ (respectively $\mu < \log r$), hence by Proposition 19.7, $\Psi(f, \mu)$ has a derivative equal to 0 and therefore is equal to a constant in $]-\infty, \log r[$. Then, by Corollary 19.6, we have $\Psi(f(x)) = \Psi(f, \Psi(x))$, hence $\Psi(f(x))$ is equal to a constant in $\Lambda$. $\square$

**Theorem 22.17.** Let $\Lambda$ be a set in one of the following forms:

(i) $\Lambda = d(0, r)$.

(ii) $\Lambda = d(0, r^{-})$.

(iii) $\Lambda = C(0, r)$.

Let $f \in H(\Lambda)$ (f not identically 0) and let $h \in H(\Lambda)$ satisfy $\|f-h\|_{\Lambda} < \|f\|_{\Lambda}$. Then $f$ and $h$ have the same number of zeros in $\Lambda$ (taking multiplicity into account).

**Proof.** Regardless of the cases (i), (ii), and (iii), we know that $\|f\|_{\Lambda} = \lambda \varphi_{0,r}(f)$ and then we have $\log \|f\|_{\Lambda} = \Psi(f, \log r)$. Since $\|f-h\|_{\Lambda} < \|f\|_{\Lambda}$, then $\Psi((f-h, \log r)) < \Psi(f, \log r)$. Hence, by Proposition 19.5, we know that $\nu^{+}(f, \log r) = \nu^{+}(h, \log s)$ and $\nu^{-}(f, \log r) = \nu^{-}(h, \log r)$. Consequently, $f$ has as many zeros as $h$ in $\Lambda$, by Theorem 22.7 if $\Lambda = d(0, r)$ and by Theorem 22.1 if $\Lambda = C(0, r)$.

We now suppose $\Lambda = d(0, r^{-})$. By Corollary 21.11, both $f, h$ are quasi-invertible in $H(\Lambda)$. Let $\rho \in ]0, r]$ be such that $d(0, \rho)$ contains all zeros of $h$ in $\Lambda$. According to the statements (i) and (ii) already proven, we see that $f$ has as many zeros as $h$ in $d(0, \rho)$ (taking multiplicity into account) and has no zero in $C(0, s)$ whenever $s \in ]\rho, r[$. This ends the proof. $\square$

**Theorem 22.18.** If $f \in H(C(a, r))$, it satisfies $|f(x)| \le D\varphi_{a,r}(f) \quad \forall x \in C(a, r)$ and the equality $|f(x)| = D\varphi_{a,r}(f)$ holds in all classes except finitely many that are the classes, where $f$ has at least one zero.
**Proof.** Let $f \in H(C(a, r))$ and $\Lambda = C(a, r)$. We can find $h \in R(\Lambda)$ such that $\|f - h\|_{\Lambda} < \|f\|_{\Lambda}$, hence by Theorem 22.17, $f$ has the same number of zeros as $h$ in each class of $\Lambda$ (taking multiplicity into account). So, in each class of $\Xi = d(b, r^{-})$ where $f$ has no zero, $|f(x)|$ is equal to $\int_{\mathcal{D}}\varphi_{b,r}(f)$. But by Lemma 13.4, we have $\int_{\mathcal{D}}\varphi_{b,r}(f) = \int_{\mathcal{D}}\varphi_{a,r}(f)$, which ends the proof. $\square$

**Corollary 22.19.** Let $f(x) \in \mathcal{A}(d(0, r^{-}))$ have infinitely many zeros in $d(0, r^{-})$. Then the set of zeros of $f$ in $d(0, r^{-})$ is a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\alpha_n| = r$. $\square$

**Proof.** Indeed, by Proposition 14.12, for each $\rho \in ]0, r[$, $f$ belongs to $H(d(0, \rho))$ and hence has finitely many zeros in $d(0, \rho)$. $\square$

**Theorem 22.20.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, r^{-})).$ Then $f$ has finitely many zeros in $d(0, r^{-})$ if and only if there exists $q \in \mathbb{N}$ such that $|a_q|r^q \ge \sup_{n \in \mathbb{N}} |a_n|r^n$. Moreover, if $t$ is the smallest of the integers $q$ such that $|a_q|r^q \ge \sup_{n \in \mathbb{N}} |a_n|r^n$, then $f$ has exactly $t$ zeros in $d(0, r^{-})$. Further, the three following statements are equivalent:

(i) $f$ has no zero in $d(0, r^{-})$.

(ii) $f$ is invertible in $\mathcal{A}(d(0, r^{-}))$.

(iii) $|f(x)|$ is a non-zero constant.

**Proof.** First, we suppose that there exists $q \in \mathbb{N}$ such that $|a_q|r^q \ge \sup_{n \in \mathbb{N}} |a_n|r^n$. Let $t \in \mathbb{N}$ be the smallest of the integers $q$ such that $|a_q|r^q \ge \sup_{n \in \mathbb{N}} |a_n|r^n$. There exists a unique $s \in ]0, r[$ such that $|a_t|s^t \ge |a_n|s^n$ for every $n < t$. Then, for all $\rho \in ]s, r[$, in $H(d(0, \rho))$, we have $\nu^{+}(f, \log \rho) = \nu^{-}(f, \log \rho) = t$, hence $f$ admits exactly $t$ zeros in $d(0, \rho)$, whenever $\rho \in ]s, r[$.

Conversely, suppose that $f$ admits exactly $t$ zeros in $d(0, r^{-})$. There then there exists $s \in ]0, r[$ such that $f$ admits exactly $t$ zeros in $d(0, s)$ and of course in each disk $d(0, \rho)$ for every $\rho \in ]s, r[$. Hence, we have $\nu^{+}(f, \log \rho) = \nu^{-}(f, \log \rho) = t$ for every $\rho \in ]s, r[$. Therefore, we have $|a_t|\rho^t > |a_n|\rho^n$ for every $n \ne t$ and for every $\rho \in ]s, r[$. Finally, we see that $|a_t|r^t \ge \sup |a_n|r^n$.

Further, the equivalence between (i) and (ii) is shown at Theorem 14.19. The equivalence of (i) with (iii) comes from the fact that $f$ has no zero if and only if $|a_0| > |a_n|s^n \forall n \in \mathbb{N}, \forall s \in ]0, r[$. $\square$
**Theorem 22.21.** Let $f(x) = \sum_{-\infty}^{+\infty} a_n x^n \in \mathcal{A}(\mathbb{K}\setminus d(0, R))$ be not identically zero, let $R' > R$, and let $q = \nu^+(f, \log(R'))$. The family of zeros of $f$ in $\mathbb{K}\setminus d(0, R'^{-})$ either is finite or is a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$.

**Proof.** By definition, $f$ is of the form $g + h$ with $g(x) = \sum_{-\infty}^{-1} a_n x^n \in \mathcal{A}(\mathbb{K}\setminus d(0, R))$ and $h(x) = \sum_{0}^{+\infty} a_n x^n \in \mathcal{A}(\mathbb{K})$. Particularly, by Theorem 14.6, for every $R'$, $R'' \in ]R, +\infty[$ (with $R' < R''$), both $g, h$ belong to $H(\Delta(0, R', R''))$. Then, so does $f$. Therefore, by Corollary 22.3, $f$ has finitely many zeros in $\Delta(0, R', R'')$ which are on the circles $C(0, r)$ such that $\nu^+(f, \log r) > \nu^-(f, \log r)$. Thus, the family of zeros of $f$ in $\mathbb{K}\setminus d(0, R'^{-})$ either is finite or is a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$. $\square$

**Theorem 22.22.** Let $R, S \in ]0, +\infty[$, $R < S$, and let

$$f(x) = \sum_{-\infty}^{+\infty} a_n x^n \in \mathcal{A}(\mathbb{K}\setminus d(0, R))$$

have infinitely many zeros in $\mathbb{K}\setminus d(0, S)$. Then, for every fixed $t \in \mathbb{N}$, we have $\lim_{r \to +\infty} \frac{|f(r)}{r^t} = +\infty$.

**Proof.** By Theorem 22.21, the sequence of zeros $(\alpha_n)_{n \in \mathbb{N}}$ is such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$. Set $r_n = |\alpha_n|$. The sequence $\nu^+(f, \log(r_n))$ is strictly increasing and hence there exists $q \in \mathbb{N}$ such that $\nu^+(f, \log(r_q)) > t$, therefore $\nu^+(f, \mu) \ge 1 \forall \mu > \log(r_q)$. Set $g(x) = \frac{f(x)}{x^t}$. Then clearly $|g(r) = \frac{|f(r)}{r^t}$ and hence $\nu^+(g, \mu) \ge 1 \forall \mu \ge \log(r_q)$. Thus, the function $(\text{in } \mu) \Psi(f, \mu)$ is a convex function, piecewise affine, whose derivative, when it is derivable, is grater than 1 whenever $\mu \ge \log(r_q)$. Consequently, $\lim_{r \to +\infty} |g(r)| = +\infty$, and therefore, $\lim_{r \to +\infty} \frac{|f(r)}{r^t} = +\infty$. $\square$

**Corollary 22.23.** Let $f(x) \in \mathcal{A}(\mathbb{K})\setminus \mathbb{K}[x]$. Then, for every fixed $t \in \mathbb{N}$, we have $\lim_{r \to +\infty} \frac{|f(r)}{r^t} = +\infty$.

**Theorem 22.24.** Let $r \in \mathbb{R}_+$, let $r_1, r_2 \in ]0, R[$ satisfy $r_1 < r_2$, and let $f \in \mathcal{A}(d(0, R^-))$. If $f$ admits exactly $q$ zeros in $d(0, r_1)$ (taking multiplicity into account) and has no zeros in $\Gamma(0, r_1, r_2)$, then $f$ satisfies

$$\Psi(f, \log r_2) - \Psi(f, \log r_1) = q(\log r_2 - \log r_1).$$
**Proof.** By Lemma 22.13, we have $\nu^+(f, \log r_1) = q$, and by Theorem 22.4, we have $\Psi(f, \mu) - \Psi(f, \log r_1) = q(\mu - \log r_1)$ for every $\mu \in [\log r_1, \log r_2[$, hence by continuity, we have $\Psi(f, \log r_2) - \Psi(f, \log r_1) = q(\log r_2 - \log r_1)$. $\square$

**Theorem 22.25.** Let $f(x) \in \mathcal{A}(d(a, r^{-}))$. If $f$ is not bounded, then $f$ has infinitely many zeros in $d(a, r^{-})$.

**Proof.** Without loss of generality, we can suppose $a = 0$. Suppose $f$ has $t$ zeros in $d(0, r^{-})$. Let $d(0, s)$ be a disk containing all zeros of $f$, with $s < r$. Then, by Theorem 22.22, for all $\rho \in [s, r[$, we have $\Psi(f, \log(\rho)) \le \Psi(f, \log(s)) + t(\log(\rho) - \log(s)) \le \Psi(f, \log(s)) + t(\log(r) - \log(s))$. So, $\psi(f, \mu)$ is bounded, hence by Theorem 14.20, $f$ is bounded in $d(0, r^{-})$, a contradiction. $\square$

**Theorem 22.26.** Let $f \in \mathcal{A}(d(0, R^{-}))$ and let $r_1, r_2 \in ]0, R[$ satisfy $r_1 < r_2$. If $f$ admits exactly $q$ zeros in $d(0, r_1)$ (taking multiplicity into account) and $t$ different zeros $\alpha_j$, of respective multiplicity order $m_j$ ($1 \le j \le t$) in $\Gamma(0, r_1, r_2)$, then $f$ satisfies

$$ \Psi(f, \log r_2) - \Psi(f, \log r_1) = \sum_{j=1}^{t} m_j(\log(r_2) - \Psi(\alpha_j)) + q(\log r_2 - \log r_1). $$

**Proof.** Let $C(0, \rho_h) = 1 \le h \le s$ be the circles containing at least one zero of $f$. For each $h = 1, \dots, s$, let $\alpha_{n(h)}, \dots, \alpha_{n(h+1)-1}$ be the zeros of $f$ in $C(0, \rho_h)$. Let $u = \sum_{j=1}^{t} m_j$.

First, by Theorem 22.24, we note that

$$ (1) \Psi(f, \log(\rho_1)) - \Psi(f, \log(r_1)) = q(\log(\rho_1) - \log(r_1)) $$

and similarly

$$ (2) \Psi(f, \log(r_2)) - \Psi(f, \log(\rho_s)) = (q + u)(\log(r_2) - \log(\rho_s)). $$

Next, for each $h = 1, \dots, s-1$, set $l_h = q + \sum_{j=1}^{n(h+1)-1} m_j$. Then, $f$ has no zero in $\Gamma(0, \rho_h, \rho_{h+1})$ and has $l_h$ zeros in $d(0, \rho_h)$, hence by Theorem 22.24, we have

$$ (3) \Psi(f, \log(\rho_{h+1})) - \Psi(f, \log(\rho_h)) = l_h(\log(\rho_{h+1}) - \log(\rho_h)). $$

Then, by (1), (2), and (3), we can check the conclusion. $\square$
**Corollary 22.27.** Let $f(x) \in \mathcal{A}(d(0, R^{-}))$ be such that $f(0) \neq 0$, let $r \in ]0, R[$ and let $a_j$, $1 \le j \le q$ be the zeros of $f$ in $d(0, r)$, of respective multiplicity $m_j$. Then,

$$ \Psi(f, \log r) = \Psi(f(0)) + \sum_{j=1}^{q} m_j(\log r - \Psi(a_j)). $$

**Corollary 22.28.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, r^{-}))$ have a set of zeros in $d(0, r^{-})$ that consists of a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\alpha_n \neq 0 \ \forall n \in \mathbb{N}$ and where each $\alpha_n$ is of order $u_n$. Then, $f$ is unbounded if and only if the sequence $(\alpha_n)_{n \in \mathbb{N}}$ satisfies $\prod_{n=0}^{\infty} (\frac{|\alpha_n|}{r})^{u_n} = 0$.

**Corollary 22.29.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, r^{-}))$ have a set of zeros in $d(0, r^{-})$ that consists of a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $\alpha_n \neq 0 \ \forall n \in \mathbb{N}$ and where each $\alpha_n$ is of order $u_n$. Then, $\|f\|_D = |f(0)|\prod_{n=0}^{\infty} (\frac{r}{|\alpha_n|})^{u_n}$.

**Corollary 22.30 (Schwarz Lemma).** Let $D = d(a, s)$ and let $f \in H(D)$ have at least (respectively at most) $q$ zeros in $d(a, r)$ with $q > 0$ and $0 < r < s$. Then, we have $\frac{\varphi_{a,s}(f)}{\varphi_{a,r}(f)} \ge (\frac{s}{r})^q$ (respectively $\frac{\varphi_{a,s}(f)}{\varphi_{a,r}(f)} \le (\frac{s}{r})^q$).

**Corollary 22.31.** Let $f \in \mathcal{A}(\mathbb{K})$. The following two statements are equivalent:

(i) $f$ is a polynomial of degree $q$.

(ii) There exists $q \in \mathbb{N}$ such that $\frac{|f(r)}{r^q}$ has a finite limit when $r$ tends to $+\infty$.

**Corollary 22.32.** Let $r, s, R \in ]0, +\infty[$ satisfy $0 < r < s < R$ and let $f \in H((0, R))$. Then,

$$ \Psi(f, \log(s)) - \Psi(f, \log(r)) \\ \le (\Psi(f, \log(R)) - \Psi(f, \log(s))) \left( \frac{\log(s) - \log(r)}{\log(R) - \log(s)} \right). $$

**Proof.** Let $q$ be the total number of zeros of $f$ in $d(0, s)$, each counted with its multiplicity. Then, by Theorem 22.30, we have
$\Psi(f, \log(s)) - \Psi(f, \log(r)) \le q(\log(s) - \log(r))$. On the other hand, $\Psi(f, \log(R)) - \Psi(f, \log(s)) \ge q(\log(R) - \log(s))$. Consequently,

$$q \le \frac{\Psi(f, \log(R)) - \Psi(f, \log(s))}{\log(R) - \log(s)},$$

and hence the proof is over. □

**Theorem 22.33.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(\mathbb{K})$ (respectively $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, r^{-}))$). All zeros of $f$ are of order one and the set of zeros of $f$ is a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that $|\alpha_n| < |\alpha_{n+1}|$ if and only if the sequence $|\frac{a_n}{a_{n+1}}|$ is strictly increasing. Moreover, if these properties are satisfied, then the sequence of zeros of $f$ in $\mathbb{K}$ (respectively in $d(0, r^{-})$) is a sequence $(\alpha_n)_{n \in \mathbb{N}^*}$ such that $\lim_{n \to +\infty} |\alpha_n| = +\infty$ (respectively $\lim_{n \to +\infty} |\alpha_n| = r$) and $|\alpha_n| = |\frac{a_n}{a_{n+1}}|$.

**Proof.** Suppose first that $f \in \mathcal{A}(d(0, r^{-}))$. First, we suppose that the set of zeros of $f$ in $d(0, r^{-})$ is an increasing distances sequence $(\alpha_n)_{n \in \mathbb{N}^*}$. Then by Corollary 22.19, we know that $\lim_{n \to +\infty} |\alpha_n| = r$. By Corollary 22.3, for each $q \in \mathbb{N}^*$, we have $\nu^+(f, \Psi(\alpha_q)) - \nu^-(f, \Psi(\alpha_q)) = 1$ and $|\alpha_0| = |\frac{a_0}{a_1}|$. Then by an immediate induction, we deduce that $|\alpha_n| = |\frac{a_n}{a_{n+1}}|$ for every $n \in \mathbb{N}^*$ and therefore the sequence $|\frac{a_n}{a_{n+1}}|$ is strictly increasing.

Conversely, we suppose that $|\frac{a_n}{a_{n+1}}|$ is a strictly increasing sequence. Hence, we have $|\frac{a_n}{a_{n+1}}| < |\frac{a_{n+1}}{a_{n+2}}|$ for every $n \in \mathbb{N}$. For each $m \in \mathbb{N}^*$, we put $s_m = \Psi(a_m) - \Psi(a_{m-1})$ and $r_m = \theta^{s_m}$. Clearly, we have $\nu^+(f, s_m) - \nu^-(f, s_m) = 1$ for every $m \in \mathbb{N}^*$ and $\nu^+(f, \mu) = \nu^-(f, \mu)$ for every $\mu \in (] -\infty, \log r[ \setminus \{s_m | m \in \mathbb{N}^*\})$. Hence, by Theorem 22.1, $f$ admits exactly one zero in each circle $C(0, r_m)$ and no other zero in $d(0, r^{-})$.

Suppose now that $f \in \mathcal{A}(\mathbb{K})$. The same proof applies with $\lim_{n \to +\infty} |\alpha_n| = +\infty$. □

**Corollary 22.34.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and suppose that the sequence $(\frac{|a_n|}{|a_{n+1}|})_{n \in \mathbb{N}}$ is strictly increasing, of limit $+\infty$ (respectively of limit $R$). Then, $f$ belongs to $\mathcal{A}(\mathbb{K})$ (respectively $\mathcal{A}(d(0, R^{-}))$). Moreover, putting $r_n = \frac{|a_n|}{|a_{n+1}|}$, $n \in \mathbb{N}$, $f$ admits a unique zero on each circle $C(0, r_n)$ and has no other zero in $\mathbb{K}$ (respectively in $d(0, R)$).
**Proof.** Indeed, thanks to the remark at the beginning of Chapter 13, the radius of convergence of $f$ is $+\infty$ (respectively $R$). Then, the conclusion comes from Theorem 22.33. □

**Remark.** We can easily construct a sequence $(r_n)_{n \in \mathbb{N}}$ satisfying the hypothesis of Corollary 22.34 and thereby the function $h$.

It is often uneasy to determine whether a function defined in an infraconnected set is an analytic element. The following example may be useful.

**Theorem 22.35.** Let $R > 0$ and let $(a_n)_{n \in \mathbb{N}}$ be a sequence of $d(a, R^{-})$ such that $|a_n| < |a_{n+1}|$, $\lim_{n \to +\infty} |a_n| = R$, $\prod_{k=0}^{+\infty} \frac{|a_n|}{R} = 0$ and set $r_n = |a_n|$, $n \in \mathbb{N}$. Let $f \in A(d(a, R^{-}))$ admit each $a_n$ as a zero of order 1 and no other zero in $d(a, r_n)$. Let $E = d(a, R^{-}) \setminus (\bigcup_{n \in \mathbb{N}} C(a, r_n))$. Then the function $\frac{1}{f}$ defined in $E$ belongs to $H(E)$.

**Proof.** Without loss of generality, we can obviously suppose $a = 0$ and $f(0) = 1$. By Corollary 22.28, $f$ belongs to $A_u(d(0, R^{-}))$. For each $n \in \mathbb{N}$, set $r_n = |a_n|$. Let $r \in ]0, R[$. If $r \neq r_n$ $\forall n \in \mathbb{N}$, then $|f(x)| = |f|(r) \forall x \in C(0, r)$. Now, let $x \in C(0, r_n)$. If $|x - a_n| = r_n$, we have $|f(x)| = r_n$. And if $x$ belongs to $E$ so that $|x - a_n| < r_n$, then $|f(x)| \ge |f|(r_n)^{\frac{r_n}{s}}$. Consequently,

$$ (1) \qquad |f(x)| = |f|(|x|) \quad \forall x \in E. $$

Let $f(x) = \sum_{n=0}^{+\infty} c_n x^n$ and for each $k \in \mathbb{N}$, let $P_k(x) = \sum_{n=0}^{k} c_n x^n$. Of course, the sequence $(P_k)_{k \in \mathbb{N}}$ converges to $f$ uniformly in every disk $d(0, r)$ with $r \in ]0, R[$. Moreover, by Theorem 22.34, each zero $a_k$ of $f$ satisfies $|a_k| = \frac{|c_{k-1}|}{|c_k|}$. Consequently, for $n > k$, $P_n$ also admits a unique zero $a_{n,k}$ in $C(0, r_n)$ and has no other zero in $E$. Therefore, we have $|P_n|(r) = |f|(r) \forall r \le r_n$. We show that the sequence $(\frac{1}{P_n})_{n \in \mathbb{N}}$ converges to $\frac{1}{f}$ in $H(E)$. Indeed, let us fix $\epsilon > 0$ and let us choose $q \in \mathbb{N}$ such that $|f|(r_q) > \frac{1}{\epsilon}$. Consider now integers $n > q$ such that $|f(x) - P_n(x)| < \epsilon \forall x \in d(0, r_q)$. Then obviously,

$$ \left| \frac{1}{f(x)} - \frac{1}{P_n(x)} \right| < \epsilon \quad \forall x \in E \cap d(0, r_q). $$
On the other hand, given $x \in E \setminus d(0, r_q)$, we have $|P_n|(r) \ge |P_n|(r_q) > \frac{1}{\epsilon}$, hence $|\frac{1}{P_n(x)} - \frac{1}{f(x)}| < \epsilon$. Therefore, $\|\frac{1}{P_n(x)} - \frac{1}{f(x)}\|_E \le \epsilon$. This finishes proving that $\frac{1}{f(x)}$ belongs to $H(E)$. $\square$

**Notation.** Let $f \in \mathcal{A}(\mathbb{K})$ and let $n \in \mathbb{N}^*$. We denote by $f^{<n}$ the function $f \circ f \circ f \circ \dots \circ f$, $n$ times.

**Theorem 22.36.** Suppose the residue characteristic $p$ is different from 0. There exists functions $\mathcal{A}(\mathbb{K})$ and points $a \in \mathbb{K}$ such that $\lim_{n \to +\infty} |f^{<n}(a)| = +\infty$.

**Proof.** Let $t(n) = n^2$, $n \in \mathbb{N}$. Let $f(x) = \sum_{n=0}^{+\infty} p^{2t(n)} x^{2n+1}$. For each $n \in \mathbb{N}$, put $r_n = \frac{p^{t(n-1)}}{p^{t(n)}}$. Then, we can check that $f$ admits exactly two zeros (taking multiplicity into account) in the circle $C(0, r_n)$ and has no zeros outside $\bigcup_{n=1}^{\infty} C(0, r_n)$, except $\{0\}$.

Consider now a number $s \in ]r_n, r_{n+1}[$ of the form $q\sqrt{p}$, with $q \in \mathbb{Q}$. Inside the disk $d(0, s)$, $f$ admits $2n + 1$ zeros (taking multiplicity into account). Consequently, $|f|(s) = |p^{2t(n)}|s^{2n+1} = \frac{p^{-2t(n)}|q|^{2n+1}p^{-2n}}{\sqrt{p}} = \frac{p^{-2t(n)-2n}|q|^{2n+1}}{\sqrt{p}}$. Thus, we can see that $|f|(s)$ is of the form $q'\sqrt{p}$ with $q' \in \mathbb{Q}$. Moreover, since $q$ belongs to $\mathbb{Q}$, $f$ has no zero in $C(0, s)$, hence $|f(x)| = |f|(s) = \frac{|q'|}{\sqrt{p}} \forall x \in C(0, s)$. Now consider $\frac{|f|(s)}{s} = \frac{|p^{2t(n)}|s^{2n+1}}{s} = p^{-2t(n)}s^{2n}$.

On the other hand, by (1), $r_n = p^{t(n)-t(n-1)}$. Since $r_n < s$, we can derive

$$p^{-2t(n)}s^{2n} > p^{-2t(n)}p^{2n(t(n)-t(n-1))}.$$

Now, $-2t(n) + 2n(t(n) - t(n-1)) = 2(-t(n) + n(t(n) - t(n-1))) = 2(-n^2 + n(2n-1)) = 4n^2 - 2n$. Consequently, when $s > r_n$, we have $\frac{|f|(s)}{s} > p^{4n^2-2n}$. Now, let us take $s_2 > r_2$, and for each $n \in \mathbb{N}$, $n \ge 2$, let us define by induction $s_{n+1} = |f|(s_n)$. So, we have $\frac{s_{n+1}}{s_n} \ge p^8$, hence $\lim_{n \to +\infty} s_n = +\infty$. But by construction, for all $x \in C(0, s_n)$, we have $|f(x)| = |f|(s_n)$. Consequently, taking $a \in C(0, s_2)$, we have $|f^{<k}(a)| = s_k$, and hence, $\lim_{n \to +\infty} |f^{<n}(a)| = +\infty$. $\square$
# Chapter 23

## Image of a Disk

In this chapter, D is just an open subset of $\mathbb{K}$.

**Theorem 23.1.** Let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in H(d(a,r))$. Then the following statements (a), (b), (c), (d), and (e) are equivalent:

(a) $|a_0| > |a_n|r^n$ for all $n > 1$.

(b) $\|f - f(a)\|_{d(a,r)} < |f(a)|$.

(c) $f$ has no zero in $d(a, r)$.

(d) $|f(x)|$ is constant and different from 0 in $d(a, r)$.

(e) $f$ is invertible in $H(d(a, r))$.

**Proof.** First, (a) and (b) are equivalent by Theorem 14.6. Second, (a), (c), and (d) are equivalent by Theorems 22.7 and 22.16. Third, by Lemma 11.3, (d) implies (e), and finally, (e) obviously implies (c). $\square$

**Corollary 23.2.** Let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in \mathcal{A}(d(a,r^-))$ (respectively let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in H((d(a,r^-)))$). Then statements (a), (b), (c), (d), and (e) are equivalent:

(a) $|a_0| \ge |a_n|r^n$ for all $n > 1$.

(b) $|f - f(a)| < |f(a)| \quad \forall x \in d(a, r^-)$.

(c) $f$ has no zero in $d(a, r^-)$.

(d) $|f(x)|$ is constant and different from 0 in $d(a, r^-)$.

(e) $f$ is invertible in $\mathcal{A}(d(a, r^-))$ (respectively $f$ is invertible in $H(d(a, r^-)$).

**Proof.** Concerning $\mathcal{A}(d(a, r^-))$, we just have to apply Theorem 23.1 to $f$ in $H(d(a, \rho))$ for every $\rho \in ]0, r[$. Concerning $H(d(a, r^-))$,
we can use Lemma 11.3 to check that an element of $H(d(0, r^{-}))$ having no zero is invertible. □

**Theorem 23.3.** Let $a, b \in \mathbb{K}$ and $r, s \in \mathbb{R}_+^*$, and let $f \in \mathcal{A}(d(a, r^{-}))$, $g \in \mathcal{A}(d(b, s^{-}))$ be such that $f(d(a, r^{-})) \subset d(b, s^{-})$. Then, $g \circ f$ belongs to $\mathcal{A}(d(a, r^{-}))$.

**Proof.** Without loss of generality, we can clearly assume $a = b = 0$. First, suppose that $f$ has no zero in $d(0, r^{-})$. Then, $|f(x)|$ is equal to a constant $c$ in $d(0, r^{-})$, with $c < s$. Hence, of course, $f(d(0, r^{-}))$ is included in $d(0, c)$. Now, let $\rho \in ]0, r[$. The restriction of $f$ to $d(0, \rho)$ belongs to $H(d(0, \rho^{-}))$ and the restriction of $g$ to $d(0, c)$ belongs to $H(d(0, c))$. Hence, the restriction of $g \circ f$ to $d(0, \rho)$ belongs to $H(d(0, \rho))$. Consequently, by Corollary 12.3, $g \circ f$ belongs to $H(d(0, \rho))$. This is true for every $\rho \in ]0, r[$, and therefore, this shows that $g \circ f$ belongs to $\mathcal{A}(d(0, r^{-}))$.

Now, we suppose that $f$ admits at least one zero in $d(0, r^{-})$. Hence, there exists $r' \in ]0, r[$ such that $f$ has at least one zero in $d(0, r')$. Therefore, by Corollary 22.30, $\|f\|_{d(0,\rho)}$ is strictly increasing in $\rho$ in the interval $[r', r[$. Now, let $\rho \in ]0, r[$ and let $\sigma = \|f\|_{d(0,\rho)}$. The restriction of $f$ to $d(0, \rho)$ belongs to $H(d(0, \rho))$, and further, $f(d(0, \rho))$ is included in $d(0, \sigma)$. Since $g$ belongs to $H(d(0, \sigma))$, $g \circ f$ belongs to $H(d(0, \rho))$. As previously, this is true for every $\rho \in ]0, r[$, hence $g \circ f$ belongs to $\mathcal{A}(d(0, r^{-}))$. □

**Theorem 23.4 (Dieudonné-Dwork).** Let $f \in \mathcal{A}(d(0, R^{-}))$ satisfies $f(0) = 1$ and have no zero in $d(0, R^{-})$. There exists a sequence $(u_k)_{k \in \mathbb{N}^*}$ in $d(0, R)$ such that $f(x) = \prod_{k=1}^{\infty} (1 - u_k x^k)$ whenever $x \in d(0, R^{-})$.

**Proof.** Since $f(0) = 1$, we can write $f(x)$ in the form $1 + \sum_{n=1}^{\infty} b_n x^n$. Since $f$ has no zero in $d(0, R^{-})$, by Corollary 23.2, we have $|b_n|R^n \le 1$ for every $n \in \mathbb{N}^*$. Now, suppose that we have already found $u_1, \dots, u_k$ such that $f(x)$ factorizes in the form

$$ (\mathcal{R}_k) \prod_{j=1}^{k} (1 - u_j x^j) \left( 1 + x^{k+1} \sum_{n=0}^{\infty} \beta_{n,k+1} x^n \right) $$
with $|\beta_{n,k+1}| \le \frac{1}{R}$ for every $n \in \mathbb{N}$.
Actually, we have $(1 + \beta_{0,k+1}x^{k+1})(1 + \sum_{n=1}^{\infty}(-\beta_{0,k+1}x^{k+1})^n) = 1$, and therefore, we can factorize $(1 + x^{k+1}\sum_{n=0}^{\infty} \beta_{n,k+1}x^n)$ in the form

$$(1+\beta_{0,k+1}x^{k+1}) \left( 1 + \sum_{n=1}^{\infty} (-\beta_{0,k+1}x^{k+1})^n \right) \left( 1 + x^{k+1} \sum_{n=0}^{\infty} \beta_{n,k+1}x^n \right).$$

Now, consider the function $g_{k+1}$ defined as

$$g_{k+1}(x) = \left( 1 + \sum_{n=1}^{\infty} (-\beta_{0,k+1}x^{k+1})^n \right) \left( 1 + x^{k+1} \sum_{n=0}^{\infty} \beta_{n,k+1}x^n \right).$$

In $g_{k+1}$, it is seen that the term in $x^{k+1}$ is equal to 0, so $g_{k+1}$ is of the form

$$\left( 1 + x^{k+2} \sum_{n=0}^{\infty} \beta_{n,k+2} x^n \right), \quad \text{with } |\beta_{n,k+2}| \le \frac{1}{R}, \quad \text{for every } n \in \mathbb{N}.$$

Now, we just put $u_{k+1} = \beta_{0,k+1}$ and then we have proven ($\mathcal{R}_{k+1}$). Since ($\mathcal{R}_0$) is trivially satisfied, by induction, we can construct a sequence $(u_k)_{k \in \mathbb{N}^*}$ in $d(0, R^-)$ and a sequence $(g_k)_{k \in \mathbb{N}^*}$ in $\mathcal{A}(d(0, R^-))$ such that for each $k \in \mathbb{N}$, $g_k$ is of the form $1 + x^{k+1} \sum_{n=0}^{\infty} \beta_{n,k+1}x^n$ and satisfies $f(x) = \prod_{j=1}^{k}(1 - u_jx^j)g_k(x)$. For each $k \in \mathbb{N}^*$, let $f_k(x) = \prod_{j=1}^{k}(1 - u_jx^j)$.

It is seen that for each $r \in ]0, R[$, we have $\|g_k - 1\|_{d(0,r)} \le r^{k+1}$. As a consequence, for every $r \in ]0, R[$, the sequence $(f_k)_{k \in \mathbb{N}^*}$ converges to $f$ in $H(d(0, r))$, and therefore, we have $f(x) = \prod_{k=1}^{\infty}(1 - u_kx^k)$ for all $x \in d(0, R^-)$. That ends the proof. $\square$

**Proposition 23.5.** Let $D$ be a set of the form $\bigcup_{i \in I} d(\alpha_i, r^-)$ with $|\alpha_i - \alpha_j| = r$ whenever $i \ne j$. Let $\ell$ be fixed in $I$, let $f \in H(D)$ be such that $f(x) = \sum_{n=0}^{\infty} a_n(x - \alpha_\ell)^n \in H(D)$ whenever $x \in d(\alpha_\ell, r^-)$, and let $s = \sup_{n \ge 1} |a_n|r^n$. For every $i \in I$, let $\beta_i = f(\alpha_i)$. Then, we have $f(d(\alpha_i, r^-)) = d(\beta_i, s^-)$ for every $i \in I$ and $|\beta_i - \beta_j| \le s$ whenever $i, j \in I$. Moreover, if $I$ is not finite, the equality $|\beta_i - \beta_\ell| = s$ holds for every $i \in I$ but finitely many.
**Proof.** We set $a = \alpha_\ell$ and $g = f - \beta_\ell$. By Theorem 22.9, for every $\rho \in ]0, r[$, we have $\Psi_a(g, \log \rho) = \sup_{n \ge 1} \Psi(a_n) + n \log \rho$, hence by continuity,

$$ \Psi_a(g, \log r) = \sup_{n \ge 1} \Psi(a_n) + n \log r = \log s. $$

Now, let $j \in I$, $j \ne \ell$ and set $b = \alpha_j$. Since $|b - a| = r$, by Proposition 19.4, we have $\Psi_a(g, \log r) = \Psi_b(g, \log r)$, hence $\Psi(g(x)) \le \Psi_a(g, \log r) = \log s$ for all $x \in d(b, r^-)$. Thus, we see that $|f(x) - \beta_\ell| \le s$ for all $x \in d(b, r^-)$. Hence, by Corollary 22.11, $f(d(b, r^-))$ is a disk $d(\beta_j, t^-)$ with $t \le s$. Since $\alpha_\ell$ and $\alpha_j$ play the same role, in the same way, we show that $s \le t$, and therefore, $s = t$. Thus, we have proven that $f(d(\alpha_i, r^-)) = d(\beta_i, s^-)$ for every $i \in I$ and $|\beta_i - \beta_j| \le s$ whenever $i, j \in I$.

Now, we suppose that $I$ is infinite. By Proposition 19.1, the equality $\Psi(g(x)) = \Psi_a(g, \log r) = \log s$ holds in all the classes of $C(a, r)$ but finitely many ones, hence we have $|\beta_j - \beta_\ell| = s$ for every $j \in I$ but finitely many and this ends the proof of Proposition 23.5. $\square$

**Theorem 23.6.** Let $D$ be an open analytic subset of $\mathbb{K}$, let $f \in H(D)$, and let $D' = f(D)$. Then, $D'$ is open and satisfies codiam($D') \ge \text{codiam}(D) \inf_{x \in D} |f'(x)|$.

**Proof.** Let $b \in f(D)$ and let $a \in D$ be such that $f(a) = b$. Since $D$ is open, there exists a disk $d(a, r)$ included in $D$. Let $f(x) = \sum_{j=0}^\infty a_n(x-a)^n \quad \forall x \in d(a, r)$. Since $D$ is an analytic set, $f$ is not identically zero in any disk included in $D$, hence by Theorem 22.9, $||f - b||_{d(a,r)}$ is a number $s = \sup_{n \ge 1} |a_n|r^n$. Then, by Theorem 22.9, $f(d(a, r))$ is the disk $d(b, s)$, hence $d(b, s) \subset D$, which proves that $f(D)$ is open. Particularly, $|f'(a)| = |a_1|$, hence $||f - b||_{d(a,r)} \ge r|f'(a)|$, therefore, $\delta(D', (\mathbb{K}\setminus D')) \ge \delta(D, (\mathbb{K}\setminus D)) \inf_{x \in D} |f'(x)|$. $\square$

**Theorem 23.7.** Let $f \in H(d(0, r))$, let $t = \nu^+(f, \log r)$, and assume $t \ge 1$. Suppose that $f'$ is not identically zero and let $\alpha_1, \dots, \alpha_q$ be the zeros of $f'$ in $d(0, r)$. For every $b \in f(d(0, r)) \setminus \{\alpha_1, \dots, \alpha_q\}$, $f - b$ admits exactly $t$ zeros of order $1$ in $d(0, r)$.

**Proof.** Let $b \in f(d(0, r)) \setminus \{\alpha_1, \dots, \alpha_q\}$. By Theorem 22.7, $f - b$ admits $t$ zeros in $d(0, r)$ (taking multiplicity into account). But for each zero $\alpha$ of $f - b$ (as $\alpha \ne \alpha_j$ whenever $j = 1, \dots, q$), we have $f'(\alpha) \ne 0$, hence the $t$ zeros of $f - b$ are of order one. $\square$
**Definition.** Let $f \in H(D)$. Then, $f$ is said to be strictly injective in $D$ if $f$ is injective and if $f'(x) \neq 0$ whenever $x \in D$.

In the same way, given $a \in \mathbb{K}$ and $r > 0$, an analytic function $f(x) \in \mathcal{A}(d(a, r^{-}))$ is said to be strictly injective in $d(a, r^{-})$ if $f$ is injective and if $f'(x) \neq 0$ whenever $x \in D$.

**Theorem 23.8.** Let $\mathbb{K}$ have characteristic zero and let $f \in H(D)$ be injective in $D$. Then, $f$ is strictly injective.

**Proof.** Suppose that $f$ is not strictly injective and let $\alpha \in D$ be a zero of $f'$. Let $d(\alpha, r)$ be a disk included in $D$. Without loss of generality, we may assume $\alpha = 0$. Hence, in $d(0, r)$, $f(x)$ is equal to a series of the form $a_0 + \sum_{n=q}^{\infty} a_n x^n$ with $q \ge 2$ and $a_q \neq 0$. Therefore, we have $\nu^+(f, \log r) \ge q \ge 2$. Let $t = \nu^+(f, \log r)$. Since $\mathbb{K}$ has characteristic zero, $f'$ is not identically zero and therefore admits finitely many zeros $\alpha_1 = 0, \alpha_2, \dots, \alpha_s$ in $d(0, r)$. Let $b \in f(d(0, r)) \setminus f(\{\alpha_1, \dots, \alpha_s\})$. By Theorem 23.7, $f-b$ admits $t$ simple zeros in $d(0, r)$ and this contradicts the hypothesis “$f$ injective in $d(0, r)$”. $\square$

**Theorem 23.9.** Let $a \in \mathbb{K}$, $r \in \mathbb{R}_+$, let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in H(d(a, r))$, and let $s = \sup_{n \ge 1} |a_n| r^n$ be $> 0$. Then the following statements are equivalent:

$$ (\alpha) \quad |a_1| > |a_n| r^{n-1} \quad \text{whenever } n > 1. $$

$$ (\beta) \quad |f(x) - f(y)| = |x - y| |a_1| \quad \text{whenever } x, y \in d(a, r). $$

$$ (\gamma) \quad f \text{ is strictly injective in } d(a, r). $$

Moreover, when conditions $(\alpha), (\beta)$, and $(\gamma)$ are satisfied, then we have $s = |a_1| r$ and $|f'(x)| = |a_1|$ whenever $x \in d(a, r)$.

**Proof.** Without loss of generality, we may obviously assume $a = 0$. First, we suppose $(\alpha)$ is satisfied and consider

$$ f(x) - f(y) = (x - y) \left( a_1 + \sum_{n=2}^{\infty} a_n \left( \sum_{j=0}^{n-1} x^j y^{n-1-j} \right) \right). $$

For every $n \ge 2$, it is seen that $|x^j y^{n-1-j}| \le r^{n-1}$. Hence, we have $|a_1| > |\sum_{n=2}^{\infty} a_n (\sum_{j=0}^{n-1} x^j y^{n-1-j})|$ and thereby $|f(x) - f(y)| = |a_1| |x - y|$. At the same time, we note that $(\alpha)$ implies $|f'(x)| = |a_1|$
whenever $x \in d(a, r)$, while, by Theorem 22.9, we have $s = |a_1|r$. So $(\alpha)$ implies $(\beta)$.

Second, we suppose $(\beta)$ is satisfied. Since $s > 0$, by Corollary 22.10, $f$ is not a constant, hence $a_1 \neq 0$. Then, by $(\beta)$, we have $f(x) \neq f(y)$ whenever $x \neq y$. Moreover, $|f'(x)| = |a_1| \neq 0$, hence $(\gamma)$ is satisfied.

Third, we suppose $(\gamma)$ is satisfied. Let $b \in f(d(a, r))$, let $g = f - b$, and let $t = \nu^+(g, \log r)$. If $t \ge 2$, either $g$ admits several different zeros or $g$ admits a zero $\alpha$ of order $t$. In both cases, we see that $g$ is not strictly injective, hence neither is $f$. Finally, we have $t = 1$ and hence $(\alpha)$ is satisfied. This ends the proof of Theorem 23.9. $\square$

Theorem 23.9 is easily applied to analytic functions inside a disk.

**Corollary 23.10.** Let $f(x) = \sum_{n=0}^{\infty} a_n (x-a)^n \in A_b(d(a, r^{-}))$ and suppose that the number $s = \sup_{n \ge 1} |a_n| r^n$ is strictly positive. Then conditions $(\alpha), (\beta), (\gamma)$, and $(\delta)$ are equivalent:

(α) $|a_1| \ge |a_n| r^{n-1}$ whenever $n > 1$.

(β) $|f(x) - f(y)| = |x - y| |a_1|$ whenever $x, y \in d(a, r^{-})$.

(γ) $f$ is strictly injective in $d(a, r^{-})$.

(δ) $s = |a_1|r$.

Moreover, when conditions $(\alpha), (\beta), (\gamma)$, and $(\delta)$ are satisfied, we have $|f'(x)| = |a_1|$ whenever $x \in d(a, r^{-})$.

**Proof.** For every $\rho \in ]0, r[$, we apply Theorem 23.9 to $f \in H(d(a, \rho))$. $\square$

**Lemma 23.11.** Let $f \in A(d(\alpha, r^{-}))$ be injective and such that $f'$ is not identically zero. Then, $f$ is strictly injective.

**Proof.** We may obviously assume that $f'(\alpha) \neq 0$ and $\alpha = 0$. Hence, $f$ is of the form $\sum_{n=0}^{\infty} a_n x^n$ with $a_1 \neq 0$. If $f'$ has a zero $\beta$, there exists an integer $q > 1$ such that

$$ (1) \quad |qa_q| |\beta|^{q-1} = |a_1|. $$

Let $g(x) = f(x) - f(0)$. Then, $g$ is also injective and has a simple zero at 0. But by (1), we have $|a_q| |\beta|^q \ge |a_1| |\beta|$, hence we have $\nu^+(g, \mu) \ge q$ when $\mu$ is close enough to $\log(r)$. Then, by Theorem 22.7, we know that $g$ has at least $q$ zeros in $d(0, r^{-})$ and then admits another zero $\gamma \neq 0$, which contradicts the fact that it is injective. Consequently, $f'$ has no zero in $d(\alpha, r^{-})$ i.e. $f$ is strictly injective. $\square$
We are now able to study the inverse functions of an analytic element inside a disk $d(a, r)$.

**Theorem 23.12.** Let $a \in \mathbb{K}$, $r \in \mathbb{R}_+$, let $f(x) \in H(d(a, r))$ be strictly injective in $d(a, r)$, let $s = |f'(a)|r$, and let $b = f(a)$. The homomorphism $\Theta$ from $H(d(b, s))$ into $H(d(a, r))$ defined as $\Theta(h) = h \circ f$ is an isometric isomorphism from $H(d(b, s))$ onto $H(d(a, r))$.

**Proof.** Without loss of generality, we may obviously suppose $a = b = f(a) = 0$ and $f'(0) = 1$. We put $E = d(0, r)$. Hence, by Theorem 23.9, in $d(0, r)$, $f(x)$ is equal to a series of the form $x + \sum_{j=2}^{\infty} a_j x^j$ with $\sup_{j \ge 2} |a_j| r^j < r$. Next, by Theorem 22.9, $r$ is equal to $s$. It is obviously seen that $\|\Theta(h)\|_E = \|h\|_E$ for every $h \in H(E)$. So, we only have to prove that $\Theta$ is surjective. Let $\lambda = \inf_{P \in \mathbb{K}[x]} \|\Theta(P) - x\|_E$ and suppose $\lambda > 0$.

Set $\lambda = \frac{\inf_{P \in \mathbb{K}[x]} (\|P(f) - x\|)}{r}$ and suppose that $\lambda > 0$. By definition, $\lambda < 1$ because $\|f - x\| < r$. Let $P \in \mathbb{K}[x]$ be such that $\|P(f) - x\| < r\lambda^{\frac{2}{3}}$. Then, we can write $x = P(f) + h(x)$ with $h \in H(d(0, r))$ and $h(x) = \sum_{k=0}^{\infty} c_k x^k$ with $|c_n| < r\lambda^{\frac{2}{3}} \forall n \in \mathbb{N}$ and $\lim_{n \to +\infty} c_n = 0$.

Let $q \in \mathbb{N}$ be such that $|c_n| < r\lambda^{\frac{2}{3}} \forall n > q$ and let $\omega(x) = \sum_{n=0}^{q} c_n x^n$. We first note that for all $n \in \mathbb{N}$, we have

$$ (1) \quad \|(P(f) + h(x))^n - (P(f) + \omega(x))^n\| \le \|h - \omega\| \le r\lambda^{\frac{2}{3}}. $$

Now, for each $n \in \mathbb{N}$, set $(P(f) + h(x))^n = (P(f) + \omega(x))^n + \omega_n(x)$. Then by (1), we have

$$ (2) \quad \|\omega_n\| \le r\lambda^{\frac{2}{3}} \forall n \in \mathbb{N}. $$

On the other hand, $\|(P(f) + \omega(x))^n - (P(f))^n\| \le \|\omega\| \le r\lambda^{\frac{2}{3}} \forall n \in \mathbb{N}$. Consequently, we can write

$$ (3) \quad \|(P(f) + \omega(x))^n = (P(f))^n + \ell_n(x) \text{ with } \|\ell_n\| \le r\lambda^{\frac{2}{3}} \forall n \in \mathbb{N}. $$

Now, we have

$$ x - P(f) = \sum_{n=0}^{q} c_n ((P(f) + \omega(x))^n + \omega_n(x)) \\ + \sum_{n=q+1}^{\infty} c_n ((P(f) + \omega(x))^n + \omega_n(x)), $$
therefore by (2) and (3), we can write

$$ (4) \qquad x - P(f) = \sum_{n=0}^{q} c_n(P(f))^n + \sum_{n=0}^{q} c_n(\ell_n(x) + \omega_n(x)) \\ \qquad \qquad + \sum_{n=q+1}^{\infty} c_n((P(f) + \omega(x))^n + \omega_n(x)) $$

with $\|c_n((P(f) + \omega(x))^n + \omega_n(x))\| < r\lambda^2 \ \forall n \ge q$ and $\|\ell_n + \omega_n\| < r\lambda^{\frac{2}{3}} \ \forall n = 0, \dots, q$, hence $\|\sum_{n=0}^q c_n(\ell_n(x) + \omega_n(x))\| \le r\lambda^{\frac{4}{3}}$. Consequently, $\|x - P(f) - \sum_{n=0}^q c_n(P(f))^n\| \le r\lambda^{\frac{4}{3}}$. Set $Q(x) = P(x) - \sum_{n=0}^q c_n(P(x))^n$. Then, the polynomial $Q$ satisfies $\|x - Q(f)\| \le r\lambda^{\frac{4}{3}}$, a contradiction to the hypothesis $\lambda > 0$. Consequently, $\lambda = 0$, and therefore, $x$ does belong to the closure of $\Theta(H(E))$. But since $\Theta$ is isometric, $\Theta(H(E))$ is obviously closed in $H(E)$, and therefore, $x$ belongs to $\Theta(H(E))$. As a consequence, $\Theta(H(E)) = H(E)$. $\square$

**Corollary 23.13.** Let $f \in H(d(a, r))$ be strictly injective and let $d(b, s) = f(d(a, r))$. Then, $f$ belongs to $H(d(b, s))$.

**Corollary 23.14.** Let $\mathbb{K}$ have characteristic 0, let $f \in H(a, r))$ be injective, and let $d(b, s) = f(d(a, r))$. Then, $f$ belongs to $H(d(b, s))$.

**Corollary 23.15.** Let $f \in \mathcal{A}(d(0, r^{-}))$ be strictly injective in $d(0, r^{-})$ and let $s = r|f'(0)|$. Then, $f$ belongs to $\mathcal{A}(d(0, s^{-}))$.

**Proof.** Indeed, by Corollary 23.13, $f$ belongs to $H(d(0, u))$ for every $u \in [0, s]$. $\square$

**Corollary 23.16.** Let $\mathbb{K}$ have characteristic 0, let $f \in \mathcal{A}(d(0, r^{-}))$ be injective in $d(0, r^{-})$, and let $s = r|f'(0)|$. Then, $f$ belongs to $\mathcal{A}(d(0, s^{-}))$.

**Definition.** An injective analytic function $f \in \mathcal{A}_b(d(a, r^{-}))$ is said to be bianalytic if $f^{-1}$ belongs to $\mathcal{A}(f(d(a, r^{-})))$.
We use the following lemma in topology:

**Lemma 23.17.** Let $E$ be a topological space and let $F$ and $G$ be subsets dense in $E$. If $F$ is open, then $F \cap G$ is dense in $E$.

**Proof.** Indeed, let $a \in E$, let $V$ be an open neighborhood of $a$, and let $u \in V \cap F$. Since $F$ is open, $V \cap F$ is a neighborhood of $u$. Hence, as $G$ is dense in $E$, there exists $x \in (V \cap F) \cap G$. Therefore, $V \cap (F \cap G) \neq \emptyset$. $\square$

**Lemma 23.18.** Let $\overline{D}$ be open. Let $a$ be a point of $\overline{D} \setminus D$ and let $f \in H(D \cup \{a\})$ be strictly injective in $D$. There exists an open set $E$ satisfying:

(i) $\overline{E}$ is open.

(ii) $D \cup \{a\} \subset E \subset \overline{D}$.

(iii) $f$ belongs to $H(E)$ and is strictly injective in $E$.

**Proof.** By Theorem 11.5, we know that $f$ is of the form $g+h$ with $g \in H(\overline{D})$ and $h \in R(\mathbb{K}\setminus(\overline{D}\setminus(D \cup \{a\})))$. Since $\overline{D}$ is open, there exists $\sigma > 0$ such that $d(a, \sigma) \subset \overline{D}$. Now, as $h \in R(D \cup \{a\})$, there exists $\tau > 0$ such that $h \in R(D \cup d(a, \tau))$. Let $\rho = \min(\sigma, \tau)$ and let $E = D \cup d(a, \rho)$. It is seen that $\overline{E} = \overline{D} \cup d(a, \rho)$, and therefore, $\overline{E}$ is open. Next, both $g, h$ belong to $H(E)$, hence so does $f$.

We suppose that $f$ is not injective in $E$. Let $b$ and $c \in E$ be such that $f(b) = f(c)$ and let $\omega = f(b)$. Let $r \in \mathbb{R}_+^*$ be such that $d(b, r) \cup d(c, r) \subset E$ whereas $d(b, r) \cap d(c, r) = \emptyset$. Then $f(d(b, r))$ is a disk $d(\omega, s)$ whereas $f(d(c, r))$ is a disk $d(\omega, t)$. We may obviously assume $s \le t$. Let $\Sigma = d(b, r) \cap D$ and let $\Lambda = d(c, r) \cap D$. Obviously, $\Sigma$ is dense in $d(b, r)$ whereas $\Lambda$ is dense in $d(c, r)$, hence $f(\Sigma)$ is dense in $d(\omega, s)$ whereas $f(\Lambda)$ is dense in $d(\omega, t)$. Hence, $f(\Lambda) \cap d(\omega, s)$ is dense in $d(\omega, s)$. Since $\Sigma$ and $\Lambda$ are open sets in $\mathbb{K}$, by Corollary 22.10, both $f(\Sigma)$ and $f(\Lambda)$ are open sets in $\mathbb{K}$ because $f$ is not a constant in $d(b, r)$ or in $d(c, r)$. Hence, both $f(\Lambda) \cap d(\omega, s)$, $f(\Sigma)$ are dense open subsets of $d(\omega, s)$. Therefore, by Lemma 23.17, we see that $f(\Sigma) \cap f(\Lambda)$ is dense in $d(\omega, s)$ and certainly is not empty. Let $y \in f(\Sigma) \cap f(\Lambda)$ and $\alpha \in \Sigma$, $\beta \in \Lambda$ satisfy $f(\alpha) = f(\beta) = y$. By definition of $\Sigma$ and $\Lambda$, we see that $\alpha$ and $\beta \in D$ whereas $\alpha \ne \beta$. This contradicts the hypothesis "f is injective in D" and finally shows f to be injective in $E$.
Now, since $f$ is strictly injective in $d(a, r^-) \cap D$, we have $f'(x) \neq 0$ whenever $x \in d(a, r^-) \cap D$, hence by Lemma 23.11, $f$ is strictly injective in $d(a, r^-)$. Finally, we have $f'(x) \neq 0$ whenever $x \in E$, and this ends the proof of Lemma 23.18. □

We remember that Condition (B) was defined in Chapter 17.

**Theorem 23.19.** Let $\bar{D}$ be open, let $D'$ satisfy $D \subset D' \subset \bar{D}$, and let $f \in H(D')$ be strictly injective in $D$. There exists an open set $D''$ satisfying $D' \subset D'' \subset \bar{D}$ such that $f$ belongs to $H(D'')$ and is strictly injective in $D''$.

**Proof.** By Lemma 23.18, for every $a \in D'$, there exists an open set $D_a$ such that $\bar{D}_a$ is open, satisfying Condition (B) such that $D \cup \{a\} \subset D_a \subset \bar{D}$ and such that $f$ belongs to $H(D_a)$ and is strictly injective in $D_a$. Let $D'' = \cup_{a \in D'} D_a$. Then, $D''$ is open and such that $D' \subset D'' \subset \bar{D}$. By Theorem 11.5, $f$ has a unique decomposition in the form $g+h$ with $g \in H(\bar{D})$ and $h \in R(\mathbb{K}\setminus(\bar{D} \setminus D'))$. Since $f \in H(D_a)$, obviously, $h \in H(D_a)$. Hence, $h$ has no pole in $D_a$ whenever $a \in D'$, therefore $h \in R(D'')$. Hence, $f$ belongs to $H(D'')$. Moreover, by Lemma 23.18, we have $f'(x) \neq 0$ whenever $x \in D_a$, hence whenever $x \in D''$.

Now, we just have to check that $f$ is injective in $D''$. Let $a, b \in D'$ satisfy $f(a) = f(b)$. Since $D_a$ satisfies Condition (B), we may apply Lemma 23.18 to $D_a$ and $b$ and then we have an open set $E$ such that $D_a \cup \{b\} \subset E \subset \bar{D}_a = \bar{D}$ and such that $f$ belongs to $H(E)$ and is strictly injective in $E$. Hence, the hypothesis $f(a) = f(b)$ is impossible, and therefore, $f$ is injective in $D'$. But the hypothesis made on $D'$ actually is satisfied on $D''$. Hence, $f$ is injective in $D''$ and this finishes proving Theorem 23.19. □

**Proposition 23.20.** Let $(d(\alpha_i, r_i^-))_{i \in I}$ be a partition of $D$. Let $h \in H(D)$ be injective in $D$ and let $f \in H(D)$ satisfy

(i) $|f'(\alpha_i) - h'(\alpha_i)| < |h'(\alpha_i)|$ whenever $i \in I$,

(ii) $\|f - h\|_{d(\alpha_i, r_i^-)} < |h'(\alpha_i)|r_i$ whenever $i \in I$.

Then $f$ is strictly injective. Furthermore, for all $i \in I$, we have $f(d(\alpha_i, r_i^-)) = h(d(\alpha_i, r_i^-))$.

**Proof.** By (i), $h'$ is not identically zero in $d(\alpha_i, r_i^-)$, hence by Lemma 23.11, $h$ is strictly injective in $d(\alpha_i, r_i^-)$. For every $i \in I$,
we put $s_i = |h'(\alpha_i)|r_i$. By Corollary 22.10, we have $h(d(\alpha_i, r_i^-)) = d(h(\alpha_i), s_i^-)$, hence by (ii), it is seen that $f(d(\alpha_i, r_i^-)) \subset d(h(\alpha_i), s_i^-)$. Let $f(d(\alpha_i, r_i^-)) = d(f(\alpha_i), t_i^-)$. Therefore, we have

$$ (1) \quad t_i \le s_i, $$

while obviously

$$ (2) \quad t_i \ge |f'(\alpha_i)|r_i. $$

But, by (i), we have $|f'(\alpha_i)| = |h'(\alpha_i)|$, hence by (1) and (2), we can derive $t_i = s_i$, hence $t_i = |f'(\alpha_i)|r_i$, and therefore, by Theorem 23.9, $f$ is strictly injective in $d(\alpha_i, r_i^-)$. Suppose that $f$ is not injective in all of $D$. Then there exists $a$ and $b \in D$ such that $f(a) = f(b)$. Since $f$ is injective in each disk $d(\alpha_i, r_i^-)$, we see that there exist $j$ and $m \in I$ with $j \ne m$ such that $a \in d(\alpha_j, r_j^-)$ and $b \in d(\alpha_m, r_m^-)$. Since we have just proven that $f(d(\alpha_i, r_i^-)) = h(d(\alpha_i, r_i^-))$ for all $i \in I$, we see that there exist $a' \in d(\alpha_j, r_j^-)$ and $b' \in d(\alpha_m, r_m^-)$ such that $h(a') = f(a), h(b') = f(b)$. This clearly contradicts the hypothesis "$h$ is injective in $D$". Consequently, $f$ is injective in all of $D$. Therefore, by Lemma 23.11, $f$ is strictly injective. $\square$

Proposition 23.21 shows the set of the strictly injective elements to be open in $H(D)$.

**Proposition 23.21.** Let $D$ be such that $\delta(D, \mathbb{K}\setminus D) = \rho > 0$. Let $\lambda \in ]0, +\infty[$. Let $h \in H(D)$ be injective and satisfy $|h'(x)| \ge \lambda$ whenever $x \in D$. For every $f \in H(D)$ such that $\|f - h\|_D < \lambda\rho$, $f$ is strictly injective in $D$ and satisfies $f(d(\alpha, \rho^-)) = h(d(\alpha, \rho^-))$ for all $\alpha \in D$.

**Proof.** Let $h \in H(D)$ satisfy $\|f - h\|_D < \lambda\rho$. Since the distance from $D$ to $\mathbb{K}\setminus D$ is $\rho > 0$, there exists a partition of $D$ in the form $(d(\alpha_i, \rho^-))_{i \in I}$ and then we have $\|f - h\|_{d(\alpha_i, \rho^-)} < \lambda\rho \le |h'(\alpha_i)|\rho$. Moreover, given any $g \in H(d(\alpha, \rho^-))$, by Theorem 18.1, we have $\|g'\|_{d(\alpha, \rho^-)} \le \frac{1}{\rho}\|g\|_{d(\alpha, \rho^-)}$, hence

$$ |f'(\alpha_i) - h'(\alpha_i)| \le \frac{1}{\rho}\|f - h\|_{d(\alpha_i, \rho^-)} < \lambda \le |h'(\alpha_i)|. $$

So, Conditions (i) and (ii) of Proposition 23.20 are clearly satisfied. $\square$

<figure><img src="image_6.png" /></figure>
**Remark.** As we know, when $D$ has holes, $\rho$ is just the lower bound of the diameters of the holes.

**Theorem 23.22.** Let $d(\alpha_i, r_i^-)_{i \in I}$ be a partition of $D$, let $h \in H(D)$, and let $u \in [0, 1[$. Let $\phi \in H(D)$ satisfy $\|\phi\|_D < 1$ and $\|h\phi\|_{d(\alpha_i, r_i^-)} \le ur_i \|h'\|_{d(\alpha_i, r_i^-)}$ for all $i \in I$. Then, for every $t \in [\max(u, \|\phi\|_D), 1[$ there exists a family $(\beta_i)_{i \in I}$ with $\beta_i \in d(\alpha_i, r_i^-)$ such that

(i) $|h(\beta_i)\phi'(\beta_i)| \le t|h'(\beta_i)|$ whenever $i \in I$,

(ii) $\|h\phi\|_{d(\alpha_i, r_i^-)} \le t|h'(\beta_i)|r_i$ whenever $i \in I$.

Let $h$ be strictly injective and let $f = h(1 + \phi)$. Then, $f$ is strictly injective and satisfies $f(d(\alpha_i, r_i^-)) = h(d(\alpha_i, r_i^-))$ for all $i \in I$.

**Proof.** Let us fix $i \in I$ and put $\alpha = \alpha_i, r = r_i$. For every $g \in H(d(\alpha, r^-))$, we have $\|g\|_{d(\alpha, r^-)} = \lim_{|x-\alpha| \to r^-} |g(x)|$, hence there clearly exists $\beta \in d(\alpha, r^-)$ such that

$$ (1) \quad |h'(\beta)| \ge \frac{u}{t} \|h'\|_{d(\alpha, r^{-})}, \text{ hence the hypothesis implies} $$

$$ (2) \quad \|h\phi\|_{d(\beta, r^{-})} \le t \frac{u}{r} |h'(\beta)|r. $$

Moreover, we know that $\|\phi'\|_{d(\alpha, r^-)} \le \frac{1}{r} \|\phi\|_{d(\alpha, r^-)}$, hence we have

$$ |h(\beta)\phi'(\beta)| \le |h(\beta)| \frac{1}{r} \|\phi\|_{d(\alpha, r^{-})} \le \frac{1}{r} \|h\|_{d(\alpha, r^{-})} \|\phi\|_{d(\alpha, r^{-})}. $$

As the norm $\| \cdot \|_{d(\alpha, r^-)}$ is multiplicative, we have $|h(\beta)\phi'(\beta)| \le \frac{1}{r}\|h\phi\|_{d(\alpha, r^-)}$, hence by the above hypothesis, $|h(\beta)\phi'(\beta)| \le u\|h'\|_{d(\alpha, r^-)}$, and finally, by (1), we obtain

$$ (3) \quad |h(\beta)\phi'(\beta)| \le t|h'(\beta)|. $$

Hence, we just have to put $\beta_i = \beta$ and do this for every $i \in I$ in order to obtain (i) and (ii) from (2) and (3). Since $\|\phi\|_D \le 1$, we may take $t \ge \|\phi\|_D$. We see that $f' - h' = h'\phi + h\phi'$, hence by Condition (i), we obtain $|f'(\beta_i) - h'(\beta_i)| \le t|h'(\beta_i)|$. Then, as $t < 1$, and as $h'(x) \ne 0$ whenever $x \in D$, we see that $|f'(\beta_i) - h'(\beta_i)| < |h'(\beta_i)|$ whenever $i \in I$. This is just Condition (i) in Proposition 23.20. Moreover, Condition (ii) implies Condition (ii) in Proposition 23.20. Hence, by Proposition 23.20, $f$ is strictly injective and satisfies $f(d(\alpha_i, r_i^-)) = h(d(\alpha_i, r_i^-))$ for all $i \in I$. $\square$
**Theorem 23.23.** Let $D$ be analytic and let $F$ be the set of the injective elements of $H(D)$. The closure of $F$ in $H(D)$ is equal to $F \cup K$.

**Proof.** Let $\bar{F}$ be the closure of $F$ and let $f \in \bar{F} \setminus K$. Suppose that $f$ is not injective and let $a, b \in D$ be such that $f(a) = f(b)$. Without loss of generality, we may obviously assume $f(a) = 0$. Now, let $r \in ]0, |a-b|[$ be such that $d(a,r) \cup d(b,r)$ is included in $D$. Suppose that $f$ is not identically zero in $D$. Since $D$ is an analytic set, the restriction of $f$ to $d(a,r)$ (respectively $d(b,r)$) is not identically zero. Let $h \in F$ satisfy $\|f-h\|_D < \min(\|f\|_{d(a,r)}, \|f\|_{d(b,r)})$. By Theorem 22.16, $h$ admits a zero in $d(a,r)$ and another in $d(b,r)$. But since $r < |a-b|$, these two zeros are different, and therefore, this contradicts the hypothesis "$h \in F$". That ends the proof. $\square$
# Chapter 24

## Quasi-Invertible Analytic Elements

Throughout this chapter D is supposed to be infraconnected.

Some of the results given here were obtained in [35], [36], [44], and [50]. We show that when an ideal of an algebra $H(D)$ contains a quasi-invertible element, this ideal is principal and generated by a polynomial.

**Lemma 24.1.** Let $T$ be a hole of $D$ and let $f \in H(D \cup T)$ be invertible in $H(D)$. If $f$ has no zero in $T$, then $f$ is invertible in $H(D \cup T)$.

**Proof.** Let $(f_n)_{n \in \mathbb{N}}$ be a sequence in $R(D \cup T)$ which converges to $f$ in $H(D \cup T)$. Let $T = d(a, r^{-})$. Since $f$ has no zero in $T$, by Theorem 22.15, we have $|f(x)| = |f(a)|$ for all $x \in T$, and therefore, $|f(x)| = {}_{D}\varphi_{a,r}(f)$ for all $x \in T$. Now, since $D$ is infraconnected, by Corollary 13.2, we have ${}_{D}\varphi_{a,r}(g) \le \|g\|_{D}$ whenever $g \in H(D)$, hence for $n$ big enough,

$$ \left| \frac{1}{f(x)} - \frac{1}{f_n(x)} \right| = \frac{|f_n(x) - f(x)|}{|f(x)|^2} \le \frac{\|f_n - f\|_D}{({}_{D}\varphi_{a,r}(f))^2} $$

whenever $x \in T$. Hence, we see that the sequence $\frac{1}{f_n}$ converges to $\frac{1}{f}$ in $H(D \cup T)$. $\square$

**Theorem 24.2.** Let $D \in \text{Alg.}$ If an ideal contains a quasi-invertible element, then it is generated by a polynomial whose zeros belong to $\overset{\circ}{D} \cap D$.
**Proof.** Let $\mathcal{J}$ be an ideal of $H(D)$ that contains a quasi-invertible element $f$ that is of the form $Pg$ with $g$ invertible in $H(D)$ and $P(x) \in \mathbb{K}[x]$, all the zeros of $P$ lying inside $D \cap \mathring{D}$. Then, $P$ belongs to $\mathcal{J}$. Consequently, the set $\mathcal{J}_0$ of polynomials that belong to $\mathcal{J}$ is not empty and hence is an ideal of $\mathbb{K}[x]$, hence $\mathcal{J}_0$ is of the form $Q(x)\mathbb{K}[x]$ with $Q \in \mathbb{K}[x]$. On the other hand, by hypothesis, $f$ factorizes in $H(D)$ in the form $Pg$ with $g$ invertible in $H(D)$ and $P(x) \in \mathbb{K}[x]$, all the zeros of $P$ lying inside $D \cap \mathring{D}$. Since $fg^{-1}$ belongs to $\mathcal{J}$, obviously $P$ belongs to $\mathcal{J}_0$. Hence, $Q$ divides $P$ and then all zeros of $Q$ lie in $D \cap \mathring{D}$. We show that $\mathcal{J} = QH(D)$.

First, we suppose that $D$ is bounded. Let $\alpha_1, \dots, \alpha_q$ be the zeros of $Q$ and suppose that there exists some $h \in \mathcal{J} \setminus QH(D)$. Since $D$ is bounded, by Theorem 11.12, $h$ is of the form $\frac{\ell}{S}$ with $\ell \in H(\bar{D})$ and $S$ a polynomial whose zeros belong to $\bar{D} \setminus D$. Hence, $\ell$ belongs to $\mathcal{J}$. Now we can find $r > 0$ such that $d(\alpha_i, r) \subset \bar{D}$, whenever $i = 1, \dots, q$. Let $\Lambda = \bigcup_{i=1}^q d(\alpha_i, r)$ and let $D' = D \cup \Lambda$. Since the zeros of $Q$ lie in $\Lambda$, there exists $\lambda > 0$ such that $|Q(x)| \ge \lambda$ whenever $x \in D \setminus \Lambda$. Now since $\bar{D}$ is closed and bounded, there exists $b \in \mathbb{K}$ such that $\|b\ell\|_D < \lambda$. We put $\phi = Q + b\ell$. Clearly, outside $\Lambda$, we have $|\phi(x)| \ge \lambda$. Next, by Theorem 22.18, in each disk $d(\alpha_i, r)$, $\phi$ has finitely many zeros, hence in $D'$, $\phi$ has finitely many zeros, all of them in $\Lambda$. Therefore, by Theorem 16.2, it factorizes in the form $V(x)W(x)$ with $W \in H(D')$, $W(x) \ne 0$ whenever $x \in D'$ and $V$ a polynomial whose zeros belong to $\Lambda$. By Theorem 22.16, $|W(x)|$ has a strictly positive lower bound in $\Lambda$ and another non-zero lower bound in $D' \setminus \Lambda$ because $|V(x)|$ is obviously bounded in $D'$. Finally, $W$ has a non-zero lower bound in $D'$, therefore it is invertible in $H(D')$. Hence, $V$ belongs to $\mathcal{J}$. But then $Q$ divides $V$ in $\mathbb{K}[x]$. But since $Q + b\ell$ is equal to $VW$, then $Q$ divides $Q + b\ell$ and hence it divides $\ell$ and $h$ too. This contradicts the hypothesis $h \in \mathcal{J} \setminus QH(D)$ and finishes proving that $Q$ generates $\mathcal{J}$ when $D$ is bounded.

Now, we suppose $D$ unbounded. We may obviously assume $D$ to have at least one hole $d(a, r^{-})$ and without loss of generality, we may assume $a = 0$. Let $\gamma(x) = \frac{1}{x}$ and let $D'' = \gamma(D)$. Then, $D''$ is a bounded set that belongs to Alg such that $0 \notin D'$. We also have $\gamma = \gamma^{-1}$ and $\gamma(D'') = D$. Let $\psi$ be the mapping from $H(D)$ onto $H(D'')$ defined as $\psi(f) = f \circ \gamma$. Then, $\psi$ is a $\mathbb{K}$-algebra isomorphism
from $H(D)$ onto $H(D'')$. Moreover, $\psi(\mathcal{J})$ is an ideal $\mathcal{J}''$ of $H(D'')$.
Let $u = \frac{1}{x}$ and $T(u) = Q(x)$. Then we can check that $\mathcal{J}'' = TH(D'')$,
which ends the proof.
□

**Definition and notation.** For any integer $n \in \mathbb{N}$, we denote by $Q_n(D)$ the set of the quasi-invertible elements $f \in H(D)$ that have exactly $n$ zeros, taking multiplicity into account and by $Q(D)$ the set $\bigcup_{n=0}^{\infty} Q_n(D)$.

Theorem 24.3 shows that if two analytic elements $f, g$ are close enough, then the zeros of $f$ and $g$ also are respectively close, once correctly ordered. It is known as the convergence of zeros theorem.

**Theorem 24.3.** Let $D$ be closed and bounded. Let $n \in \mathbb{N}$, let $f \in Q_n(D)$, and let $\alpha_1, \dots, \alpha_n$ be the zeros of $f$ (taking multiplicity into account). For every $\epsilon > 0$, there exists $\eta > 0$ such that for every $h \in H(D)$ satisfying $\|f - h\|_D \le \eta$, $h$ belongs to $Q_n(D)$ and the zeros $\beta_1, \dots, \beta_n$ of $h$, once correctly ordered, satisfy $|\alpha_i - \beta_i| \le \epsilon$ ($1 \le i \le n$).

**Proof.** Let $f = Pg \in Q_n(D)$ with $g$ invertible in $H(D)$ and $P$ an $n$-degree monic polynomial whose zeros are interior to $D$. Let $\gamma_1, \dots, \gamma_q$ be the different zeros of $P$, each $\gamma_j$ of order $s_j$ (with obviously $\sum_{j=1}^q s_j = n$). Let $\xi = \inf_{j \ne \ell} |\gamma_j - \gamma_\ell|$, let $\epsilon \in ]0, \xi[$, let $\Lambda_j(\epsilon) = d(\gamma_j, \epsilon)$, and let $\Lambda(\epsilon) = \bigcup_{j=1}^q d(\gamma_j, \xi)$. It is easily seen that $|P(x)|$ has a non-zero lower bound in $D \setminus \Lambda(\epsilon)$. Since $D$ is closed and bounded, $|g(x)|$ has a non-zero lower bound in $D$. Hence, $|f(x)|$ has a lower bound $\lambda > 0$ in $D \setminus \Lambda(\epsilon)$. Let $\eta = \min(\lambda, \min_{1 \le j \le q} \|f\|_{\Lambda_j(\epsilon)})$ and let $h \in H(D)$ satisfy

$$ (1) \quad \|f - h\|_D < \eta. $$

Obviously, we have $|f(x)| = |h(x)| \ge \lambda$ whenever $x \in D \setminus \Lambda(\epsilon)$. But then by (1) and by Theorem 22.17, we see that $h$ has exactly $s_i$ zeros like $f$ in $\Lambda_i(\epsilon)$ ($1 \le i \le q$) taking multiplicity into account. Thus, we have already proven the statement when all zeros of $f$ have order 1.

Now, extending this to the general case is just a question of writing. We may assume the $\alpha_i$ to be ordered in such a way that

$$ \alpha_1 = \dots = \alpha_{s_1} = \gamma_1, \quad \alpha_{s_1+1} = \dots = \alpha_{s_1+s_2} = \gamma_2, \\ \alpha_{s_1+\dots+s_{q-1}+1} = \dots = \alpha_{s_1+\dots+s_q} = \gamma_q. $$
Thus, for every $j = 1, \dots, q$ in $\Lambda_j(\epsilon)$, we can check that $\gamma_j$ is equal to $\alpha_{s_1+\dots+s_{j-1}+k}$ whenever $k = 1, \dots, s_j$. Since $f$ admits $s_j$ zeros in $\Lambda_j(\epsilon)$, as does $h$, we may denote them by $\beta_{s_1+\dots+s_{j-1}+1}, \dots, \beta_{s_1+\dots+s_j}$ (some of them being eventually equal). So, we obtain $|\alpha_i - \beta_i| \le \epsilon$ whenever $i = 1, \dots, n$. $\square$

**Corollary 24.4.** Let $D$ be closed and bounded. For every $n \in \mathbb{N}$, $Q_n(D)$ is open in $H(D)$ and so is $Q(D)$.

**Lemma 24.5.** Let $a \in \bar{D}$ satisfy $a \notin \bar{D}^o$. There exists a quasi-minorated element $f \in H_b(D)$ which is not semi-invertible, satisfying
$$ \lim_{x \to a} f(x) = 0 \quad \text{and} \quad \limsup_{x \to D} \left| \frac{f(x)}{x} \right| = +\infty. $$

**Proof.** Without loss of generality, we may obviously assume $a = 0$. Then the Cauchy filter $\mathcal{F}$ of base $\{d(0, r) \cap D \mid r > 0\}$ is pierced. Let $(T_m)_{m \in \mathbb{N}}$ be a sequence of holes of $D$ that runs $\mathcal{F}$ and let $D' = \mathbb{K} \setminus (\bigcup_{m=0}^{\infty} T_m)$. By Corollary 17.5, there exists $f \in H_b(D')$ such that $\lim_{x \to 0} f(x) = 0$ and such that $\limsup_{x \to D'} \left| \frac{f(x)}{x} \right| = +\infty$. We check that $D'$ has no monotonous pierced filter because its only holes are the $T_m$. Hence, by Corollary 21.3, $f$ is quasi-minorated. If $0$ belongs to $D$, it is seen that $f(0) = 0$ while $\limsup_{x \to 0} \left| \frac{f(x)}{x} \right| = +\infty$, hence $f$ can't factorize in the form $xg(x)$ with $g \in H(D)$, and therefore, $f$ is not semi-invertible.

Now, suppose $0 \notin D$ and that $f$ semi-invertible. Then it factorizes in the form $P(x)g(x)$ with $P$ the polynomial of the zeros of $f$ in $\bar{D}$ and $g$ an invertible element in $H(D)$. Since $0 \notin D$, we have $P(0) \ne 0$, hence $\lim_{x \to 0} g(x) = 0$. But since $\lim_{x \to 0} |g(x)| = +\infty$, by Corollary 11.9, $\frac{1}{g(x)}$ admits a pole at $0$. Let $n$ be its order. Then, by Corollary 11.9, $\frac{x^n}{g(x)}$ has a finite limit different from zero at $0$. But since $\frac{f(x)}{x}$ is unbounded in any set $d(0, r) \cap (D' \setminus \{0\})$, so is $\frac{g(x)}{x}$ and therefore we have $\liminf_{x \to 0} \left| \frac{x^n}{g(x)} \right| = 0$. Hence, $g$ can't be invertible. This finally shows that $f$ is not semi-invertible and finishes the proof of Lemma 24.5. $\square$

**Lemma 24.6.** Let $D$ be such that $\tilde{D} \setminus \bar{D}$ is not bounded. Then there exists a quasi-minorated element $f \in H_b(D)$ satisfying

$$ (1) \quad \lim_{\substack{|x| \to \infty \\ x \in D}} f(x) = 0 $$
and

$$ (2) \limsup_{\substack{|x| \to \infty \\ x \in D}} |xf(x)| = +\infty. $$

Moreover, $xf$ does not belong to $H(D)$.

**Proof.** Since $D$ has holes, we may obviously assume that $0$ belongs to a hole. Let $\gamma(x) = \frac{1}{x}$ and let $D' = \gamma(D)$. Then $D'$ is bounded and $0$ belongs to $\overline{D'} \setminus \overline{D'}$. By Lemma 24.5, there exists a quasi-minorated element $h \in H_b(D')$ satisfying

$$ (3) \lim_{\substack{x \to 0 \\ x \in D}} f(x) = 0 $$

and

$$ (4) \limsup_{\substack{x \to 0 \\ x \in D}} \left| \frac{f(x)}{x} \right| = +\infty. $$

Then we set $f = h \circ \gamma$. By (3), $f$ satisfies (1), and by (4), $f$ satisfies (2). Now, by Lemma 16.7, $f$ is quasi-minorated. Finally, we check that $xf$ does not belong to $H(D)$. Indeed, suppose $xf \in H(D)$. By Theorem 11.5, $xf$ is of the form $g(x)+P(x)$, with $g \in H_b(D)$ and $P \in \mathbb{K}[x]$. Let $q = \deg(P)$. Since $xf$ is not bounded, we have $q > 0$. Then, $x^{1-q}f$ has a limit different from 0 when $|x|$ tends to $+\infty$ and this contradicts (1). This ends the proof of Lemma 24.6. $\square$

**Theorem 24.7.** If $D$ does not belong to Alg, there exist invertible elements $f,g \in H(D)$ such that $fg$ belongs to $H(D)$ but is not semi-invertible.

**Proof.** First, we suppose that there exists $a \in (\overline{D} \setminus D) \setminus \overline{D}$. Without loss of generality, we assume $a = 0$. By Lemma 24.5, there exists $f \in H_b(D)$ such that $\lim_{x \to 0} f(x) = 0$ while $\frac{f(x)}{x}$ is not bounded in any set $D \cap d(0, r)$ ($r > 0$). Since $f \in H_b(D)$, we can find $A \in \mathbb{K}$ such that $|A| > \|f\|_D$. Let $g = A + f$. Then $g$ is invertible in $H(D)$. Let $T = d(b, \rho^{-})$ be a hole of $D$ and let $F = \frac{x}{(x-b)g}$. Then both $\frac{x}{x-b}$, $g^{-1}$ belong to $H_b(D)$, hence so does $F$. But by definition, $F$ is the product of two invertible elements of $H(D)$. We also note that $F$ has no zero in $D$. Next, we note that $\frac{f(x)(x-b)}{x}$ is not bounded in any set $D \cap d(0, r)$ ($r > 0$), although $\lim_{x \to 0} f(x)(x-b) = 0$.
Thus, $\frac{f(x)(x-b)}{x}$ cannot admit a pole at 0, and therefore, by Corollary 11.9, $\frac{f(x)(x-b)}{x}$ does not belong to $H(D)$. But then, since $\frac{A(x-b)}{x}$ does belong to $H(D)$, we see that $F^{-1}$ does not belong to $H(D)$. Since $F$ has no zero in $D$, it is not semi-invertible although both $\frac{x}{x-b}$, $g^{-1}$ are invertible in $H(D)$.

Now, we suppose that $\tilde{D}\setminus\overline{D}$ is not bounded. Since $D$ has holes, we may obviously assume that 0 belongs to a hole. Let $\gamma(x) = \frac{1}{x}$ and let $D' = \gamma(D)$. Then $D'$ is bounded, and 0 belongs to $\in \tilde{D}'\setminus\overline{D}'$. Hence, as we just saw, there exist invertible elements $h, g \in H_b(D')$ such that $hg$ belongs to $H_b(D')$ and has no zero in $D'$ but is not invertible in $H(D')$. Then we put $\tau = h \circ \gamma$, $\psi = g \circ \gamma$, $\phi = (hg) \circ \gamma$. By Theorem 12.7, both $\tau, \psi$ are invertible in $H(D)$, and $\phi$ belongs to $H(D)$ and has no zero in $D$. Since $hg$ is not invertible in $H(D')$, by Theorem 12.7 again, $\phi$ is not invertible in $H(D)$. Since it has no zero in $D$, it is not semi-invertible in $H(D)$ and that finishes the proof of Theorem 24.7. $\square$

**Theorem 24.8.** The following three statements are equivalent:

(i) $D$ belongs to Alg and $\overline{D}$ is open.

(ii) $\tilde{D}\setminus\overline{D}$ is bounded and $\overline{D}$ is open.

(iii) The set of the quasi-minorated elements of $H(D)$ is equal to the set of the quasi-invertible elements.

**Proof.** By Theorem 17.9, we know that (i) implies (ii). Conversely, suppose (ii) is satisfied. Particularly, $\overline{D}$ is open. Suppose (i) is not satisfied. Then $D$ does not belong to Alg. Since $\tilde{D}\setminus\overline{D}$ is bounded, there must exist $a \in \overline{D}\setminus D$ that does not belong to $\overline{D}$, and therefore, this contradicts (ii). Hence, (i) and (ii) are equivalent. Now, since (ii) implies (i), we can apply Theorem 21.7, hence (ii) implies (iii). Finally, it just remains to show that if $\tilde{D}\setminus\overline{D}$ is not bounded or if $\overline{D}$ is not open, then there exist quasi-minorated elements that are not quasi-invertible.

On one hand, if $\overline{D}$ is not open, by Lemma 24.5, such an element does exist. On the other hand, if $\tilde{D}\setminus\overline{D}$ is not bounded, then $D$ does not belong to Alg, and therefore, by Theorem 24.7, there exist invertible elements $f, g$ in $H(D)$ such that $fg$ is not semi-invertible and hence is not quasi-invertible. But, by Theorem 21.6, both $f, g$
are quasi-minorated, and then, by Theorem 21.5, $fg$ also is quasi-minorated. That ends the proof of Theorem 24.8. $\square$

**Theorem 24.9.** Let $D$ be closed, let $T$ be a hole of $D$, and let $f \in H(D \cup T)$ have no zero in $T$. There exists a bounded closed infraconnected set $E$ such that $T \subset E \subset D \cup T$, $T \neq E$ and such that the restriction of $f$ to $E$ is invertible in $H(E)$.

**Proof.** Let $T = d(a, r^{-})$ and let $\lambda =_{D} \varphi_{a,r}(f)$. By Theorem 22.16, we know that $|f(x)| = \lambda$ for all $x \in T$. Moreover, since the restriction of $f$ to $T$ is not identically zero, we have $\lambda > 0$. Let $\ell = \frac{\lambda}{2}$.

First, we suppose $C(a, r) \cap D \neq \emptyset$. Let $b \in D \cap C(a, r)$. Then we have $\lim_{|x-b| \to r, \ x \in D} |f(x)| =_{D} \varphi_{b,r}(f) =_{D} \varphi_{a,r}(f) = \lambda$. Hence, there exists $s \in ]0, r[$ such that $|f(x)| \ge \ell$ for every $x \in d(b, r^{-}) \cap D$. Then, the set $E = T \cup (d(b, r^{-}) \cap D)$ is clearly infraconnected, closed, and bounded and we have $|f(x)| \ge \ell$ for all $x \in E$. Hence, the restriction of $f$ to $E$ is invertible in $H(E)$.

Now, we suppose $C(a, r) \cap D = \emptyset$. There exists $s > r$ such that $|f(x)| \ge \ell$ for every $x \in \Gamma(a, r, s) \cap D$. So, we consider the set $E = T \cup (\Gamma(a, r, s) \cap D)$. It is infraconnected, closed, and bounded and we have $|f(x)| \ge \ell$ for all $x \in E$. Hence the restriction of $f$ to $E$ is invertible in $H(E)$.

In both cases, we can see that $T$ is strictly included in $E$. That ends the proof of Theorem 24.9. $\square$

We can now briefly examine the ideals of an algebra $H(D)$ when all elements are quasi-invertible.

**Theorem 24.10.** Let $D \in \text{Alg}$. If an ideal contains a quasi-invertible element, then it is generated by a polynomial whose zeros belong to $\frac{\circ}{D} \cap D$.

**Proof.** Let $\mathcal{H}$ be an ideal of $H(D)$ that contains a quasi-invertible element $f$, and let $\mathcal{H}_0$ be the set of polynomials that belong to $\mathcal{H}$. By hypothesis, $f$ factorizes in $H(D)$ in the form $Pg$ with $g$ invertible in $H(D)$ and $P(x) \in K[x]$, all the zeros of $P$ lying inside $D \cap \frac{\circ}{D}$. Since $fg^{-1}$ belongs to $\mathcal{H}$, obviously $P$ belongs to $\mathcal{H}_0$. Hence, $T$ divides $P$, and then all the zeros of $T$ lie in $D \cap \frac{\circ}{D}$. We show that $\mathcal{H} = TH(D)$.
First, we suppose that $D$ is bounded. It is clearly seen that $\mathcal{H}_0$ is an ideal of $\mathbb{K}[x]$, hence there exists $T(x) \in K[x]$ such that $\mathcal{H}_0 = T(x)K[x]$.

Let $\alpha_1, \dots, \alpha_q$ be the zeros of $T$. Now, we suppose that there exists some $h \in \mathcal{H} \setminus TH(D)$. Since $D$ is bounded, by Theorem 11.12, $h$ is of the form $\frac{\ell}{S}$ with $\ell \in H(\bar{D})$ and $S$ a polynomial whose zeros belong to $\bar{D} \setminus D$. Hence, $\ell$ belongs to $\mathcal{H}$. Now, we can find $r > 0$ such that $d(\alpha_i, r) \subset \bar{D}$, whenever $i = 1, \dots, q$. Let $\Lambda = \bigcup_{i=1}^q d(\alpha_i, r)$ and let $D' = D \cup \Lambda$. Since the zeros of $T$ lie in $\Lambda$, there exists $\lambda > 0$ such that $|T(x)| \ge \lambda$ whenever $x \in D \setminus \Lambda$. Now, since $\bar{D}$ is closed and bounded, there exists $b \in K$ such that $\|b\ell\|_D < \lambda$. We put $\phi = T+b\ell$. Clearly, outside $\Lambda$, we have $|\phi(x)| \ge \lambda$. Besides, by Theorem 22.18, in each disk $d(\alpha_i, r)$, $\phi$ has finitely many zeros, hence in $D'$, $\phi$ has finitely many zeros, all of them in $\Lambda$. Hence, it factorizes in the form $Q(x)W(x)$ with $W \in H(D')$, $W(x) \ne 0$ whenever $x \in D'$ and $Q$ a polynomial whose zeros belong to $\Lambda$. By Theorem 22.16, $|W(x)|$ has a strictly positive lower bound in $\Lambda$ and another non-zero lower bound in $D' \setminus \Lambda$ because $|Q(x)|$ is obviously bounded in $D'$. Finally, $W$ has a non-zero lower bound in $D'$, therefore it is invertible in $H(D')$. Hence, $Q$ belongs to $\mathcal{H}$. But then $T$ divides $Q$ in $\mathbb{K}[x]$. Since $T+b\ell$ is equal to $WQ$, then $T$ divides $T+b\ell$, and $\ell$, and $h$ too. This contradicts the hypothesis $h \in \mathcal{H} \setminus TH(D)$ and finishes proving that $T$ generates $\mathcal{H}$ when $D$ is bounded.

Now, we suppose $D$ unbounded. We may obviously assume $D$ to have at least one hole $d(a, r^{-})$, and without loss of generality, we may assume $a = 0$. Let $\gamma(x) = \frac{1}{a_i}$, and let $D' = \gamma(D)$. Then $D'$ is a bounded set that belongs to $\text{Alg}$ such that $0 \notin D'$. We also have $\gamma = \gamma^{-1}$, and $\gamma(D') = D$. Let $\psi$ be the mapping from $H(D)$ into $H(D')$ defined as $\psi(f) = f \circ \gamma$. Then $\psi$ is a $\mathbb{K}$-algebra isomorphism from $H(D)$ onto $H(D')$. Besides, $\psi(\mathcal{H})$ is an ideal $\mathcal{H}'$ of $H(D')$. Let $P(x) = \prod_{j=1}^q (x - a_i)$, for every $j = 1, \dots, q$, let $a_i' = \frac{1}{a_i}$, and let $B(u) = \prod_{j=1}^q (u - a_i')$. Clearly, $\psi(P) = \frac{B(u)}{u^q}$. As $0 \notin D'$, $u$ is invertible in $H(D')$, and $B$ belongs to $\mathcal{H}'$. Hence, $\mathcal{J}'$ is generated by a polynomial whose zeros lie inside $D' \cap \bar{D}'$. Now, let $W(u) = \prod_{j=1}^t (u - c_j)$. Since the $c_j$ lie in $D'$, they are different from 0. For every $j = 1, \dots, t$, let $e_j = \frac{1}{c_j}$, and let $S(x) = \prod_{j=1}^t (x - e_j)$. It is seen that for each
$j = 1, \dots, t$, $e_j$ does belong to $D \cap \frac{1}{D}$. Now, let $h \in \mathcal{H}$. Then $\psi(h)$ belongs to $\mathcal{H}'$ and is of the form $W(u)G(u)$, with $G \in H(D')$. Putting $F = \psi^{-1}(G)$, in $H(D)$, we have $h = (-1)^t S(x) \frac{F(x)}{x^t \prod_{i=1}^t e_j}$. As $\frac{F(x)}{x^t}$ is an invertible element of $H(D)$, this finishes showing that $S$ generates $\mathcal{H}$. $\square$

**Corollary 24.11.** Suppose $D \in \text{Alg}$ and all elements are quasi-invertible, except 0. Then $H(D)$ is a principal ring and each ideal is generated by a polynomial whose zeros lie in the opening of $D$ and every maximal ideal of $H(D)$ is of the form $(x-a)H(D)$ with $a \in D$.

By Corollaries 21.9, 21.10, and 24.11, we can derive again Corollaries 24.12 and 24.13:

**Corollary 24.12.** Suppose $D \in \text{Alg}$ be closed and let $\mathcal{T}$ be the set of holes of $D$. If $\{\tilde{T} | T \in \mathcal{T}\}$ is finite, $H(D)$ is principal, each ideal is generated by a polynomial whose zeros lie in $D$, and every maximal ideal of $H(D)$ is of the form $(x-a)H(D)$, with $a \in D$.

**Corollary 24.13.** If $D$ is a disk $d(a,r)$ or $d(a,r^{-})$, or if $D$ is an annulus $\Gamma(a, r_1, r_2)$ (with $0 < r_1 < r_2$) or $\Delta(a, r_1, r_2)$ (with $0 < r_1 < r_2$) or a circle $C(a,r)$, then $H(D)$ is principal, each ideal is generated by a polynomial whose zeros lie in $D$ and every maximal ideal of $H(D)$ is of the form $(x-b)H(D)$ with $b \in D$.
Chapter 25

Logarithm and Exponential
in a *p*-adic Field

*In this chapter, the field K is supposed to have characteristic zero.*

We define the *p*-adic logarithm and the *p*-adic exponential and shortly study them, in connection with the study of the roots of 1 made in Chapter 24. Both functions are also defined in [1]. Here, as in [50], we compute the radius of convergence of the *p*-adic exponential by using results on injectivity seen in Chapter 23.

**Lemma 25.1.** $\mathbb{K}$ is supposed to have residue characteristic $p \neq 0$. Let $r \in ]0, 1[$, and for each $n \in \mathbb{N}$, let $h_n(x) = (1+x)^{p^n}$. The sequence $h_n$ converges to 1 with respect to the uniform convergence on $d(0, r)$.

**Proof.** Without loss of generality, we may assume $|p| = \frac{1}{p}$. Let $E = d(0, r)$, and for each $n \in \mathbb{N}$, let $u_n = p^n$ and let $q_n$ be the integral part of $\frac{n}{2}$. Now, we put $t_n = p^{q_n}$ and we denote by $h$ the identical function on $E$. Then, $h_n - 1 = \sum_{j=1}^{u_n} \binom{u_n}{j} h^j$. By Lemma 7.1, we have $\left|\binom{u_n}{j}\right| \le \frac{p^{-n}}{|j|}$, hence

$$
(1) \quad \left| \binom{u_n}{j} \right| \le j p^{-n} \le t_n p^{-n} \le p^{-\frac{n}{2}} \quad \text{whenever } j = 1, \dots, t_n.
$$

Next, we have

$$
h_n - 1 = \sum_{j=1}^{t_n} \binom{u_n}{j} h^j + \sum_{j=t_n+1}^{u_n} \binom{u_n}{j} h^j. \text{ By (1), it is seen that}
$$
$$ (2) \qquad \left\| \sum_{j=1}^{t_n} \binom{u_n}{j} h^j \right\|_E \le p^{-\frac{n}{2}} $$

while

$$ (3) \qquad \left\| \sum_{j=t_n+1}^{u_n} \binom{u_n}{j} h^j \right\|_E \le \|h\|_E^{t_n+1}. $$

Now, by (2) and (3), we see that $\lim_{n \to \infty} \|h_n - 1\|_E = 0$. $\square$

<figure><img src="image_7.png" /></figure>

**Definition and notation.** As previously defined, for each $q \in \mathbb{N}^*$, we denote by $r_q$ the positive number such that $\log_p(r_q) = -\frac{1}{p^{q-1}(p-1)}$. We denote by $f(x)$ the series $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}$.

**Theorem 25.2.** *f has a radius of convergence equal to 1. If the residue characteristic of $K$ is $p \neq 0$, then $f$ is unbounded in $d(0, 1^{-})$. If the residue characteristic is zero, then $|f(x)|$ is bounded by 1 in $d(0, 1^{-})$. The function defined in $d(1, 1^{-})$ as $\text{Log}(x) = f(x-1)$ has a derivative equal to $\frac{1}{x}$ and satisfies $\text{Log}(ab) = \text{Log}(a) + \text{Log}(b)$ whenever $a, b \in d(1, 1^{-})$.*

**Proof.** It is clearly seen that the radius of $f$ is 1 because $|n| \ge \frac{1}{n}$. As in the Archimedean case, the property $\text{Log}(ab) = \text{Log}(a) + \text{Log}(b)$ comes from the fact that both $\text{Log}$ and the function $h_a$ defined as $h_a(x) = \text{Log}(ax)$ have the same derivative. The other statements are immediate. $\square$

<figure><img src="image_8.png" /></figure>

In Chapter 7, when $K$ has residue characteristic $p \neq 0$, we have introduced the group $W$ of the $p^s$th roots of 1, i.e. the set of the $u \in K$ satisfying $u^{p^s} = 1$ for some $s \in \mathbb{N}$.

**Theorem 25.3.** *$K$ is supposed to have residue characteristic $p \neq 0$ (respectively 0). All zeros of $\text{Log}$ are of order 1. The set of zeros of the function $\text{Log}$ is equal to $W$ (respectively 1 is the only zero of $\text{Log}$). The restriction of $\text{Log}$ to the disk $d(1, (R_1)^{-})$ (respectively $d(1, 1^{-})$) is injective and is a bijection from $d(1, (R_1)^{-})$ onto $d(0, (R_1)^{-})$ (respectively from $d(1, 1^{-})$ onto $d(0, 1^{-})$).*

**Proof.** It is obvious that the zeros of $\text{Log}$ are of order 1 because the derivative of $\text{Log}$ has no zero. First, we suppose $K$ to have residue
characteristic $p \neq 0$. Each root of 1 in $d(1, 1^{-})$ is a zero of $\text{Log}$. Moreover, by Theorem 7.8, we know that the only roots of 1 in $d(1, 1^{-})$ are the $p^n$th roots. Now, we can check that $\text{Log}$ admits no zero other than the roots of 1. Indeed, suppose that $a$ is a zero of $\text{Log}$ but is not a root of 1, and for each $n \in \mathbb{N}$, let $b_n = a^{p^n}$. Since $b_n$ belongs to $d(1, 1^{-})$, by Lemma 25.1, we have $\lim_{n \to \infty} b_n = 1$. But obviously $\text{Log}(b_n) = 0$ for every $n \in \mathbb{N}$, hence this contradicts the fact that 1 is an isolated zero of $\text{Log}$.

Thus, $\text{Log}$ has no zero in the disk $d(1, (R_1)^{-})$, except 1, and therefore, by Theorem 22.7, the series $f(x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n^2}$ satisfies $\nu^+(f, \log_p r) = 1$ for every $r \in]0, R_1[$, hence $r > \frac{r^n}{|n|}$ for all $r \in]0, R_1[$, for every $n \in \mathbb{N}^*$. Therefore, by Corollary 23.10, it is injective in $d(0, R_1^{-})$. Then, by Corollary 22.10, we see that $\text{Log}(d(1, R_1^{-})) = d(0, R_1^{-})$.

Now, we suppose that $\mathbb{K}$ has residue characteristic zero. Then, the function $f(x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}$ satisfies $\nu^+(f, \log_p r) = 1$ for every $r \in]0, 1[$, hence $r > \frac{r^n}{n}$ for all $r \in]0, 1[$, for every $n \in \mathbb{N}^*$. Therefore, $f$ has no zero different from 1 in $d(0, 1^{-})$ and, by Corollary 23.10, is injective in $d(0, 1^{-})$. Then, by Corollary 22.10, we see that $\text{Log}(d(1, 1^{-})) = d(0, 1^{-})$. This ends the proof. $\square$

**Corollary 25.4.** $\mathbb{K}$ is supposed to have residue characteristic 0. There is no root of 1 in $d(1, 1^{-})$, except 1.

**Proof.** Indeed, any root of 1 should be a zero of $\text{Log}$ in $d(1, 1^{-})$. $\square$

**Definition and notation.** If $\mathbb{K}$ has residue characteristic $p \neq 0$, we denote by $\exp$ the inverse (or reciprocal) function of the restriction of $\text{Log}$ to $d(1, R_1^{-})$ which obviously is a function defined in $d(0, R_1^{-})$, with values in $d(1, R_1^{-})$. If $\mathbb{K}$ has residue characteristic 0, we denote by $\exp$ the inverse function of $\text{Log}$ which is obviously defined in $d(0, 1^{-})$ and takes values in $d(1, 1^{-})$.

**Theorem 25.5.** $\mathbb{K}$ is supposed to have residue characteristic $p \neq 0$ (respectively $p = 0$). The function $\exp$ belongs to $A_b(d(0, R_1^{-}))$ (respectively $A_b(d(0, 1^{-}))$), is a bijection from $d(0, R_1^{-})$ onto $d(1, R_1^{-})$ (respectively from $d(0, 1^{-})$ onto $d(1, 1^{-}))$, and satisfies $\exp(x) = \exp'(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ whenever $x \in d(0, R_1^{-})$ (respectively $x \in d(0, 1^{-})$). Moreover, the disk of convergence of its series is equal to
$d(0, R_1^-)$ (respectively $d(0, 1^-)$). Further, if $p \neq 0$, then $exp$ does not belong to $H(d(0, R_1^-))$.

**Proof.** By Corollary 23.15, we know that the function $exp$ belongs to $\mathcal{A}_b(d(0, R_1^-))$ (respectively $\mathcal{A}_b(d(0, 1^-)$)) and is obviously a bijection from $d(0, R_1^-)$ onto $d(1, R_1^-)$ (respectively from $d(0, 1^-)$ onto $d(1, 1^-)$). As it is the reciprocal of $Log$, it must satisfy $exp(x) = exp'(x)$ for all $x \in d(0, R_1^-)$ (respectively $x \in d(0, 1^-)$), and therefore, $exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ whenever $x \in d(0, R_1^-)$ (respectively $x \in d(0, 1^-)$). Thus, the radius of convergence $r$ is at least $R_1$ (respectively 1). If the residue characteristic is 0, it is obviously seen that the series cannot converge for $|x| = 1$, hence the disk of convergence is $d(0, 1^-)$.

Now, we suppose that the residue characteristic is $p \neq 0$. Suppose that the power series of $exp$ converges in $d(0, R_1)$. Then $exp$ has continuation to an element of $H(d(0, R_1))$. On the other hand, since $\nu(f, \log_p r) = 1$ for all $r \in ]0, R_1[$, we have $\nu^-(f, \log_p R_1) = 1$, and then by Theorem 22.9, $Log(d(1, R_1))$ is equal to $d(0, R_1)$. Hence, we can consider $exp(Log(x))$ in all the disk $d(0, R_1)$. By Corollary 12.3, this is an element of $H(d(1, R_1))$. But this element is equal to the identity in all of $d(1, R_1^-)$ and therefore in all of $d(1, R_1)$. Of course, this contradicts the fact that $Log$ is not injective in the circle $C(1, R_1)$. This finishes proving that the disk of convergence of $exp$ is just $d(0, R_1^-)$.

Let us show that $exp$ does not belong to $H(d(0, R_1^-))$. Indeed, suppose $exp$ belongs to $H(d(0, R_1^-))$. Consider the Mittag-Leffler decomposition of $exp$ on the infraconnected set $d(0, R_1^-)$. It is of the form $\sum_{n=0}^{\infty} g_n$ with $g_0 \in H(d(0, R_1)$ and $g_n \in H_0(\mathbb{K}\setminus d(a_n, R_n^-))$ with $a_n \in C(0, R_1)$. Set $T_n = d(a_n, R_1^-)$, $n \in \mathbb{N}^*$ and $S = \bigcup_{n=1}^{\infty} T_n$.

Let $\mathcal{K}$ be the residue class field of $\mathbb{K}$. By Theorem 8.4, we can consider a complete algebraically closed extension $\tilde{\mathbb{K}}$ of $\mathbb{K}$ whose residue class field $\tilde{\mathcal{K}}$ is not countable. Thus, we can find a class $G$ of $\tilde{C}(0, R_1)$ that has an empty intersection with $S$, and then, by Theorem 15.1, $exp$ has continuation to an element of $H(d(0, R_1^-) \cup G)$. Let $c \in G$. Since $|c| = R_1$, by Theorem 25.3, the function $h(x) = Log(1+x) - c = -(c + \sum_{n=1}^{\infty} \frac{(-x)^n}{n})$ satisfies $\nu^-(h, \log_p(R_1) = 0$, $\nu^+(h, \log_p(R_1) > 1$, hence $h$ admits a zero $a \in \tilde{C}(0, R_1)$. Then, $a$ does not belong to $\mathbb{K}$ because if $a \in \mathbb{K}$, then $Log(1 + a) \in \mathbb{K}$, a contradiction. Now, let $\zeta$ be a $p$th root of 1 different from 1 and let $t = \zeta(1 + a)$.
Since $|\zeta - 1| = R_1$, $t$ is of the form $1 + b$ with $b \in \hat{C}(0, R_1)$. We then have $\text{Log}(1+a) = \text{Log}(1+b)$. Set $E = \hat{d}(a, R_1^{-})$, $F = \hat{d}(b, R_1^{-})$, $D' = d(0, R_1^{-}) \cup E \cup F$, and $D'' = d(0, R_1^{-}) \cup G$. Since the image of $d(0, R_1^{-})$ by $\text{Log}(1+x)$ is $d(0, R_1^{-})$ and since the derivative of $\text{Log}(1+x)$ has no zero, by Corollary 22.10, we can check that for each $u \in \hat{C}(0, R_1)$, the image of $d(u, R_1^{-})$ by $\text{Log}(1+x)$ is $d(\text{Log}(1+u), R_1^{-})$. Consequently, both images of $E$ and $F$ by $\text{Log}(1+x)$ are equal to $G$. Now, $\text{Log}(1+x)$ belongs to $\hat{H}(D')$, the image of $D'$ by the function $f(x) = \text{Log}(1+x)$ is $D''$, and $\text{exp}$ belongs to $\hat{H}(D'')$. Consequently, by Corollary 12.3, $\text{exp} \circ \text{Log}(1+x)$ belongs to $\hat{H}(D')$ and we have $\text{exp} \circ \text{Log}(1+x) = 1+x \quad \forall x \in d(0, R_1^{-})$. Finally, since $D'$ has no pierced filter, by Proposition 20.5, it is an analytic set. Consequently, the equality $\text{exp} \circ \text{Log}(1+x) = 1+x \quad \forall x \in d(0, R_1^{-})$ holds in all $D'$, a contradiction since $\text{Log}(1+x)$ is not injective in $D'$. That finishes showing that $\text{exp}$ does not belong to $H(d(0, R_1^{-}))$. $\square$

**Notation.** Henceforth, we put $e^x = \text{exp}(x)$.

**Theorem 25.6.** Suppose that $p \neq 0$. Let $x \in d(0, R_1^{-})$. Then $e^x$ is algebraic over $\mathbb{Q}_p$ if and only if so is $x$. Let $u \in d(0, 1^{-})$. Then $\text{log}(1+u)$ is algebraic over $\mathbb{Q}_p$ if and only if so is $u$.

**Proof.** By Theorem 14.24, if $x$ is algebraic over $\mathbb{Q}_p$, so is $e^x$. Similarly, if $u$ is algebraic over $\mathbb{Q}_p$, so is $\text{log}(1+u)$. Consequently, suppose that $e^x$ is algebraic over $\mathbb{Q}_p$. Then $e^x$ is of the form $1+t$ with $|t| < 1$, hence $\text{log}(1+t)$ is algebraic over $\mathbb{Q}_p$. But then, $\text{log}(1+t) = \text{log}(e^x) = x$, hence $x$ is algebraic over $\mathbb{Q}_p$. Now, more generally, suppose $\text{log}(1+u)$ is algebraic over $\mathbb{Q}_p$, with $|u| < 1$. Take $q \in \mathbb{N}$ such that $|p^q \text{log}(1+u)| < R_1$. We have $p^q \text{log}(1+u) = \text{log}((1+u)^{p^q})$. Since $|p^q \text{log}(1+u)| < R_1$, we have $|\text{log}((1+u)^{p^q})| < R_1$, hence $\text{exp}(\text{log}((1+u)^{p^q})) = (1+u)^{p^q}$. Consequently, $(1+u)^{p^q}$ is algebraic over $\mathbb{Q}_p$ and hence so is $u$. $\square$

We can show a similar result when $p=0$:

**Theorem 25.7.** Suppose that $p=0$. Let $x \in d(0, 1^{-})$. Then $e^x$ is algebraic over $\mathbb{Q}_p$ if and only if so is $x$. Let $u \in d(0, 1^{-})$. Then $\text{log}(1+u)$ is algebraic over $\mathbb{Q}_p$ if and only if so is $u$.
# Chapter 26

## Problems on *p*-adic Exponentials

The author is grateful to Michel Waldschmidt for his advices. On the other hand, most of results first proven in the field $\mathbb{K}$ also hold (with slide changes) in an ultrametric field $\mathcal{K}$ of residue characteristic 0, as, for example, the Levi-Civita field [88].

**Notation.** 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 $b\zeta(r)$.

Throughout this chapter, the field $\mathbb{K}$ is supposed to have characteristic 0 and residue characteristic $p \neq 0$ and we now denote by $\mathbb{I}$ an algebraically closed complete field with respect to an ultrametric absolute value, with residue characteristic 0.

Next, we denote by $\Psi_p$ the function $\Psi$ on the field $\mathbb{C}_p$ and by $\Psi_0$ the function $\Psi$ on the field $\mathbb{I}$.

Proposition 26.1 is used in the proof of the next theorems and is proven by induction, similar to (1.4.2) in [92].

**Proposition 26.1.** Let $P_1, \dots, P_q \in \mathbb{K}[X]$ be different from 0 and let $w_1, \dots, w_q \in \mathbb{K}$ be pairwise distinct. Let $F(x) = \sum_{j=1}^q P_j(x)e^{w_j x}$. Then $F$ is not identically zero.

**Corollary 26.2.** Let $b_1, \dots, b_n \in D_1$ (respectively in $D_0$). The functions $x, e^{b_1 x}, \dots, e^{b_n x}$ are algebraically independent over $\mathbb{K}$ (respectively over $\mathbb{I}$) if and only if $b_1, \dots, b_n$ are $\mathbb{Q}$-linearly independent.
**Theorem 26.3.** Let $\lambda > 0$ and let $F(x) = \sum_{i=1}^{k} f_i(x)e^{\omega_i x}$, where each $\omega_i$ satisfies $-\Psi_p(\omega_i) \ge \frac{1}{p-1} + \lambda$, each $f_i$ lies in $\mathbb{C}_p[x]$, and $\deg(f_i) = m_i - 1$. Let $n = \sum_{i=1}^{k} m_i$.

Then the number $N$ of zeros of $F$ in $d(0,1)$ satisfies $N \le (n-1)(1 + \frac{1}{\lambda(p-1)})$.

**Proof.** Denoting by $\mathcal{D}$ the derivation $\frac{d}{dx}$, then $F$ satisfies a relation of the form

$$ (1) \qquad \mathcal{D}^n F = c_1 \mathcal{D}^{n-1} F + c_2 \mathcal{D}^{n-2} F + \dots + c_n F, $$

where

$$ (2) \qquad \prod_{i=1}^{k} (x - \omega_i)^{m_i} = x^n - c_1 x^{n-1} - c_2 x^{n-2} - \dots - c_n. $$

Expanding $F$ as a power series, we have

$$ F(x) = \sum_{h=0}^{\infty} a_h x^h = \sum_{h=0}^{\infty} \frac{b_h}{h!} x^h, \quad |x| < 1. $$

Moreover, by Proposition 26.1, $F$ is not identically zero. From the differential equation (1), we have the recurrence

$$ (3) \qquad b_{h+n} = c_1 b_{h+n-1} + \dots + c_n b_h, \quad h = 0, 1, 2, \dots $$

Since $-\Psi_p(\omega_i) \ge \frac{1}{p-1} + \lambda$, the series $F$ converges in $d(0,1)$, hence it lies in $H(d(0,1))$, therefore there exists $t \in \mathbb{N}$ such that $|a_s| = \max_{h \in \mathbb{N}} |a_h|$, and hence without loss of generality, we can suppose that $\max_{1 \le i \le k} \Psi_p(a_h) = 0$. Let $t$ be the biggest of the $s \in \mathbb{N}$ such that $|a_s| = \max_{h \in \mathbb{N}} |a_h|$. Thus we have $|a_t| = 1$ and $|a_h| < 1 \forall h > t$. Consequently, by Theorem 22.7, the number of zeros of $F$ in $d(0,1)$ is $t$.

Let $q = \frac{1}{p-1} + \lambda$, hence $-\Psi_p(\omega_i) \ge q$ ($i = 1, \dots, k$). From (2), we can see that $-\Psi_p(c_j) > jq$ ($j = 1, 2, \dots, n$).
Now, since by assumption we have $\Psi_p(a_h) \le 0$ and $\Psi_p(a_h) = \Psi_p(b_h) - \Psi_p(h!)$, this implies that $\Psi_p(b_h) \le 0$ ($h = 0, 1, \dots$). By (3), we have

$$-\Psi_p(b_n) \ge \min_{1 \le j \le n} (-\Psi_p(c_j) - \Psi_p(b_{n-j}) > \min_{1 \le j \le n} (qj - \Psi_p(b_{n-j})) \ge q.$$

Then, by (3), with $h = 1, 2, \dots$, we obtain $-\Psi_p(b_{h+n-1}) > hq$, hence

$$-\Psi_p(a_{h+n-1}) > hq + \Psi_p((n-1+h)!) > hq - \frac{(n-1+h)}{p-1} \\ = h\lambda - \frac{n-1}{p-1}.$$

Finally, it follows that if $h \ge \frac{(n-1)}{\lambda(p-1)}$, then necessarily $\Psi_p(a_{h+n-1}) < 0$, hence $|a_{h+n-1}| < |a_t| \forall h \ge \frac{(n-1)}{\lambda(p-1)}$, $\forall h \ge 1$. Thus, we have $|a_j| < 1 = |a_t| \forall j \ge (n-1)(1+\frac{1}{\lambda(p-1)})$, therefore $t < (n-1)(1+\frac{1}{\lambda(p-1)})$ and hence the number of zeros of $F$ in $d(0, 1)$ is bounded by $(n-1)(1+\frac{1}{\lambda(p-1)})$. $\square$

**Theorem 26.4.** Let $F(x) = \sum_{i=1}^k f_i(x)e^{\omega_i x}$, where each $\omega_i \in \mathbb{I}$ satisfies $\Psi_0(\omega_i) < 0$, each $f_i$ lies in $\mathbb{I}[x]$, and $\deg(f_i) = m_i - 1$. Let $n = \sum_{i=1}^k m_i$.

Then the number $N$ of zeros of $F$ in $d(0, 1)$ satisfies $N \le n - 1$.

**Proof.** Denoting by $\mathcal{D}$ the derivation $\frac{d}{dx}$, then $F$ satisfies a relation of the form

$$(1) \qquad \mathcal{D}^n F = c_1 \mathcal{D}^{n-1} F + c_2 \mathcal{D}^{n-2} F + \dots + c_n F,$$

where

$$(2) \qquad \prod_{i=1}^{k} (x - \omega_i)^{m_i} = x^n - c_1 x^{n-1} - c_2 x^{n-2} - \dots - c_n.$$

Expanding $F$ as a power series, we have

$$F(x) = \sum_{h=0}^{\infty} a_h x^h, \quad |x| < 1.$$
Let $b_h = a_h h!$, $h \in \mathbb{N}$. Then, $F(x) = \sum_{h=0}^{\infty} \frac{b_h}{h!} x^h$. By Proposition 26.1, $F$ is not identically zero. From the differential equation (1), we have the recurrence

$$ (3) \qquad b_{h+n} = c_1 b_{h+n-1} + \dots + c_n b_h, \quad h = 0, 1, 2, \dots $$

Since $\Psi_0(\omega_i) < 0 \forall i = 1, \dots, k$, the series $F$ converges in $d(0, 1)$, hence it lies in $H(d(0, 1))$, therefore there exists $t \in \mathbb{N}$ such that $|a_s| = \max_{h \in \mathbb{N}} |a_h|$, and hence without loss of generality, we can suppose that $\max_{1 \le i \le k} \Psi_0(a_h) = 0$. Let $t$ be the biggest of the $s \in \mathbb{N}$ such that $|a_s| = \max_{h \in \mathbb{N}} |a_h|$. Thus we have $|a_t| = 1$ and $|a_h| < 1 \forall h > t$. Consequently, by Theorem 22.7, the number of zeros of $F$ in $d(0, 1)$ is $t$. Let $\epsilon = \min_{1 \le i \le k} (-\Psi_p(\omega_i))$. From (2), we can see that $\Psi_0(c_j) \le j\epsilon$ ($j = 1, 2, \dots, n$).

By assumption, we have $\Psi_0(a_h) \le 0$ and $\Psi_0(a_h) = \Psi_0(b_h) - \Psi_0(h!) = \Psi_0(b_h)$ because $\Psi_0(h!) = 0 \forall h$, hence $\Psi_0(b_h) \le 0$ ($h = 0, 1, \dots$). By (3), we have

$$ -\Psi_0(b_n) \ge \min_{1 \le j \le n} (-\Psi_0(c_j) - \Psi_0(b_{n-j})) > \min_{1 \le j \le n} (\epsilon j - \Psi_0(b_{n-j})) \ge \epsilon. $$

Then, by (3), with $h = 1, 2, \dots$ we have $-\Psi_0(b_{h+n-1}) > h\epsilon$, hence

$$ -\Psi_0(a_{h+n-1}) > h\epsilon + \Psi_0((n-1+h)!) = h\epsilon \quad \forall h \ge 1. $$

Thus, for every $h \ge 1$, we have $-\Psi_0(a_{h+n-1}) > 0$, hence $t \le n-1$. Consequently, the number of zeros of $F$ in $d(0, 1)$ is bounded by ($n-1$), which ends the proof. $\square$

Hermite-Lindemann's theorem is well known in complex analysis. The same holds in $p$-adic analysis. We need Siegel's lemma in all the following theorems of this chapter. We choose a particular form of this lemma [92]:

**Lemma 26.5 (Siegel).** Let $E$ be a finite extension of $\mathbb{Q}$ of degree $q$ and let $\lambda_{i,j}$ $1 \le i \le m$, $1 \le j \le n$ be elements of $E$ integral over $\mathbb{Z}$. Let $M = \max(|\lambda_{i,j}| \ 1 \le i \le m, \ |1 \le j \le n|$) and let $(S)$ be the linear system $\{\sum_{j=1}^n \lambda_{i,j} x_j = 0, \ 1 \le i \le m\}$. There exists solutions $(x_1, \dots, x_n)$ of $(S)$ such that $x_j \in \mathbb{Z} \ \forall j = 1, \dots, n$ and

$$ \log(|x_j|_{\infty}) \le \log(M) \frac{qm}{n-qm} + \frac{\log(2)}{2} \quad \forall j = 1, \dots, n. $$
Lemma 26.6 is necessary in the proof of the following theorems and is easily proved in [92] since its proof implies no change in the field $K$ since it only concerns algebraic numbers.

**Lemma 26.6.** Let $a_1, \dots, a_q \in K$ be algebraic over $\mathbb{Q}$, let $P(X_1, \dots, X_q) \in \mathbb{Z}[X_1, \dots, X_q]$ be such that $\deg_{X_j}(P) \le r_j$ $1 \le j \le q$, and let $\beta = P(a_1 \cdots a_q)$. Then $\beta$ is algebraic over $\mathbb{Q}$, $d(a_1)^{r_1} \cdots d(a_q)^{r_q}$ is a multiple of $\text{den}(\beta)$, and we have

$$s(\beta) \le \log H(P) + \sum_{j=1}^{q} (r_j s(a_j) + \log(r_j) + 1).$$

The $p$-adic version of Hermite-Lindemann's theorem was proved by K. Mahler [73]. Here we give another proof, using specific ultra-metric tools.

**Notation.** We denote by $D_0$ the disk $d(0, 1^{-})$, and if the residue characteristic of $K$ is $p > 0$, we put $r_1 = p^{\frac{1}{p-1}}$ and denote by $D_1$ the disk $d(0, r_1^{-})$.

Given a positive real number $a$, we denote by $[a]$ the biggest integer $n$ such that $n \le a$.

**Remark.** In particular, Levi-Civita's fields have residue characteristic 0 [88].

**Theorem 26.7.** Let $\alpha \in D_1 \subset K$ be algebraic. Then $e^{\alpha}$ is transcendental.

**Proof.** We suppose that $\alpha$ and $e^{\alpha}$ are algebraic. Let $h = |\alpha|$. Let $E$ be the field $\mathbb{Q}[\alpha, e^{\alpha}]$, let $q = [E : \mathbb{Q}]$, and let $w$ be a common denominator of $\alpha$ and $e^{\alpha}$. We construct a sequence of polynomials $(P_N(X, Y))_{N \in \mathbb{N}}$ in two variables such that $\deg_X(P_N) = [\frac{N}{\log(N)}]$, $\deg_Y(P_N) = [(\log N)^3]$ and such that the function $F_N(x) = P_N(x, e^x)$ satisfy further, for every $s = 0, \dots, N-1$ and for every $j = 0, \dots, [\log(N)]$,

$$\frac{d^s}{dx^s} F_N(j\alpha) = 0.$$
According to computations in the proof of Hermite Lindemann's theorem in the complex context (Theorem 3.1.1 in [92]), we have

$$ (1) \quad \frac{d^M F_N(\gamma_N)}{dx^M} = \sum_{l=0}^{u_1(N)} \sum_{m=0}^{u_2(N)} b_{l,m,N} \sum_{\sigma=0}^{u_1(N)} \left( \frac{u_1(N)!}{\sigma!(u_1(N)-\sigma)!} \right) \times \left( \frac{l!}{(u_1(N)-\sigma)!} \right) m^{u_1(N)-\sigma} \\ j^{u_1(N)-\sigma} \cdot (\alpha)^{u_1(N)-\sigma} \cdot (e^{\alpha})^{ju_2(N)}. $$

We put $u_1(N) = \deg_X(P_N)$, $u_2(N) = \deg_Y(P_N)$. We solve the system

$$ w^{u_1(N)+u_2(N)} \frac{d^s}{dx^s} F_N(j\alpha) = 0, \quad 0 \le s \le N-1, \quad j = 0, \dots, [\log(N)], $$

where the undeterminates are the coefficients $b_{l,m,N}$ of $P_N$. We then write the system under the form

$$ \sum_{l=0}^{u_1(N)} \sum_{m=0}^{u_2(N)} b_{l,m,N} \sum_{\sigma=0}^{\min(s,l)} \left( \frac{s!}{\sigma!(s-\sigma)!} \right) \left( \frac{l!}{(l-\sigma)!} \right) m^{s-\sigma} \cdot j^{l-\sigma}, \\ (2) \quad (w\alpha)^{l-\sigma} (we^{\alpha})^{jm} \cdot w^{u_1(N)-(l-\sigma)+u_2(N)-jm} = 0. $$

That represents a system of $N[\log(N)]$ equations of at least $N([\log(N)])^2$ undeterminates, with coefficients in $E$, integral over $\mathbb{Z}$.

According to computations of Hermite-Lindemann's theorem in the complex context (Theorem 3.1.1 in [92]), it appears that in system (2), each factor $(\frac{s!}{\sigma!(s-\sigma)!})$, $(\frac{l!}{(l-\sigma)!})$, $m^{s-\sigma}$, $j^{l-\sigma}$, $(w\alpha)^{l-\sigma}$, $(we^{\alpha})^{jm}$, $w^{u_1(N)-(l-\sigma)+u_2(N)-jm}$ admits a bounding of the form $SN(\log(\log(N)))$, when $N$ goes to $+\infty$. On one hand, $w^{u_1(N)+u_2(N)}$ is a common denominator and we have

$$ \log(w^{u_1(N)+u_2(N)}) \le \log(\omega) \left( \frac{N}{\log(N)} + (\log(N))^3 \right), $$

and hence we have a constant $T > 0$ such that

$$ (3) \quad \log(w^{u_1(N)+u_2(N)}) \le \frac{TM}{\log M}. $$
Next, we note that

$$ (4) \quad \begin{aligned} \log \left( \frac{u_1(N)!}{\sigma!(u_1(N)-\sigma)!} \right) &\le u_1(N) \log(u_1(N)) \\ &\le \frac{N}{\log(N)} \log \left( \frac{N}{\log(N)} \right) \le N, \end{aligned} $$

and similarly,

$$ (5) \quad \log \left( \frac{l!}{(u_1(N) - \sigma)!} \right) \le u_1(N) \log(u_1(N)) \le N, $$

and

$$ (6) \quad \log(m^{u_1(N)-\sigma}) \le \frac{3N}{\log(N)} \log(\log(N)). $$

Now, we check that

$$ \log \left( j^{u_1(N)-\sigma} \cdot (|\bar{\alpha}|)^{u_1(N)-\sigma} \cdot (|\overline{e^\alpha}|)^{ju_2(N)} \right) \\ \le N + \frac{N}{\log(N)} \log(|\bar{\alpha}|) + \log(N)(\log(N))^3 \log(|\overline{e^\alpha}|), $$

and hence there exists a constant $L > 0$ such that

$$ (7) \quad \log \left( j^{u_1(N)-\sigma} \cdot (|\bar{\alpha}|)^{u_1(N)-\sigma} \cdot (|\overline{e^\alpha}|)^{ju_2(N)} \right) \le LN. $$

Therefore, by (2), (3), (4), (5), (6), and (7), we have a constant $C > 0$ such that each coefficient $a$ of the system satisfies

$$ (8) \quad s(a) \le CN(\log(\log(N))). $$

By Siegel's lemma 26.5 and by (8), there exist integers $b_{l,m,N}$, $0 \le l \le u_1(N)$, $0 \le m \le u_2(N)$ in $\mathbb{Z}$ such that

$$ (9) \quad \begin{aligned} & 0 < \max_{l \le u_1(N), m \le u_2(N)} \log(|b_{l,m,N}|_\infty) \\ & \le \frac{qN \log(N)}{N(\log(N))^2 - qN \log(N)} (CN \log(\log(N))) \end{aligned} $$
and such that the function

$$ (10) \qquad F_N(x) = \sum_{l=0}^{u_1(N)} \sum_{m=0}^{u_2(N)} b_{l,m;N} x^l e^{mx} $$

satisfies

$$ \frac{d^s}{dx^s} F_N(j\alpha) = 0, \quad 0 \le s \le N-1, \quad j = 0, 1, \dots, [\log(N)]. $$

Now, by (9), we can check that there exists a constant $G > 0$ such that

$$ (11) \qquad \max_{l \le u_1(N), m \le u_2(N)} (\log(|b_{l,m,N}|_\infty) \le \frac{GN \log(\log(N))}{\log(N)}. $$

The function $F_N$ we have defined in (10) belongs to $\mathcal{A}(D_1)$ and by Proposition 26.1 is not identically zero, hence at least one of the numbers $\frac{d^s}{dx^s}F_N(0)$ is not null. Let $M$ be the biggest of the integers such that $\frac{d^s}{dx^s}F_N(j\alpha) = 0 \forall s = 0, \dots, M-1, j = 0, 1, 2, \dots, [\log(N)]$. Thus, we have $M \ge N$ and there exists $j_0 \in \{0, 1, \dots, [\log(N)]\}$ such that $\frac{d^M}{dx^M}F_N(j_0\alpha) \ne 0$. We put $\gamma_N = \frac{d^M}{dx^M}F_N(j_0\alpha)$.

Let us now give an upper bound of $s(\gamma_N)$. On one hand, $w^{u_1(N)+u_2(N)}$ is a common denominator, and by (2), we have a constant $T > 0$ such that

$$ \log(w^{u_1(N)+u_2(N)}) \le \frac{TM}{\log M}. $$

On the other hand, by (1), we have

$$ \frac{d^M F_N(\gamma_N)}{dx^M} = \sum_{l=0}^{u_1(N)} \sum_{m=0}^{u_2(N)} b_{l,m,N} \sum_{\sigma=0}^{u_1(N)} \left( \frac{u_1(N)!}{\sigma!(u_1(N)-\sigma)!} \right) \times \left( \frac{l!}{(u_1(N)-\sigma)!} \right) m^{u_1(N)-\sigma}. $$

$$ j^{u_1(N)-\sigma} \cdot (\alpha)^{u_1(N)-\sigma} \cdot (e^{\alpha})^{ju_2(N)}. $$

Now, by (2), (3), (6), (7), (8), and (10) and taking into account that the number of terms is bounded by $N(\log N)^2$, we can check that
there exists a constant $B$ such that

$$ (12) \qquad s(\gamma_N) \le BN. $$

Let us now give an upper bound of $|\gamma_N|$. For convenience, we first suppose that $j_0 = 0$, hence $\frac{d^M}{dx^M}F_N(0) \ne 0$. Set $h = |\alpha|$. Then, by Theorem 18.1, we have $|\gamma_N| \le \frac{|F_N|(h)}{h^M}$. Moreover, we note that $F_N$ admits at least $M[\log(M)]$ zeros in $d(0, h)$, and therefore, by Corollary 22.30, we have $|F_N|(h) \le \left(\frac{h}{R_1}\right)^{M[\log(M)]}$ because $|F_N|(r) \le 1 \forall r < R_1$. Consequently, $|\gamma_N| \le \frac{h^{M(\log(M)-1)}}{(R_1)^{M\log M}}$, and hence

$$ \log(|\gamma_N|) \le M(\log(M) - 1)(\log(h)) - M \log(M)(\log(R_1)). $$

Let $\lambda = \log(h) - \log(R_1)$. Then $\lambda < 0$. And we have $\log(|\gamma_N|) \le \lambda M \log(M) - M \log(h)$, therefore there exists a constant $A > 0$ such that

$$ (13) \qquad \log(|\gamma_N|) \le -AM \log(M). $$

Let us now stop assuming that $j_0 = 0$. Putting $z = x - j\alpha$ and $g(z) = f(x)$, since all points $j\alpha$ belong to $d(0, h)$, it is immediate to go back to the case $j_0 = 0$, which confirms (13) in the general case. But now, by Theorem 9.10, relations (12) and (13) make a contradiction to the relation $-2qs(\gamma_N) \le \log(|\gamma_N|)$ satisfied by algebraic numbers and shows that $\gamma_N$ is transcendental. But then, so is $e^\alpha$. $\square$

In $\mathbb{L}$, we have a similar version:

**Theorem 26.8.** Let $\alpha \in \mathbb{L}$ be algebraic such that $|\alpha| < 1$. Then $e^\alpha$ is transcendental over $\mathbb{Q}$.

**Proof.** Everything works in $\mathbb{L}$ as in a field of residue characteristic $p \ne 0$ up to Relation (8) in the proof of Theorem 26.7. Here we can replace $r_1$ by 1 and therefore the conclusion is the same as in Theorem 26.7. $\square$

**Theorem 26.9 (Gelfond–Schneider).** Let $\ell \in D_1$, $\ell \ne 0$, and let $b \notin \mathbb{Q}$ belong to $\mathbb{K}$ be such that $b\ell \in D_1$. Then at least one of the three numbers $a = e^\ell$, $b$, $e^{b\ell}$ is transcendental.
**Proof.** A large part of the proof does not involve the topology of the field $\mathbb{K}$ and hence is similar to the proof in the field $\mathbb{C}$ [92] where we can copy many technical relations. We suppose that $a = e^\ell$, $b$ and $e^{b\ell}$ are algebraic over $\mathbb{Q}$. Let $E = \mathbb{Q}[e^\ell, b, e^{b\ell}]$ and let $\delta = [E : \mathbb{Q}]$ and let $d$ be a common denominator of $b, e^\ell, e^{b\ell}$.

Put $S = \max(1, |b|)$, $T \in ]S, \frac{R_1}{|b|}[$, $\sigma = \log(\frac{T}{S})$, $\tau = \log T$, $\Lambda = d(0, S)$, and $\Delta = d(0, T)$. We consider integers $N$ of the form $q^2$, with $q \in \mathbb{N}$, and we first show that there exists a non-identically zero polynomial $P_N(X, Y) \in \mathbb{Z}[X, Y]$ such that $\deg_X(P_N) \le N^{\frac{3}{2}}$, and $\deg_Y(P_N) \le 2\delta N^{\frac{1}{2}}$ such that the function $F_N(x)$ defined in $\Delta$ by $F_N(x) = P_N(x, e^{\ell x})$ satisfy

$$F_N(i + jb) = 0 \quad \forall i = 1, \dots, N, \quad \forall j = 1, \dots, N.$$

In order to find $P_N$, let us write

$$\sum_{h=0}^{N^{\frac{3}{2}}-1} \sum_{k=0}^{2\delta N^{\frac{1}{2}}-1} C_{h,k}(N)X^hY^k$$

with $C_{h,k}(N) \in \mathbb{Z}$ and consider the system of equations where the $C_{h,k}(N)$ are the undeterminates:

$$d^{(4\delta+1)N^{\frac{3}{2}}}. F_N(i + jb) = 0 \quad (1 \le i \le N; 1 \le j \le N).$$

Thus, we obtain a system of $N^2$ equations of $2\delta N^2$ undeterminates in $\mathbb{Z}$, with coefficients in $E$. By Lemma 26.4, these coefficients have size bounded by

$$N^{\frac{3}{2}} \log(N) + N^{\frac{3}{2}}(8\delta + 2) \log(d) + \log(1 + |b|) + 2\delta \log(|e^\ell + b\ell|) \\ \le \frac{3}{2} N^{\frac{3}{2}} \log(N).$$

By Lemma 26.3, we can find in $\mathbb{Z}$ a family of integers not all equal to zero, $(C_{h,k}(N), 0 \le N^{\frac{3}{2}} - 1, 0 \le k \le 2\delta N^{\frac{1}{2}} - 1)$ satisfying

$$\log \left( \max_{h,k} |C_{h,k}(N)|_\infty \right) \le 2N^{\frac{3}{2}} \log N \left( \frac{\delta N^2}{2\delta N^2 - \delta N^2} \right) = 2N^{\frac{3}{2}} \log N$$

such that the function $F_N$ defined by $F_N(x) = P_N(x, e^{\ell x})$ satisfies $F_N(i + jb) = 0 \quad \forall i = 1, \dots, N, \quad j = 1, \dots, N$.
Now, we can check the function $F_N$ is an analytic element in every disk of the form $d(0, r)$ such that $r|\ell| < R_1$ and hence in $\Delta = d(0, T)$. Since the power of $x$ in the various terms is at most $N^{3/2}$ and since all coefficients are integers, we can check that $\log(|F_N|(T)) \le \tau N^{3/2}$. On the other hand, since the polynomial $P_N$ is not identically zero, by Proposition 26.1, $F_N$ is not identically zero and then $F_N$ has finitely many zeros in $\Lambda$. Particularly, there exists a point of the form $i+jb$ such that $F_N(i+jb) \ne 0$. Consequently, there exists $M \ge N$ such that $F_N(i+jb) = 0 \ \forall i \le M, \forall j \le M$ and there exists a point $\gamma_N$ of the form $i_0+j_0b$ such that $F_N(\gamma_N) \ne 0$ with $M < i_0 \le M+1, M < j_0 \le M+1$. Consequently, the number of zeros of $F_N$ in $\Lambda$ is at least $M^2$. Then, by Corollary 22.30, we have $\log(|F_N(\gamma_N)|) \le \tau N^{3/2} - \sigma M^2$, hence there exists $\lambda > 0$ such that

$$ (1) \qquad \log(|F_N(\gamma_N)|) \le -\lambda M^2 \quad \forall N \in \mathbb{N}. $$

By definition, neither $\sigma$ nor $\tau$ depend on $N$, hence neither does $\lambda$.

On the other hand, by Lemma 26.4, we can check that $s(F_N(\gamma_N))$ satisfies an inequality of the form $s(F_N(\gamma_N)) \le AM^{3/2} \log(M)$ which by (1) contradicts the inequality $-2\delta s(F_N(\gamma_N)) \le \log(|F_N(\gamma_N)|)$ and this ends the proof. $\square$

**Theorem 26.10 (Gelfond–Schneider in zero residue characteristic).** Let $\ell \in D_0$, $\ell \ne 0$, and let $b \notin \mathbb{Q}$ belong to $\mathbb{L}$ and be such that $b\ell \in D_0$. Then at least one of the three numbers $a = e^\ell$, $b$, $e^{b\ell}$ is transcendental.

**Proof.** The proof is identical to the proof of Theorem 26.9 except that $T$ now belongs to $]S, \frac{1}{|\ell|}[$. $\square$

The six-exponential problem is well known on $\mathbb{C}$ and was solved by Serge Lang [71] and K. Ramachandra. The problem is the following: let $a_1, a_2, a_3$ (respectively $b_1, b_2 \in \mathbb{C}$) be $\mathbb{Q}$-linearly independent. Then at least one of the six numbers $e^{a_ib_j}$ is transcendental. Next, consider the same problem with only four exponentials: let $a_1, a_2$ (respectively $b_1, b_2 \in \mathbb{C}$) be $\mathbb{Q}$-linearly independent. The question is whether one of the numbers $e^{a_ib_j}$ is transcendental: this is the four exponentials conjecture on $\mathbb{C}$ due to Serge Lang.
The problem, however, has a solution somewhat similar to that of the six exponentials problem, in the particular case when one of the ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$ is algebraic.

The same problems make sense on a p-adic field such as $\mathbb{C}_p$ (provided the numbers $a_ib_j$ lie in the disk of convergence of the exponential). Here we give the solution of the six p-adic exponentials problem on the field $\mathbb{C}_p$ and this of the four p-adic exponentials problem when one of the ratios $\frac{a_1}{a_2}$ or $\frac{b_1}{b_2}$ is algebraic. This was described by Jean-Pierre Serre [87].

**Theorem 26.11.** Let $a_1, a_2, a_3$ (respectively $b_1, b_2 \in \mathbb{C}_p$) be $\mathbb{Q}$-linearly independent and such that $\max_{i=1,2,3} \max_{j=1,2} |a_ib_j| < r_1$. Then at least one of the numbers $e^{a_ib_j}$ is transcendental.

**Proof.** Assume that all numbers $e^{a_ib_j}$ are algebraic, put $E = \mathbb{Q}[(e^{a_ib_j})_{i=1,2,3}, j=1,2]$ and $q = [E : \mathbb{Q}]$. Without loss of generality, we can assume that $a_1 = 1, |a_2|, |a_3| \le 1$ and that $\max(|b_1|, |b_2|) \le \frac{1}{p^2}$.

Let $t \in \mathbb{N}^*$ be such that $te^{a_ib_j}$ is integral over $\mathbb{Z}$ for every $i = 1, 2, 3$ and every $j = 1, 2$ and let $B = \log(t \max\{|e^{a_ib_j}|, i = 1, 2, 3, j = 1, 2\})$. Let $\ell \in \mathbb{N}$ be such that $\ell > 9\sqrt{2}B(q+1)$.

Consider now the linear system of $\ell^2 N^2$ equations with coefficients in $E$:

$$ (S_N) \quad \sum_{\substack{1 \le m \le N, 1 \le n \le N, 1 \le s \le N \\ 1 \le i \le \ell N}} c_{m,n,s,N} e^{(ma_1+na_2+sa_3)(ib_1+jb_2)} = 0, $$

$1 \le j \le \ell N$. We note that the coefficients $e^{(ma_1+na_2+sa_3)(ib_1+jb_2)}$ of $(S_N)$ satisfy

$$ \log\left(\frac{|e^{(ma_1+na_2+sa_3)(ib_1+jb_2)}|}{6B\ell N^2}\right) \le 6B\ell N^2 $$

$1 \le j \le \ell N$. Now, by Siegel's lemma 26.5, there exists a family of solutions $(c_{m,n,s,N})_{1 \le m \le N, 1 \le n \le N, 1 \le s \le N}$ in $\mathbb{Z}$ such that

$$ (1) \quad \log |c_{m,n,s,N}|_{\infty} \le \frac{\log 2}{2} + 6B \frac{(q\ell^2 N^2)\ell N^2}{(N^3 - q\ell^2 N^2)}. $$
Let $f_N(x) = \sum_{1 \le m \le N, 1 \le n \le N} c_{m,n,s,N} e^{(ma_1+na_2+sa_3)x}$. By Proposition 26.1, $f_N$ is not identically zero. Then, by definition of ($S_N$), we have $f_N(ib_1+jb_2) = 0 \forall i = 1, \dots, \ell N, j = 1, \dots, \ell N$, hence $f_N$ admits at least $\ell^2 N^2$ zeros in the disk $d(0, \frac{1}{p^2})$. Let $u$ be a point of the form $ib_1+jb_2$ with $i, j \in \mathbb{N}$ such that $f_N(ib_1+jb_2) \neq 0$ and such that $i+j$ is minimum and let $h$ be this minimum: say $h = i_0 + j_0$. Thus we can check that when $i$ and $j$ are two positive integers such that $i+j < h$, then $f_N(ib_1+jb_2) = 0$. Consequently, by construction, we have $h > 2\ell N$ and the number of zeros of $f_N$ in $d(0, \frac{1}{p^2})$ is at least $\frac{(h-1)^2}{2}$. We note that $\|f_N\| \le 1$ because $|e^x| = 1 \forall x \in d(0, R_1^-)$ and $c_{m,n} \in \mathbb{Z} \forall m, n \in \mathbb{N}$. Consequently, by Corollary 22.30, we have $\log(|f_N|(\frac{1}{p})) \le -(h-1)N^2$ and therefore

$$ (2) \qquad \log|f_N(u)| \le -\frac{(h-1)^2}{2}. $$

Consider now some $c_{m,n,s,N} e^{(ma_1+na_2+sa_3)(ib_1+jb_2)}$ at the point $u$. By (1), we have $\log(|c_{m,n,s,N}|_\infty \le \frac{6B(q\ell^2 N^2)\ell N^2}{N^3-q\ell^2 N^2} + \text{and} \log|e^{(ma_1+na_2+sa_3)(i_0b_1+j_0b_2)}| \le 6BNh$. Consequently, by (1), we can derive

$$ \log \overline{c_{m,n,s,N} e^{(ma_1+na_2+sa_3)(i_0b_1+j_0b_2)}} \le 6BNh + \frac{\log 2}{2} + \frac{6Bq\ell^3 N^4}{N^3 - q\ell^2 N^2}, $$

therefore

$$ (3) \qquad \log(\overline{|f_N(u)|}) \le 6BNh + \frac{\log 2}{2} + \frac{6Bq\ell^3 N^4}{N^3 - q\ell^2 N^2} + 3 \log N. $$

Here we note that the denominator of $F_N(u)$ is bounded by $t^{3Nh}$ because $t^{3Nh}$ is clearly a multiple of the denominator of each term $e^{(ma_1+na_2+sa_3)(ib_1+jb_2)}$ whenever $i + j \le h$. Therefore, by Corollary 9.12, we can derive

$$ \log(|f_N(u)|) \ge -3Nh(q+1)\log(t) -q\left(6BNh + \frac{\log 2}{2} + \frac{6B\ell^3 N^4}{N^3 - q\ell^2 N^2} + 3\log(N)\right). $$
Consequently, by (2) and (3), we obtain

$$ \frac{(h-1)^2}{2} \le (q+1) \left( 9BNh + \frac{\log 2}{2} + \frac{6B\ell^3 N^4}{N^3 - q\ell^2 N^2} + 3\log(N) \right), $$

therefore $\frac{(h-1)^2}{2} \le (q+1)(9BNh + O(N))$, and hence

$$ h - 2 \le \frac{(h-1)^2}{h} \le 18B(q+1)N + O(1), $$

and hence $h - 1 \le 18B(q + 1)N + O(1)$. Now, since $\frac{(h-1)^2}{2} \ge \ell^2 N^2$, we have $\sqrt{2}\ell N \le 18B(q + 1)N + O(1)$, hence $\ell \le 9\sqrt{2}B(q + 1)$ when $N$ is big enough, a contradiction to the hypothesis on $\ell$. $\square$

And similarly, we have the following:

**Theorem 26.12.** Let $a_1, a_2, a_3$ (respectively $b_1, b_2 \in \mathbb{I}$) be $\mathbb{Q}$-linearly independent and such that $\max_{i=1,2,3} \max_{j=1,2} |a_ib_j| < 1$. Then at least one of the numbers $e^{a_ib_j}$ is transcendental over $\mathbb{Q}$.

As explained above, when reducing to 4 exponentials $e^{a_ib_j}$, $i = 1, 2, j = 1, 2$, the transcendence of one of the four numbers is just a conjecture in the general case. Here we give a proof in the particular case when one of the ratios $\frac{a_1}{a_2}$ or $\frac{b_1}{b_2}$ is algebraic.

**Theorem 26.13.** Let $a_1, a_2$ (respectively $b_1, b_2 \in \mathbb{C}_p$) be $\mathbb{Q}$-linearly independent and such that $\max_{i=1,2} \max_{j=1,2} |a_ib_j| < r_1$ and such that $\frac{a_1}{a_2}$ is algebraic. Then at least one of the numbers $e^{a_ib_j}$ is transcendental.

**Proof.** Assume that all numbers $e^{a_ib_j}$ are algebraic. Without loss of generality, we can assume that $|b_i| \le \frac{1}{p^2}$, $i = 1, 2$. Put $a = \frac{a_1}{a_2}$. Let $t \in \mathbb{N}^*$ be such that all the $te^{a_ib_j}$ and $ta$ are integral over $\mathbb{Z}$ for every $i = 1, 2$ and every $j = 1, 2$ and let $B = \log(t\max(|a|, \max\{|e^{a_ib_j}|, i = 1, 2, j = 1, 2\})$. Without loss of generality, we can assume that $a_1 = 1$, hence $a = a_2$. Since $a$ is algebraic, we can put $E = \mathbb{Q}[a, (e^{a_ib_j})_{i=1,2, j=1,2}]$ and $q = [E : \mathbb{Q}]$. We can find integers $s, l \in \mathbb{N}$ satisfying

$$ (1) \qquad s > (q+1)Bl \left( 12 + \frac{qs}{l^2 - qs} \right). $$
Consider now the linear system of $sN^2$ equations with coefficients in $E$:

$$ (S_N) \quad \sum_{\substack{1 \le m \le lN, 1 \le n \le lN \\ 1 \le i \le N}} c_{m,n,N} (m+na)^k e^{(m+na)(ib_1+jb_2)} = 0, $$

1 ≤ j ≤ N. We note that the coefficients $(m + na)^k e^{(m+na)(ib_1+jb_2)}$ of $(S_N)$ satisfy

$$ (2) \quad \log \left( \left| (m + na)^k e^{(m+na)(ib_1+jb_2)} \right| \right) \le 2BlN^2 + k \log(2N + B) $$

$$ 1 \le m \le lN, \quad 1 \le n \le lN, \quad 1 \le i \le N, \quad 1 \le j \le N, \quad 1 \le k \le s. $$

By Siegel's lemma 26.5 and by (2), there exists a family of solutions $(c_{m,n,N})_{1 \le m \le lN, 1 \le n \le lN}$ in $\mathbb{Z}$ such that

$$ \log |c_{m,n,N}|_{\infty} \le \frac{\log 2}{2} + (BlN^2 + s(\log 2N + B)) \frac{qsN^2}{(l^2N^2 - qsN^2)}, $$

therefore

$$ (3) \quad \log |c_{m,n,N}|_{\infty} \le \frac{qsBlN^4}{l^2N^2 - qsN^2} + O(\log(N)). $$

Now, let $f_N(x) = \sum_{m \le lN, n \le N} c_{m,n,N} e^{x(m+na)}$ and let $h$ be the smallest integer such that $(f_N)^{(h)}(ib_1+jb_2) \ne 0$ for some pair $(i_0, j_0)$ such that $i_0 \le N, j_0 \le N$. By Proposition 26.1, $f_N$ is not identically zero, and by definition, $s < h$. Consequently, the number of zeros of $f_N$ in $d(0, \frac{1}{p^2})$ is at least $(h-1)N^2$, taking multiplicity into account. Set $u = i_0b_1 + j_0b_2$. Consider now some $c_{m,n,N} e^{(m+na)(ib_1+jb_2)}$ at the point $u$. First, we have

$$ \log |e^{(m+na)(i_0b_1+j_0b_2)}| \le 4BlN^2. $$

Consequently, by (3), when $N$ is big enough, we can derive

$$ \log \overline{|c_{m,n,N} e^{(m+na)(i_0b_1+j_0b_2)}|} \le \frac{qsBlN^4}{l^2N^2 - qsN^2} + 4BlN^2 + O(\log(N)), $$
hence and therefore, by (2),

$$ \log |f_N^{(h)}(u)| \le \frac{qsBlN^4}{l^2N^2 - qsN^2} + 4BlN^2 + O(\log(N)). $$

On the other hand, we can check that

$$ \log(\mathrm{den}((m + na)^h e^{(m+na)(i_0b_1+j_0b_2)})) \le 2BlN^2 + hB + O(\log(N)), $$

hence $\log(\mathrm{den}(f_N^{(h)}(u))) \le 2BlN^2 + sB + O(\log(N))$. Consequently, by Corollary 9.11, we can derive

$$ (4) \quad \log(|f_N^{(h)}(u)|) \ge -(q+1) \left( \frac{qsBlN^4}{l^2N^2 - qsN^2} + 4BlN^2 + 8BlN^2 + 2hB + O(\log(N)) \right). $$

As in Theorem 26.11, we have $\|f_N\| \le 1$, hence, by Theorem 18.1, we can derive $\|f_N^{(h)}\| \le \frac{1}{ph}$. Next, $c_{m,n,N} \in \mathbb{Z} \forall m, n \in \mathbb{N}$, consequently, by Corollary 22.30, we have $\log(|f_N|(\frac{1}{p^2})) \le -(h-1)N^2$, and by Theorem 18.1, $\log(|f_N^{(h)}|(\frac{1}{p^2})) \le -(h-1)N^2 + h$. Therefore, by (2) and (4), we obtain

$$ (5) \quad (h-1)N^2 - h \le (q+1) \left( \frac{qsBlN^4}{l^2N^2 - qsN^2} + 12BlN^2 + 2hB \right) + O(\log(N)). $$

Now, by (1), we have $s > (q+1)Bl(12 + \frac{qs}{l^2-qs})$, and since $h > s$, we can see that (5) is impossible when $N$ is big enough, which ends the proof. □

Similarly, we have the following:

**Theorem 26.14.** Let $a_1, a_2$ (respectively $b_1, b_2 \in \mathbb{I}$) be $\mathbb{Q}$-linearly independent and such that $\max_{i=1,2} j=1,2 |a_ib_j| < 1$ and such that $\frac{a_1}{a_2}$ is algebraic. Then at least one of the numbers $e^{a_ib_j}$ is transcendental over $\mathbb{Q}$.
# Chapter 27

## Divisors of Analytic Functions

In this chapter, we define divisors in $K$ or in a disk $d(a, R^{-})$. We then define the divisor of an analytic function and of an ideal.

**Definition.** We call a divisor in $K$ (respectively a divisor in a disk $d(a, R^{-})$) a mapping $T$ from $K$ (respectively from $d(a, R^{-})$) to $N$ whose support is countable and has a finite intersection with each disk $d(a, r)$, $\forall r > 0$ (respectively $\forall r \in ]0, R[$). Thus, a divisor on $K$ (respectively of $d(a, R^{-})$) is characterized by a sequence $(a_n, q_n)_{n \in N}$ with $a_n \in K$, $\lim_{n \to \infty} |a_n| = \infty$ (respectively $a_n \in d(a, R^{-})$, $\lim_{n \to \infty} |a_n - a| = R$), $|a_n| \le |a_{n+1}|$, and $q_n \in N^* \forall n \in N$. So, we frequently denote a divisor by the sequence $(a_n, q_n)_{n \in N}$ which characterizes it.

The set of divisors on $K$ (respectively on $d(a, R^{-})$) is provided with a natural additive law that makes it a semi-group. It is also provided with a natural order relation: given two divisors $T$ and $T'$, we can set $T \le T'$ when $T(\alpha) \le T'(\alpha) \forall \alpha \in d(a, R^{-})$. Moreover, if $T, T'$ are two divisors such that $T(\alpha) \ge T'(\alpha) \forall \alpha \in d(0, R^{-})$, we can define the divisor $\frac{T}{T'}$.

Given $f \in A(K)$ (respectively $f \in A(d(a, R^{-}))$), we can define the divisor of $f$, denoted by $D(f)$ on $K$ (respectively of $d(a, R^{-})$), as $D(f)(\alpha) = 0$ whenever $f(\alpha) \ne 0$ and $D(f)(\alpha) = s$ when $f$ has a zero of order $s$ at $\alpha$.

Similarly, given an ideal $I$ of $A(K)$ (respectively of $A(d(a, R^{-}))$), we denote by $D(I)$ the lower bound of the $D(f)$ $f \in I$ and $D(I)$ is called the divisor of $I$.
Conversely, given a divisor $E$ of $\mathbb{K}$ (respectively of $d(a, R^{-}))$, we denote by $\mathcal{T}(E)$ (respectively by $\mathcal{T}_{a,R}(E)$) the ideal of $\mathcal{A}(\mathbb{K})$ (respectively of $\mathcal{A}(d(a, R^{-}))$) of the $f \in \mathcal{A}(\mathbb{K})$ (respectively of the $f \in \mathcal{A}(d(a, R^{-}))$) such that $\mathcal{D}(f) \ge E$.

As we see, given a divisor $T$ on $\mathbb{K}$, there is no problem to construct an entire function whose divisor is $T$. But given a divisor $T$ on a disk $d(a, R^{-})$, if $\mathbb{K}$ is not spherically complete, it is not always possible to find an analytic function (in that disk) whose divisor is $T$ due to Lazard's problem [72].

Finally, given a divisor $T = (a_n, q_n)_{n \in \mathbb{N}}$, we denote by $\bar{T}$ the divisor $(a_n, 1)_{n \in \mathbb{N}}$. Let $T = (a_n, q_n)_{n \in \mathbb{N}}$ be a divisor on $\mathbb{K}$ (respectively of $d(a, R^{-}))$. For every $r > 0$ (respectively $r \in [0, R]$), we set $|T|(r) = \prod_{|a_j| \le r} \left(\frac{r}{|a_j|}\right)^{q_j}$. The divisor $T$ on $d(a, R^{-})$ is said to be bounded if $\lim_{r \to R} |T|(r) < \infty$ and then we put $\|T\| = \lim_{r \to R} |T|(r)$.

The $\mathbb{K}$-algebra $\mathcal{A}(\mathbb{K})$ (respectively $\mathcal{A}(d(a, R^{-}))$) is provided with the following topology of $\mathbb{K}$-algebra: given $\epsilon \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^{-}))$), the neighborhoods of $f$ are the sets $\mathcal{W}(f, r, \epsilon) = \{h \in \mathcal{A}(\mathbb{K}) \mid |f - h|(r) \le \epsilon\}$ (respectively $\mathcal{W}(f, r, \epsilon) = \{h \in \mathcal{A}(d(a, R^{-})) \mid |f - h|(r) \le \epsilon\}$, with $0 < r < R, \epsilon > 0$).

**Remark.** Let $f \in \mathcal{A}(d(a, R^{-}))$ and let $(a_n, q_n)_{n \in \mathbb{N}} = \mathcal{D}(f)$. Then $\omega_{a_n}(f) = q_n \quad \forall n \in \mathbb{N}$ and $\omega_{\alpha}(f) = 0 \quad \forall \alpha \in d(a, R^{-}) \setminus \{a_n \mid n \in \mathbb{N}\}$.

Theorem 27.1 is immediate:

**Theorem 27.1.** Let $a \in \mathbb{K}$, $R > 0$. Let $f, g \in \mathcal{A}(\mathbb{K})$ (respectively $f, g \in \mathcal{A}((a, R^{-}))$) be such that $\mathcal{D}(f) \ge \mathcal{D}(g)$. Then there exists $h \in \mathcal{A}(\mathbb{K})$ (respectively $h \in \mathcal{A}(d(a, R^{-}))$) such that $f = gh$.

**Proof.** Let $T = \mathcal{D}(g) = (a_n, q_n)_{n \in \mathbb{N}}$. Let us fix $r > 0$ (respectively $r \in [0, R]$), and let $s \in \mathbb{N}$ be such that $|a_n| \le r \quad \forall n \le s$ and $|a_n| > r \quad \forall n > s$. Let $P_r(x) = \prod_{n=0}^{s} \left(1 - \frac{x}{a_n}\right)^{q_n}$. We can factorize $f$ in the form $P_r \hat{f}$, and similarly, we can factorize $g$ in the form $P_r \hat{g}$, hence $\frac{\hat{f}}{\hat{g}} = \frac{\hat{g}}{f}$. Since $\hat{g}$ has no zero in $d(0, r)$, it is invertible in $H(d(0, r))$, hence $\frac{\hat{f}}{\hat{g}}$ belongs to $H(d(0, r))$. This is true for all $r > 0$ (respectively for all $r \in [0, R]$, and hence $\frac{\hat{f}}{\hat{g}}$ belongs to $\mathcal{A}(\mathbb{K})$ (respectively to $\mathcal{A}(d(a, R^{-}))$)). $\square$

**Corollary 27.2.** Let $a \in \mathbb{K}$, $R > 0$. Let $I$ be an ideal of $\mathcal{A}(\mathbb{K})$ (respectively an ideal of $\mathcal{A}(d(a, R^{-}))$) and suppose that there exists
$g \in I$ such that $\mathcal{D}(g) = \mathcal{D}(I)$. Then $I = g\mathcal{A}(\mathbb{K})$ (respectively $I = g\mathcal{A}(d(a, R^{-}))$).

As an immediate application of the definitions, by Theorem 22.26, we have Lemma 27.3:

**Lemma 27.3.** Let $R \in \mathbb{R}_+^*$ and let $f, g \in \mathcal{A}(d(0, R^-))$ be such that $\mathcal{D}(f) \le \mathcal{D}(g)$. Then, given $r, s \in ]0, R[$ such that $r < s$, we have $\Psi(f, \log s) - \Psi(f, \log r) \le \Psi(g, \log s) - \Psi(g, \log r)$.

In the whole field $\mathbb{K}$, given a divisor $T$, it is always possible to find an entire function admitting $T$ for divisor.

**Theorem 27.4.** Let $T = (\alpha_n, q_n)_{n \in \mathbb{N}}$ be divisor of $\mathbb{K}$. The infinite product $\prod_{n=1}^{\infty} (1 - \frac{x}{\alpha_n})^{q_n}$ is uniformly convergent in all bounded subsets of $\mathbb{K}$ and defines an entire function $f \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(f) = T$. Moreover, given $g \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(g) = T$, then $g$ is of the form $\lambda f$.

**Proof.** We assume that $|\alpha_n| \le |\alpha_{n+1}| \forall n \in \mathbb{N}$. Let us fix $R > 0$ and set $f_m(x) = \prod_{n=1}^{m} (1 - \frac{x}{\alpha_n})^{q_n}$. Consider $N \in \mathbb{N}$ such that $|\alpha_N| > R$ and $m \ge N$. On the one hand, $|f_m|(R) = |f_N|(R)$. Set $M = |f_N|(R)$. On the other hand, we check that

$$ 
\begin{aligned}
|f_{m+1}(x) - f_m(x)| &= \left| \prod_{n=1}^{m} \left(1 - \frac{x}{\alpha_n}\right)^{q_n} \right| \left| \left(1 - \frac{x}{\alpha_{m+1}}\right)^{q_{m+1}} - 1 \right| \\
&\le M \left| \sum_{k=1}^{q_{m+1}} (-1)^k \binom{q_{m+1}}{k} \left(\frac{x}{\alpha_{m+1}}\right)^k \right| \\
&\le M \frac{R}{|\alpha_{m+1}|} \quad \forall x \in d(0, R).
\end{aligned}
$$

Consequently, $|f_{m+1} - f_m|(R) \le M \frac{R}{|\alpha_{m+1}|}$, which shows that the sequence $(f_m)_{m \in \mathbb{N}}$ is uniformly converging in $d(0, R)$ to an element of $H(d(0, R))$, hence to a power series. This is true for all $R > 0$, hence the limit $f$ defined in $\mathbb{K}$ belongs to $\mathcal{A}(\mathbb{K})$. Now, for each $m \in \mathbb{N}$, let $r_m = |\alpha_m|$. By construction, the zeros of $f_m$ in $d(0, r_m)$ are the $\alpha_n$ with $1 \le n \le m$, each with multiplicity $q_n$. And next, we note that $|(1 - \frac{x}{\alpha_n})^{q_n}| = 1 \forall n > m$, $\forall x \in d(0, r_m)$. Consequently, the zeros of $f$ in $d(0, r_m)$ are exactly those of $f_m$. Now, consider $g \in \mathcal{A}(\mathbb{K})$ such
that $\mathcal{D}(g) = T$. The function $h \in \mathcal{A}(\mathbb{K})$ such that $f = gh$ has no zero in $\mathbb{K}$ and hence is a constant. $\square$

**Corollary 27.5.** For every divisor $T$ on $\mathbb{K}$, there exists $f \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(f) = T$. Moreover, if $f(0) = 1$, $f$ satisfies $|f|(r) = |T|(r) \quad \forall r > 0$.

**Corollary 27.6.** Let $T$ be a divisor on $\mathbb{K}$, let $g \in \mathcal{A}(\mathbb{K})$ be such that $\mathcal{D}(g) = T$, and let $f \in \mathcal{A}(\mathbb{K})$ be such that $\mathcal{D}(f) \ge T$. Then there exists $h \in \mathcal{A}(\mathbb{K})$ such that $f = gh$.

**Proof.** Indeed, let $E = \frac{\mathcal{D}(f)}{T}$ and let $h \in \mathcal{A}(\mathbb{K})$ be such that $\mathcal{D}(h) = \frac{\mathcal{D}(f)}{T}$. Then $\mathcal{D}(f) = \mathcal{D}(gh)$, hence by Theorem 27.4, $\frac{f}{gh}$ is a constant and we can choose $h$ such that the constant is 1. $\square$

**Theorem 27.7.** Let $f \in \mathcal{A}(\mathbb{K})$ have a divisor of the form $(a_n, sq_n)$ with $s \in \mathbb{N}^*$. Then, there exists $g \in \mathcal{A}(\mathbb{K})$ such that $f = g^s$.

**Proof.** By Corollary 27.5, there exists $h \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(h) = (a_n, q_n)$. Then, $\frac{l}{h^s}$ has no zero and no pole, and therefore, it is a constant $\lambda$. Let $l \in \mathbb{K}$ be such that $l^s = \lambda$ and let $g = lh$. Then, $g^s = f$. $\square$

So, by Theorem 27.4, given a divisor $T$ on $\mathbb{K}$, we can find an entire function whose divisor is just $T$. It is natural to consider the same problem inside a disk $d(a, r^{-})$. Indeed, in $\mathbb{C}$, it is known that a similar problem always admits a solution: in the whole field $\mathbb{C}$ as well as inside an open disk. Actually, in the general context of a complete ultrametric algebraically closed field $\mathbb{K}$, the problem has no solution when $\mathbb{K}$ is not spherically complete.

This problem was first considered by M. Lazard [72] and we detail the solutions he gave. First, we construct a function $f$ whose divisor is bigger than the given divisor but narrows it.

We deal with the problem by showing that given a sequence $(a_n)_{n \in \mathbb{N}}$ such that $|a_n - a| < R$ for all $n \in \mathbb{N}$ and $\lim_{n \to \infty} |a_n - a| = R$, a sequence of integers $(q_n)_{n \in \mathbb{N}}$ and a number $\epsilon > 0$, there exists an analytic function $f \in \mathcal{A}(d(0, R^{-}))$ that admits each $a_n$ as a zero of
order $t_n \ge q_n$ and such that $|f|(r) \le (1 + \epsilon) |T|(r)$. First, we need Lemma 27.8.

**Lemma 27.8.** Let $T = (a_n, q_n)_{n \in \mathbb{N}}$ be a divisor on $d(a, R^{-})$ and let $f \in \mathcal{A}(d(a, R^{-}))$ satisfy $f(0) = 1$, $\mathcal{D}(f) \ge T$ and $|f|(r) = |T|(r) \quad \forall r \in [0, R[$. Then, $\mathcal{D}(f) = T$.

**Proof.** Since $f(0) = 1$, we may write $f$ in the form $\prod_{j=0}^{\infty} \left(1 - \frac{x}{a_j}\right)^{s_j}$ with $q_j \le s_j \quad \forall j \in \mathbb{N}$. By hypothesis, we have $q_j \le s_j \quad \forall j \in \mathbb{N}$. Suppose that $s_k > q_k$ for some index $k$ and let $r_n = |a_n|$, $n \in \mathbb{N}$. Since $|f|(r_k) = |T|(r_k)$, when $r \in]r_k, r_{k+1}[$, we have $|f|(r) = |f|(r_k)(\frac{r}{r_k})^{s_k}$, $|T|(r) = |f|(r_k)(\frac{r}{r_k})^{q_k}$, and since $|f|(r) = |T|(r) \quad \forall r \in]0, R[$, clearly $s_k = q_k$. $\square$

**Notation.** For each divisor $E$ of $\mathbb{K}$, we denote by $\mathcal{T}(E)$ the set of $f \in \mathcal{A}(\mathbb{K})$ such that $E \le \mathcal{D}(f)$. Similarly, for each divisor $E$ of $d(a, R^{-})$, we denote by $\mathcal{T}_{a,R}(E)$ the set of $f \in \mathcal{A}(d(a, R^{-}))$ such that $E \le \mathcal{D}(f)$.

**Theorem 27.9.** For every divisor $E$ of $\mathbb{K}$, $\mathcal{T}(E)$ is a closed ideal of $\mathcal{A}(\mathbb{K})$. Moreover, $\mathcal{T}$ is a bijection from the set of divisors of $\mathbb{K}$ onto the set of closed ideals of $\mathcal{A}(\mathbb{K})$. Further, given a closed ideal $I$ of $\mathcal{A}(\mathbb{K})$, then $I = \mathcal{T}(\mathcal{D}(I))$.

Similarly, we have Theorem 27.10:

**Theorem 27.10.** Let $a \in \mathbb{K}$ and take $R > 0$. For every divisor $E$ of $d(a, R^{-})$, $\mathcal{T}_{a,R}(E)$ is a closed ideal of $\mathcal{A}(d(a, R^{-}))$. Moreover, $\mathcal{T}_{a,R}$ is a bijection from the set of divisors of $d(a, R^{-})$ onto the set of closed ideals of $\mathcal{A}(d(a, R^{-}))$. Further, given a closed ideal $I$ of $\mathcal{A}(d(a, R^{-}))$, then $I = \mathcal{T}_{a,R}(\mathcal{D}(I))$.

**Proof.** (Theorems 27.9 and 27.10) Let $E$ be a divisor of $\mathbb{K}$ (respectively of $d(a, R^{-}))$. First, let us check that $\mathcal{T}(E)$ (respectively $\mathcal{T}_{a,R}(E)$) is a closed ideal of $\mathcal{A}(\mathbb{K})$ (respectively of $\mathcal{A}(d(a, R^{-}))$. Let $E = (a_n, q_n)_{n \in \mathbb{N}}$ and let $(f_m)_{m \in \mathbb{N}}$ be a sequence of elements of $\mathcal{T}(E)$ (respectively of $\mathcal{T}_{a,R}(E)$) converging to a limit $f$ in $\mathcal{A}(\mathbb{K})$ (respectively in $\mathcal{A}(d(a, R^{-}))$. For every $n \in \mathbb{N}$, each $f_m$ admits $a_n$ as a zero of order at least $q_n$, hence by Lemma 16.1, so does $f$. Consequently, $f$ belongs to $\mathcal{T}(E)$ (respectively $f$ belongs to $\mathcal{T}_{a,R}(E)$).

Now, let us show that $\mathcal{T}$ (respectively $\mathcal{T}_{a,R}$) is injective. Let $E, F$ be two distinct divisors of $\mathbb{K}$ (respectively of $d(a, R^{-}))$. Without loss
of generality, we can suppose that $E$ admits a pair $(b, s)$ with $s > 0$ and that $F$ either does not admit any pair $(b, m)$ or admits a pair $(b, m)$ with $m < s$. Let $f \in \mathcal{T}(F)$ (respectively let $f \in \mathcal{T}_{a,R}(F)$) and suppose that $\omega_b(f) \ge s$. Then, by Lemma 16.1, $f$ factorizes in the form $(x-b)^{s-m}g$ with $g \in \mathcal{A}(\mathbb{K})$ (respectively $g \in \mathcal{A}(d(a, R^-))$) and of course $g$ belongs to $\mathcal{T}(F)$ (respectively to $\mathcal{T}_{a,R}(F)$). But by construction, $g$ does not belong to $\mathcal{T}(E)$ (respectively to $\mathcal{T}_{a,R}(E)$) because $\omega_b(g) < s$. Therefore, $\mathcal{T}(E) \ne \mathcal{T}(F)$ (respectively $\mathcal{T}_{a,R}(E) \ne \mathcal{T}_{a,R}(F)$). So, $\mathcal{T}$ (respectively $\mathcal{T}_{a,R}$) is injective.

Now, let us show that $\mathcal{T}$ (respectively $\mathcal{T}_{a,R}$) is injective. Let $E, F$ be two distinct divisors of $\mathbb{K}$ (respectively of $d(a, R^-)$). By Theorem 27.4, there exists $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^-))$) such that $\mathcal{D}(f) = F$, hence $\mathcal{D}(f) \ne E$. Therefore, $\mathcal{D}(f) \notin \mathcal{T}(E)$, and hence $\mathcal{T}(E) \ne \mathcal{T}(F)$. So, $\mathcal{T}$ is injective (respectively $\mathcal{T}_{a,R}(E) \ne \mathcal{T}_{a,R}(F)$). So, $\mathcal{T}_{a,R}$ is injective).

Let us show that it is also surjective. Let $I$ be a closed ideal of $\mathcal{A}(\mathbb{K})$ (respectively of $\mathcal{A}(d(a, R^-))$) and let $E = \mathcal{D}(I)$. Then $E$ is of the form $(a_n, q_n)_{n \in \mathbb{N}}$ with $|a_n| \le |a_{n+1}|$ and $\lim_{n \to +\infty} |a_n| = +\infty$ (respectively $\lim_{n \to +\infty} |a_n| = R$), hence there is a unique $s \in \mathbb{N}$ such that $a_n \in d(0, r) \ \forall n \le s$ and $a_n \notin d(0, r) \ \forall n > s$.

Let $J = \mathcal{T}(E)$ (respectively $J = \mathcal{T}_{a,R}(E)$). Then, of course, $I \subset J$. Let us show that $J \subset I$. Let $f \in J$ and take $r > 0$. Denoting by $P_r$ the polynomial $\prod_{i=0}^s (X - a_i)^{q_i}$ by Theorem 24.2, $I \cap H(d(0, r)) = P_r(x)H(d(0, r))$. But now all functions $g \in J \cap H(d(0, r))$ also are of the form $P_r(x)h(x)$ with $h \in H(d(0, r))$. Consequently, in $H(d(0, r))$, we can write $f$ in the form $f = \sum_{j=1}^m g_j h_j$ with $g_j \in I$ and $h_j \in H(d(0, r))$. Let $\epsilon > 0$ be fixed.

For each $j = 1, \dots, m$, narrowing each $h_j$ by a polynomial $\ell_j$ in $H(d(0, r))$ we can find $\ell_j \in \mathbb{K}[x]$ such that $|g_j(h_j - \ell_j)|(r) \le \epsilon$. Now, let $\phi_r = \sum_{j=1}^m g_j \ell_j$. Then $\phi_r$ belongs to $I$ and satisfies $|\phi_r - f|(r) \le \epsilon$. This is true for each $r > 0$ and for every $\epsilon > 0$. Consequently, since $I$ is closed, $f$ belongs to $I$. This finishes proving that $\mathcal{T}$ (respectively $\mathcal{T}_{a,R}$) is surjective. Further, we have proven that $I = \mathcal{T}(\mathcal{D}(I))$ (respectively $I = \mathcal{T}_{a,R}(\mathcal{D}(I)))$. $\square$

**Theorem 27.11.** Every closed ideal of $\mathcal{A}(\mathbb{K})$ is principal.

**Proof.** Indeed, consider a closed ideal $I$ and let $E = \mathcal{D}(I)$. By Theorem 27.9, $I$ is of the form $\mathcal{T}(E)$ with $E = \mathcal{D}(I)$. By Theorem 27.4, there exists $g \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(g) = E$ and of course, $g$
belongs to I. Hence, $g\mathcal{A}(\mathbb{K}) \subset I$. Now, let $f \in I$. Then, $\mathcal{D}(f) \ge E$, hence by Theorem 27.1, $f$ factorizes in the form $gh$ with $h \in \mathcal{A}(\mathbb{K})$, hence $I = g\mathcal{A}(\mathbb{K})$. $\square$

**Theorem 27.12.** Let $r \in |\mathbb{K}^*|$, let $f \in H(C(0,r))$, and let $P \in \mathbb{K}[x]$ have all its zeros in $d(0,r)$. There exists $g \in H(C(0,r))$ and $L \in \mathbb{K}[x]$ unique such that $f = Pg + L$, $\deg(L) < \deg(P)$, $\Psi(R, \log r) \le \Psi(f, \log r)$, $\Psi(g, \log r) \le \Psi(f, \log r) - \Psi(P, \log r)$. Moreover, if $f$ belongs to $H(d(0,r))$, then so does $g$.

**Proof.** Since $r \in |K|$, without loss of generality, we may assume that $r=1$. Similarly, we may also assume that $\Psi(P,0) = \Psi(f,0)=0$, so $P$ is quasi-monic. Thus, the problem now consists of finding $g \in H(C(0,1))$ and $L \in \mathbb{K}[x]$, each unique, satisfying the statements.

Let $f(x) = \sum_{m=-\infty}^{\infty} a_m x^m$, and for each $n \in \mathbb{N}$, let $f_n(x) = \sum_{m=-n}^{n} a_m x^m$. For each $n \in \mathbb{N}$, set $u_n = \sup_{|j|_{\infty}>n} |a_j|$. Then, $\lim_{n \to +\infty} a_n = \lim_{n \to -\infty} a_n = 0$. Next, we note that $x^n P$ is quasi-monic like $P$. By applying Lemma 5.2 to $x^n f_n$, we have a Euclidean division of $x^n f_n$ by $x^n P$ in the form $x^n f_n = x^n Pg_n + S_n$ with $S_n \in \mathbb{K}[x]$, $\deg(S_n) < n + \deg(P)$, $\Psi(S_n, 0) \le 0$, $\Psi(g_n, 0) \le 0$. Now, by construction, $S_n = x^n(f_n - Pg_n)$, hence $S_n$ is of the form $x^n L_n$, with $L_n \in \mathbb{K}[x]$, $\deg(L_n) < \deg(P)$, and $\Psi(L_n, 0) = \Psi(S_n, 0) \le 0$. So, $f_n = Pg_n + L_n$.

Consequently, $f_{n+1} - f_n = P(g_{n+1} - g_n) + L_{n+1} - L_n$. By applying again Lemma 5.2 to $x^{n+1}(f_{n+1} - f_n)$, we can check that $\Psi(g_{n+1} - g_n, 0) \le \log(u_n)$ and $\Psi(L_{n+1} - L_n, 0) \le \log(u_n)$, hence both sequences $(g_n)_{n \in \mathbb{N}}$, $(L_n)_{n \in \mathbb{N}}$ converge in $H(C(0,1))$, and more precisely, the sequence $(L_n)$ converges to a polynomial $L$ of degree $< \deg(P)$. Moreover, setting $g = \lim_{n \to \infty} g_n$, clearly we have $f = Pg + L$ and $\Psi(g, 0) \le 0$, $\Psi(L, 0) \le 0$, which shows the existence of $g$ and $L$ in the first claim.

Now, let us check that they are unique satisfying these relations. Suppose we have $h \in H(C(0,1))$ and $S \in \mathbb{K}[x]$ satisfying the same properties, with particularly $f = Ph + S$, $\deg(S) < \deg(P)$. Then, $P(g - h) = S - L$. Since $\deg(S - L) < \deg(P)$, $S - L$ is an element of $H(C(0,1))$ having strictly less zeros than $P$ in $C(0,1)$, a contradiction, except if $g = h$, hence $L = S$.

Now, assume that $f$ lies in $H(d(0,1))$. Then, by Corollary 21.9, $f - L$ is a quasi-invertible element of $H(d(0,1))$ hence is of the form $Q\phi$ with $\phi$ invertible in $H(d(0,1))$ and $Q \in \mathbb{K}[x]$, having all its zeros
in $d(0, 1)$. Hence, $Pg = Q\phi$. This holds in $H(C(0, 1))$. But since $P$
has all its zeros in $C(0, 1)$, $P$ must divide $Q$: say $Q = PV$, with
$V \in \mathbb{K}[x]$ having all its zeros in $d(0, 1)$. So, $Pg = V\phi$, hence $g = V\phi$,
and by hypothesis, both $V$, $\phi$ lie in $H(d(0, 1))$, hence so does $g$, which
completes the proof. $\square$

**Definition.** Given $r \in |\mathbb{K}^*|$, the division of an element $f$ of $H(C(0,r))$ by a polynomial $P$ having all its zeros in $d(0,r)$, as defined in Theorem 27.12, is called *Euclidean division of $f$ by $P$* or *r-Euclidean division of $f$ by $P$*.

**Lemma 27.13.** Let $(u_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{R}_+$ such that $\sum_{n=0}^{\infty} u_n < +\infty$. Let $A = \sum_{n=0}^{\infty} u_n$ and let $B > A$. There exists an increasing sequence $(q_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to \infty} q_n = +\infty$ and such that $\sum_{n=0}^{\infty} q_n u_n \le B$.

**Proof.** Let $E = B - A$. For every $n \in \mathbb{N}$, we denote by $s_n$ the small-
est integer such that $\sum_{j=s_n}^{\infty} u_j \le 4^{-n}E$. Then, for every $j \in \mathbb{N}$ such
that $s_n \le j < s_{n+1}$, we set $q_j = 2^n$. We have $\sum_{j=s_n}^{s_{n+1}-1} q_j u_j \le 2^{-n}E$,
hence $\sum_{j=0}^{s_{n+1}-1} q_j u_j \le A + E \sum_{k=1}^{n} 2^{-k}$, and finally, $\sum_{j=0}^{\infty} q_j u_j \le B$. $\square$

**Theorem 27.14.** Let $T = (a_n, q_n)_{n \in \mathbb{N}}$ be a divisor on the disk $d(a, R^{-})$ with $a_n \neq 0 \forall n \in \mathbb{N}$ and let $\epsilon > 0$. There exists $f \in \mathcal{A}(d(a, R^{-}))$ such that $\mathcal{D}(f) \ge T$, $f(a) = 1$ and $|f|(r) \le |T|(r)(1+\epsilon) \forall r \in [0, R[$.

**Proof.** Without loss of generality, we can assume $a = 0$. The set $\{|a_j| \mid j \in \mathbb{N}\}$ is obviously equal to the image of a strictly increasing sequence of limit $R$ that we denote by $(r_m)_{m \in \mathbb{N}}$. For each $m \in \mathbb{N}$, the set $S_m$ of the $a_j$ lying in $C(0, r_m)$ is of the form $\{a_{h_m}, a_{h_{m+1}}, \dots, a_{k_m}\}$. We set $B = 1 + \epsilon$ and $P_m = \prod_{j=h_m}^{k_m} \left(1 - \frac{x}{a_j}\right)^{q_j}$.

We can construct a polynomial whose divisor is $(a_n, q_n)_{n<t}$. For
every $m \in \mathbb{N}$, we set $\mu_m = \log r_m$, and $\lambda = \log(B)$.

Now, by Lemma 27.13, there exists an increasing sequence $(t_n)_{n \in \mathbb{N}}$
in $\mathbb{N}$ such that $\lim_{n \to \infty} t_n = +\infty$ and such that

$$
(1) \quad \sum_{j=0}^{\infty} t_j (\mu_{j+1} - \mu_j) \le \lambda.
$$
For every $s \in \mathbb{N}$, we put $\tau(s) = \sum_{j=0}^{s} t_j(\mu_{j+1} - \mu_j)$, and $g_s = \prod_{m=0}^{s} P_m$. We note that $\Psi(g_q, \mu) = \Psi(g_s, \mu)$ whenever $\mu \le \mu_q$ and $q \le s$. So, we can define the function $\ell$ from $]-\infty, \log S[$ into $\mathbb{R}$ as $\ell(\mu) = \lim_{s \to \infty} \Psi(g_s, \mu)$. Then $\ell$ is an increasing function in $\mu$ that satisfies

$$ (2) \quad \ell(\mu) = \sum_{j=0}^{k(m)} q_j(\mu - \Psi(a_j)) \quad \text{whenever } \mu \in [\mu_m, \mu_{m+1}]. $$

We construct a sequence $(f_s)_{s \in \mathbb{N}}$ in $K[x]$ satisfying the following relations $(\alpha_s), (\beta_s), (\gamma_s)$, and $(\delta_s)$ for every $s \in \mathbb{N}$ and $(\epsilon_s), (\varphi_s)$ for every $s \in \mathbb{N}^*$:

$$ (\alpha_s) \quad f_s(0) = 1. $$

$$ (\beta_s) \quad P_j \text{ divides } f_s \text{ for every } j \le s. $$

$$ (\gamma_s) \quad \Psi(f_s, \mu_{s+1}) \le \ell(\mu_{s+1}) + \tau(s). $$

$$ (\delta_s) \quad \Psi(f_s, \mu) \le \ell(\mu) + \lambda \quad \text{whenever } \mu < \log S. $$

$$ (\epsilon_s) \quad \Psi(f_s - f_{s-1}, \mu) \le \ell(\mu) + t_s(\mu - \mu_s) + \lambda \quad \text{whenever } \mu \le \mu_s. $$

$$ (\varphi_s) \quad \Psi(f_s - f_{s-1}, \mu) \le \ell(\mu_s) + t_s(\mu - \mu_s) + \tau(s) \quad \text{whenever } \mu \in [\mu_s, \log S]. $$

We proceed by induction and prove that when $(\alpha_s), (\beta_s), (\gamma_s)$, and $(\delta_s)$ are satisfied for $s \in \mathbb{N}$, then we can derive $(\alpha_{s+1}), (\beta_{s+1}), (\gamma_{s+1}), (\delta_{s+1}), (\epsilon_{s+1})$, and $(\varphi_{s+1})$. By taking $f_0 = P_0$, we check that $(\alpha_0), (\beta_0), (\gamma_0)$, and $(\delta_0)$ are obviously satisfied. We now suppose already constructed $f_m$ satisfying $(\alpha_m), (\beta_m), (\gamma_m)$, and $(\delta_m)$ for every $m = 0, \dots, s$ and $(\epsilon_m), (\varphi_m)$ for every $m = 1, \dots, s$. We define $f_{s+1}$ satisfying $(\alpha_{s+1}), (\beta_{s+1}), (\gamma_{s+1}), (\delta_{s+1}), (\epsilon_{s+1})$, and $(\varphi_{s+1})$. It is seen that each polynomial $P_s$ has all its zeros in $C(0, r_s)$.

Let $R_{s+1}$ be the rest of the Euclidean division of $f_s$ by $P_{s+1}$ in $H(C(0, r_{s+1}))$. Let $Q_{s+1} = x^{t_{s+1}} g_s$. We have

$$ (3) \quad \Psi(Q_{s+1}, \mu) = \mu t_{s+1} + \Psi(g_s, \mu) \quad \text{whenever } \mu \in \mathbb{R}. $$

We note that $Q_{s+1}$ admits no zero in $C(0, r_{s+1})$ and then is invertible in $H(C(0, r_{s+1}))$. As a consequence, according to Theorem 27.12, we can perform the $r_{s+1}$-Euclidean division of $R_{s+1}/Q_{s+1}$ by $P_{s+1}$ in $H(C(0, r_{s+1}))$. Let $V_{s+1}$ be the rest of this division. Thus, $R_{s+1}/Q_{s+1}$ is of the form $T_{s+1}P_{s+1} + V_{s+1}$ with $T_{s+1} \in H(C(0, r_{s+1}))$ and

$$ (4) \quad \Psi(Q_{s+1}V_{s+1}, \mu_{s+1}) \le \Psi(R_{s+1}, \mu_{s+1}), $$
we have $R_{s+1} = Q_{s+1}(T_{s+1}P_{s+1} + V_{s+1})$. Now we put $f_{s+1} = f_s - Q_{s+1}V_{s+1}$. Of course, $f_{s+1}$ satisfies $(\alpha_{s+1})$. We check that $f_{s+1}$ satisfies $(\beta_{s+1})$. By definition, each $P_j$ divides $g_s$ for every $j = 0, \dots, s$, and hence it divides $Q_{s+1}$. Next, by $(\beta_s)$, $P_j$ also divides $f_s$. Consequently, $P_j$ divides $f_{s+1}$ for every $j = 0, \dots, s$. Moreover, $P_{s+1}$ divides both $f_s - R_{s+1}$ and $Q_{s+1}T_{s+1}P_{s+1}$. Hence it also divides $f_{s+1}$ and thereby $(\beta_{s+1})$ is satisfied.

Now, we prove $(\varphi_{s+1})$. By (4), $R_{s+1}$ satisfies $\Psi(R_{s+1}, \mu_{s+1}) \le \Psi(f_s, \mu_{s+1})$. Hence, by Relation $(\gamma_s)$, we have

$$ (5) \quad \Psi(R_{s+1}, \mu_{s+1}) \le \ell(\mu_{s+1}) + \tau(s) = \Psi(g_{s+1}, \mu_{s+1}) + \tau(s). $$

Since $\deg(R_{s+1}) < \deg(P_{s+1}) < \deg(g_{s+1})$ and since all zeros of $g_{s+1}$ lie in $d(0, r_{s+1})$, $g_{s+1}$ has more zeros than $R_{s+1}$ in $d(0, r_{s+1})$, and therefore, by Theorem 22.26, we have

$$ \Psi(R_{s+1}, \mu) - \Psi(R_{s+1}, \mu_{s+1}) \le \Psi(g_{s+1}, \mu) - \Psi(g_{s+1}, \mu_{s+1}) $$

whenever $\mu \in ]\mu_{s+1}, \log S]$, and therefore, by (5), we obtain

$$ (6) \quad \Psi(R_{s+1}, \mu) \le \Psi(g_{s+1}, \mu) + \tau(s) \quad \text{whenever } \mu \in ]\mu_{s+1}, \log S]. $$

Since $\frac{R_{s+1}}{Q_{s+1}} = T_{s+1}P_{s+1} + V_{s+1}$, by Theorem 27.12, we have $\Psi(Q_{s+1}V_{s+1}, \mu_{s+1})) \le \Psi(R_{s+1}, \mu_{s+1})$, and hence, by (5), $\Psi(Q_{s+1}V_{s+1}, \mu_{s+1}) \le \ell(\mu_{s+1}) + \tau(s)$. But $\ell(\mu_{s+1}) = \Psi(g_{s+1}, \mu_{s+1})$, hence by (6), we obtain

$$ (7) \quad \Psi(Q_{s+1}V_{s+1}, \mu_{s+1}) \le \Psi(g_{s+1}, \mu_{s+1}) + \tau(s). $$

We note that $\deg(Q_{s+1}V_{s+1}) < \deg(g_{s+1}) + t_{s+1}$ and that all zeros of $g_{s+1}$ lie in $d(0, r_{s+1})$. Hence, by Theorem 22.26 and by (3) and (7), we have

$$ (8) \quad \Psi(Q_{s+1}V_{s+1}, \mu) \le \Psi(g_{s+1}, \mu) + t_{s+1}(\mu - \mu_{s+1}) + \tau(s) \quad \text{for every } \mu \in ]\mu_{s+1}, \log S]. $$

Actually, by definition of $f_{s+1}$, we have $\Psi(Q_{s+1}V_{s+1}, \mu) = \Psi(f_{s+1} - f_s, \mu)$ and $\Psi(g_{s+1}, \mu) \le \ell(\mu)$ for every $\mu \in ]\mu_{s+1}, \log S]$, hence by (7), we have proved $\varphi_{s+1}$. We deduce $(\epsilon_{s+1})$.
In particular, when $\mu = \mu_{s+2}$, we obtain

$$ \Psi(Q_{s+1}V_{s+1}, \mu_{s+2}) \le \ell(\mu_{s+2}) + t_{s+1}(\mu_{s+2} - \mu_{s+1}) + \tau(s). $$

But we note that $t_{s+1}(\mu_{s+2} - \mu_{s+1}) + \tau(s) = \tau(s+1)$, hence

$$ (9) \quad \Psi(Q_{s+1}V_{s+1}, \mu_{s+2}) \le \ell(\mu_{s+2}) + \tau(s+1). $$

And, by (8) and (1), we obtain

$$ (10) \quad \Psi(f_{s+1} - f_s, \mu) \le \ell(\mu) + \lambda \quad \text{whenever } \mu \in [\mu_{s+1}, \log S]. $$

Now we take $\mu \le \mu_{s+1}$. It is seen that

$$ \Psi(Q_{s+1}, \mu) - \Psi(Q_{s+1}, \mu_{s+1}) = \ell(\mu) - \ell(\mu_{s+1}) + t_{s+1}(\mu - \mu_{s+1}). $$

Therefore, we have

$$ \Psi(Q_{s+1}V_{s+1}, \mu) \le \Psi(Q_{s+1}V_{s+1}, \mu_{s+1}) + \ell(\mu) - \ell(\mu_{s+1}) + t_{s+1}(\mu - \mu_{s+1}). $$

But by (8), we have $\Psi(Q_{s+1}V_{s+1}, \mu_{s+1}) \le \ell(\mu_{s+1}) + \lambda$, hence we obtain $\Psi(Q_{s+1}V_{s+1}, \mu) \le \ell(\mu) + \lambda + t_{s+1}(\mu - \mu_{s+1})$ whenever $\mu \le \mu_{s+1}$, and this is $(\epsilon_{s+1})$. In particular, we have $\Psi(f_{s+1} - f_s, \mu) \le \ell(\mu) + \lambda$ whenever $\mu \le \mu_{s+1}$, and therefore, by $(\delta_s)$, we obtain $(\delta_{s+1})$.

Now, we show $(\gamma_{s+1})$. Obviously, we have

$$ (10) \quad \Psi(f_0, \mu_{s+2}) \le \ell(\mu_{s+2}). $$

Next, by Relations $(\varphi_m)_{1 \le m \le s+1}$ for every $m \in \mathbb{N}^*$, we have

$$ \Psi(f_m - f_{m-1}, \mu_{s+2}) \le \ell(\mu_m) + [t_m(\mu_{s+2} - \mu_m) + \tau(m)]. $$

But as the sequence $(t_m)_{m \in \mathbb{N}}$ is increasing, it is seen that

$$ \tau(m) + t_m(\mu_{s+2} - \mu_m) \le \sum_{j=0}^{s+1} t_j(\mu_{j+1} - \mu_j) = \tau(s+1). $$

Obviously, $\ell(\mu_{s+2}) \ge \ell(\mu_m)$, hence we obtain $\Psi(f_m - f_{m-1}, \mu_{s+2}) \le \ell(\mu_{s+2}) + \tau(s+1)$ whenever $m = 1, \dots, s+1$. Finally, by (10), $f_{s+1}$ satisfies $\Psi(f_{s+1}, \mu_{s+2}) \le \ell(\mu_{s+2}) + \tau(s+1)$ and this is $(\gamma_{s+1})$.
We note that $(\epsilon_s)$ and $(\varphi_s)$ are not used to prove

$$ (\alpha_{s+1}), (\beta_{s+1}), (\gamma_{s+1}), (\delta_{s+1}), (\epsilon_{s+1}), (\varphi_{s+1}). $$

Consequently, $(\epsilon_1)$ and $(\varphi_1)$ are clearly proven by $(\alpha_0)$, $(\beta_0)$, $(\gamma_0)$, and $(\delta_0)$, and therefore, we are now done with the recurrence. Therefore, we can now construct the sequence $(f_s)_{s \in \mathbb{N}}$ satisfying $(\alpha_s)$, $(\beta_s)$, $(\gamma_s)$, $(\delta_s)$, $(\epsilon_s)$, and $(\varphi_s)$. By Relations $(\epsilon_s)$, the sequence is easily seen to converge in each algebra $H(d(0, u))$ whenever $u \in ]0, S[$. Indeed, given $u \in ]0, S[$ and $N \in \mathbb{N}$ such that $\mu_N < \log u$, by $(\epsilon_{s+1})$, we have

$$ \log(\|f_{s+1} - f_s\|_{d(0,u)}) = \Psi(f_{s+1} - f_s, \log u) \\ \le \ell(\log u) - t_s(\log u - \mu_{s+1}) + \lambda, $$

hence $\log(\|f_{s+1} - f_s\|_{d(0,u)}) \le \ell(\log u) - t_s(\log u - \mu_N) + \lambda$, whenever $s > N$. As $\lim_{s \to +\infty} t_s = +\infty$, it is seen that $\lim_{s \to +\infty} \|f_{s+1} - f_s\|_{d(0,u)} = 0$.

Let $f$ be the function defined in $d(0, R^{-})$ as the limit of the sequence $(f_s)_{s \in \mathbb{N}}$ in each disk $d(0, u)$. Obviously, as an element of $H(d(0, u))$ for every $u \in ]0, S[$, $f$ belongs to $\mathcal{A}(d(0, R^{-}))$. By Relations $(\alpha_s)$, $f$ satisfies $f(0) = 1$.

We check $|f|(r) \le B|T|(r)$. Let $u \in ]0, R[$ be such that $\mu_N \le \log u \le \mu_{N+1}$. We have $\log \|f\|_{d(0,u)} = \Psi(f, \log u)$.

When $s$ is big enough, $\Psi(f_s, \log u)$ is clearly equal to $\Psi(f, \log u)$, hence $f$ satisfies $\log \|f\|_{d(0,u)} = \Psi(f_s, \log u) \le \ell(\log u) + \lambda$. Hence, by (2), we obtain $|f|(r) \le B|T|(r)$. Now, we just have to check that every $a_j$ is a zero of $f$ of order $z_j \ge q_j$. Let $m$ be such that $h_m \ge j$. For every $s \ge m$, $(1 - \frac{x}{a_j})^{q_j}$ divides $f_s$ in $H(d(0, u))$ (for every $u \in ]0, S[$), hence by Lemma 16.1, $(1 - \frac{x}{a_j})^{q_j}$ divides $f$ in $H(d(0, u))$ and this finishes the proof of Theorem 27.14. $\square$

We can obtain a small improvement of Theorem 27.14:

**Theorem 27.15.** Let $T = (a_n, q_n)_{n \in \mathbb{N}}$ be a divisor on the disk $d(a, R^{-})$ with $a_n \ne 0 \forall n \in \mathbb{N}$, and let $\epsilon > 0$ and $\rho \in ]0, R[$. There exists $g \in \mathcal{A}(d(a, R^{-}))$ such that $\mathcal{D}(g) \ge T$, $g(a) = 1$ and $|g|(r) \le |T|(r)(1 + \epsilon) \forall r \in ]0, R[$ and $\mathcal{D}(g)(\alpha) = T(\alpha) \forall \alpha \in d(a, \rho)$.
**Proof.** By Theorem 27.14, we have a function $f \in \mathcal{A}(d(a, R^{-}))$ such that $\mathcal{D}(f) \ge T$, $f(a) = 1$ and $|f|(r) \le |T|(r)(1 + \epsilon) \quad \forall r \in ]0, R[$. Now, we can construct a polynomial $P(x)$ such that $P(0) = 1$ and admitting in $d(a, R^{-})$ a divisor $\mathcal{D}(P)$ satisfying $\mathcal{D}(P)(\alpha) = \frac{\mathcal{D}(f)(\alpha)}{T(\alpha)} \quad \forall \alpha \in d(a, \rho)$ and $\mathcal{D}(P)(\alpha) = 0 \quad \forall \alpha \in d(a, R^{-}) \setminus d(a, \rho)$. Then the function $g = \frac{f}{P}$ satisfies $\mathcal{D}(g)(\alpha) = T(\alpha) \quad \forall \alpha \in d(a, \rho)$, $T \le \mathcal{D}(g) \le \mathcal{D}(f)$, and hence $|g|(r) \le |f|(r) \le |T|(r)(1 + \epsilon) \quad \forall r \in ]0, R[$. $\square$

**Remark.** Here we may note that $H(d(0, R^{-}))$ is much smaller than $\mathcal{A}_b(d(0, R^{-}))$. Indeed, by Theorem 27.15, there exist functions $f \in \mathcal{A}_b(d(0, R^{-}))$ having infinitely many zeros in $d(0, R^{-})$. But, by Theorem 21.8, any element of $H(d(0, R^{-}))$ is quasi-invertible and hence has finitely many zeros.
# Chapter 28

## Michel Lazard's Problem

This chapter is aimed at studying the following problem mentioned in Chapter 26 and first considered by M. Lazard in a tremendous work [72]. Let $T$ be a divisor on a disk $d(a, R^{-})$. Does there exist a function $f \in \mathcal{A}(d(a, R^{-}))$ such that $\mathcal{D}(f) = T$? The answer depends on whether or not $\mathbb{K}$ is spherically complete.

Theorems 28.1 and 28.4 were first proved in [72]. Proofs are long and much technical. Here we try to give an easier presentation of the proofs which is due to Labib Haddad. More precisely, when $\mathbb{K}$ is spherically complete, Theorem 28.4 shows the following result, as it was done in [72]: let $T = (a_n, q_n)$ be a divisor on $d(a, R^{-})$, with $|a_n| \le |a_{n+1}|$ and $|a_{u(m)}| < |a_{u(m)+1}|$. For each $m \in \mathbb{N}$, let $P_m(x) = \prod_{j=u(m)+1}^{u(m+1)} (x - a_j)$ and let $(Q_m) \in \mathbb{K}[x]$ be such that $|Q_m|(\rho_m) \le |T|(\rho_m)$. Then there exists a function $f \in \mathcal{A}(d(a, R^{-}))$ such that $P_m$ divides $f - Q_m$. Hence, in particular, given a divisor $T$ on $d(a, R^{-})$, there exists a function $f$ analytic in $d(a, R^{-})$, whose divisor is $T$.

**Theorem 28.1.** Let $\mathbb{K}$ be not spherically complete and let $(D_n)_{n \in \mathbb{N}}$ be a decreasing sequence of disks $d(u_n, \rho_n)$ such that $\bigcap_{n=0}^{\infty} D_n = \emptyset$. Let $R = \frac{1}{\lim_{n \to \infty} \text{diam}(D_n)}$. There exists sequences $(c_n)_{n \in \mathbb{N}}$ of $d(0, R^{-})$ such that $\lim_{n \to \infty} |c_n| = R$ and such that no function $f \in \mathcal{A}(d(0, R^{-}))$ admits for divisor the divisor $T = (c_n, 1)_{n \in \mathbb{N}}$.

**Proof.** Without loss of generality, we may obviously assume that $R > 1$, hence $\frac{1}{R} < 1 < R$. Consequently, we may assume that $D_0 \subset d(0, R^{-})$. For each $n \in \mathbb{N}$, we set $\rho_n = \text{diam}(D_n)$, hence $\rho_n > \rho_{n+1}$.
For every $n \in \mathbb{N}$, we can take $\alpha_n \in D_n \setminus D_{n+1}$. Let $\beta_n = \alpha_{n+1} - \alpha_n$, $n \in \mathbb{N}$, hence

$$ (1) \quad \rho_{n+1} < |\beta_n| \le \rho_n. $$

Consider the divisor $T = (\frac{1}{\beta_n}, 1)_{n \in \mathbb{N}}$ and suppose that there exists $f \in \mathcal{A}(d(0, R^{-}))$ whose divisor is exactly $T$. Without loss of generality, we can assume that $f(x) = \prod_{n \in \mathbb{N}} (1 - \beta_n x)$. Then $f(x)$ is a series of the form $1 + \sum_{n=1}^{\infty} a_n x^n$. We show that $\alpha_1 - a_1 \in D_n \forall n \ge 1$.

Let us fix $n \in \mathbb{N}$. We can check that $\alpha_n - \alpha_1 = \sum_{j=1}^{n-1} \beta_j$, hence $\alpha_n - (\alpha_1 - a_1) = a_1 + \sum_{j=1}^{n-1} \beta_j$. Since $\alpha_n \in D_n$, then $\alpha_1 - a_1$ lies in $D_n$ if and only if $|a_1 + \sum_{j=1}^{n-1} \beta_j| \le \rho_n$.

For all $n \in \mathbb{N}$, we set $t_n = |(\beta_n)^{-1}|$. By (1), the sequence $(t_n)$ is strictly increasing and satisfies

$$ (2) \quad |a_n| = \prod_{j=1}^{n} t_j \quad \forall n \ge 1. $$

Particularly, $a_n \ne 0 \forall n \ge 1$. For each $s \ge 1$, we put $f_s = \sum_{n=0}^{s} a_n x^n$. Then, $\deg(f_s) = s$. By (2), we see that the $s$ zeros of $f_s$ are distinct and are of the form $\gamma_j^s$, $1 \le j \le s$, with $|\gamma_j^s| = t_j$. Thus, $f_s$ is of the form $\prod_{j=1}^{s} (1 - \beta_{s,j} x)$, with $|\beta_{s,j}| = |\beta_j| = t_j^{-1}$. By identification of coefficients, we obtain $a_1 = - \sum_{j=1}^{s} \beta_{s,j}$. Consequently, when $n \le s$, we have

$$ (3) \quad \sum_{j=1}^{s} \beta_j - \beta_{s,j} = a_1 + \sum_{j=1}^{s} \beta_j. $$

Let us fix $n \in \mathbb{N}^*$. By Theorem 24.3, we can see that for each $j = 1, \dots, n$, we have $\lim_{s \to \infty} \beta_{s,j} = \beta_j$. Consequently, when $s$ is big enough, we can see that $|\sum_{j=1}^{n-1} \beta_j - \sum_{j=1}^{s} \beta_{s,j}| \le \rho_n$. Therefore, by (3), we have $|a_1 + \sum_{j=1}^{n-1} \beta_j| \le \rho_n$.

Consequently, $\alpha_1 - a_1$ lies in $D_n$. This is true for every $n \in \mathbb{N}$, a contradiction to the hypothesis $\bigcap_{n=0}^{\infty} D_n = \emptyset$. $\square$

In order to prove Theorem 28.4, we must introduce a set of notations that hold throughout this chapter.