Consider now the sum $\sum_{m=0}^{q} \frac{P^{(m)}(b_{q-1})}{m!} (h_q)^m$. Since the sequence $|a_m|_p$ is strictly decreasing, we have $|h_q|_p = |a_q|_p$, hence $\log_p(|h_q|_p) = -r_q$. We note that $\mathbb{Q}_p[a_1, \dots, a_{q_1}]$ is an algebraic extension of $\mathbb{Q}_p$ of degree at most $\prod_{j=1}^{q-1} v_j$. Consequently, by Corollary 6.8, the extension $\mathbb{Q}_p[a_1, \dots, a_{q_1}]$ has a value group of the form $\frac{s}{t}\mathbb{Z}$ with $t \le \prod_{j=1}^{q-1} v_j$. On the other hand, due to the hypothesis $r_q = \frac{u_q}{v_q}$, it appears that $v_q$ is a prime integer, prime to $u_q$ and bigger than $q$ and $\prod_{j=1}^{q-1} v_j$. Consequently, we can apply Lemma 6.19 with $h_q$ in the role of $c$ and $b_{q-1}$ in the role of $a$. Therefore, for each $m = 0, \dots, q-1$, all the $\left| \frac{P^{(m)}(b_{q-1})}{m!} (h_q)^m \right|_p$ are pairwise distinct. Consequently, we have

$$ (5) \left| \sum_{m=0}^{q} \frac{P^{(m)}(b_{q-1})}{m!} (h_q)^m \right|_p = \max_{1 \le m \le q} \left| \frac{P^{(m)}(b_{q-1})}{m!} (h_q)^m \right|_p. $$

Next, since $-\Psi_p(h_q) = r_q < (q+1)^\omega$, for each integer $m = 1, \dots, q$, we have $\Psi_p(h_q)^m = -mr_q > -(m(q+1)^\omega) \ge -q(q+1)^\omega$, hence

$$ (6) \Psi_p((h_q)^m) \ge -q(q+1)^\omega > -(q+1)^{1+\omega} \quad \forall m \le q. $$

Consequently, by (4), (5), and (6), the polynomial $Q(X) = \sum_{m=0}^{q} \frac{P^{(m)}(b_{q-1})}{m!} (X)^m$ has all coefficients in $d(0, 1^{-})$ and hence we have $\|Q\| < 1$. But since $|b_{q-1}|_p < 1$, by Theorem 4.5, we have $\|P\| = \|Q\| < 1$, a contradiction to the hypothesis $\|P\| = 1$. Therefore, Relation (3) is proven for every polynomial $P \in \mathbb{Q}_p[X]$ of degree $q \ge t(\omega)$ such that $\|P\| = 1$. Consequently, by (3), we obviously have a constant $C > 0$, not depending on $b$, such that $-\Psi_p(P(b)) \le C(\deg(P))^{1+\omega}$ for every $P \in \mathbb{Q}_p[X]$ such that $\deg(P) \ge t(\omega)$ and $\|P\| = 1$.

Particularly, $b$ is transcendental over $\mathbb{Q}_p$ because if it were algebraic, the degrees of polynomial $P \in \mathbb{Q}_p[X]$ such that $P(b) = 0$ wouldn't be bounded. Finally, it is easily seen that there exists a constant $m > 0$ such that $|Q(b)|_p \ge m$ for every polynomial $Q \in \mathbb{Q}_p[X]$ of degree $q \le t(\omega)$ and $\|Q\| = 1$. Therefore, $b$ is clearly of order $\le 1+\omega$. $\square$

**Corollary 9.3.** $\Omega_p$ is not complete.

The transcendence type is defined in $\mathbb{C}_p$ in the same way as in $\mathbb{C}$ [41].
**Definition and notation.** Given a complex number $z$, we denote by $|z|_{\infty}$ its modulus. Throughout this chapter, a number $a \in \mathbb{C}_p$ is just said to be *algebraic* (respectively *transcendental*) if it is algebraic (respectively transcendental) over $\mathbb{Q}$. When $a$ is algebraic or transcendental over $\mathbb{Q}_p$, we precise this. Throughout this chapter, we denote by $\Omega$ the field of algebraic numbers and by $\mathbf{A}$ the ring of algebraic integers.

Let $a \in \Omega$. We call *denominator of $a$* any strictly positive integer $n$ such that $na$ and we denote by $\text{den}(a)$ the smallest denominator of $a$. Let $a_2, \dots, a_n$ be the conjugates of $a$ over $\mathbb{Q}$ in $\mathbb{C}$ and put $a_1 = a$. For convenience, we use the logarithm of base $p$ denoted by $\log_p$. We set $|\overline{a}| = \max_{j=1,\dots,n}(|a_j|_{\infty})$ and $s(a) = \max(\log_p(|\overline{a}|, \log(\text{den}(a)))$.

The following relations are classical and immediate:

**Lemma 9.4.** Let $a, b \in \Omega$ and let $m \in \mathbb{N}$. Then, $\text{den}(ab) \le \text{den}(a)$
$\text{den}(b)$, $\text{den}(a+b) \le \text{den}(a)\text{den}(b)$, $\text{den}(ma) \le m\text{den}(a)$, $\text{den}(a^m) \le$
$(\text{den}(a))^m$ and $|\overline{ab}| \le |\overline{a}|\cdot|\overline{b}|$, $|\overline{a}+b| \le |\overline{a}|+|\overline{b}|$, $|\overline{ma}| \le m|\overline{a}|$,
$|\overline{a}^m| = (|\overline{a}|)^m$.

**Definition and notation.** Let

$$P(X_1, \dots, X_n) = \sum_{i_1, \dots, i_n} a_{i_1, \dots, i_n} (X_1)^{i_1} \dots (X_n)^{i_n} \in \mathbb{C}[X_1, \dots, X_n].$$

We put $H(P) = \max_{i_1, \dots, i_n} |a_{i_1, \dots, i_n}|_{\infty}$ and $t(P) = \max(\log_p(H(P)),$
$\deg(P) + 1)$.

A number $a \in \mathbb{C}_p$ is said to have transcendence type less than $\alpha$ if there exists a constant $C_a > 0$ such that, for every $Q \in \mathbb{Z}[X]$, we have either $Q(a) = 0$ or $-\Psi_p(Q(a)) \le C_a(t(Q))^{\alpha}$. We denote by $\mathcal{T}(\alpha)$ the set of numbers $a \in \mathbb{C}_p$ having a transcendence type less than or equal to $\alpha$.

If a number $a \in \mathbb{C}_p$ does not belong to $\mathcal{T}(\alpha)$ for all $\alpha > 0$, we say
that $a$ is of infinite type.

By Lemma 6.17, Lemmas 9.5, 9.6, and 9.7 are immediate:

**Lemma 9.5.** Let $P \in \mathbb{Z}[X]$. Then, $-\Psi_p(P, 0) \le \log_p(H(P))$.

**Lemma 9.6.** Let $P \in \mathbb{Z}[X]$ be of degree $k$ and let $a \in \Omega$. Then,

$$|P(a)| \le H(P)(k+1)(\max(|a|, 1))^k.$$
**Lemma 9.7.** Let $\alpha_1, \alpha_2 \in ]0, +\infty[$ satisfy $\alpha_1 \le \alpha_2$. Then, $\mathcal{T}(\alpha_1) \subset \mathcal{T}(\alpha_2)$.

There exists a link between transcendence order over $\mathbb{Q}_p$ and transcendence type over $\mathbb{Q}$ [41], [50].

**Theorem 9.8.** Let $\alpha \in [1, +\infty[$. Then, $S(\alpha) \subset \mathcal{T}(\alpha)$.

**Proof.** Let $a \in S(\alpha)$. By hypothesis, there exists $C > 0$ such that

$$-\Psi_p(Q(a)) \le -\Psi_p(Q, 0) + C(\deg(Q))^{\alpha} \quad \forall Q \in \mathbb{Q}_p[X].$$

Hence, by Lemma 9.5, we have $-\Psi_p(Q(a)) \le \log_p(H(Q)) + C(\deg(Q))^{\alpha} \quad \forall Q \in \mathbb{Q}[X]$. Then, taking $C \ge 1$, we can derive

$$-\Psi_p(Q(a)) \le C(\log_p(H(Q))+(\deg(Q))^{\alpha}) \le 2C(t(Q))^{\alpha} \quad \forall Q \in \mathbb{Q}[X],$$

which proves that $a \in \mathcal{T}(\alpha)$. $\square$

<figure><img src="image_4.png" /></figure>

By Theorem 9.2, we can now state the following corollary [41], [49]:

**Corollary 9.9.** There exists $b \in \mathbb{C}_p$, transcendental over $\mathbb{Q}_p$, such that $b \in \mathcal{T}(1 + \epsilon)$ for every $\epsilon > 0$.

**Proof.** Indeed, in Theorem 9.2, we saw that there exists $b \in \mathbb{C}_p$ that belongs to $S(1 + \epsilon)$ for all $\epsilon > 0$. $\square$

<figure><img src="image_5.png" /></figure>

By Lemma 6.17, we can immediately derive the following inequality:

**Theorem 9.10.** Let $a \in \Omega^*$ be integral of degree $q$, over $\mathbb{Z}$. Then, $|a|_p \ge \frac{1}{(|a|)^q}$.

**Proof.** Let $Q(X) = irr(a, \mathbb{Q})$ and let $a_1, \dots, a_q$ be the conjugates of $a$ over $\mathbb{Z}$, with $a_1 = a$. Then, $\prod_{j=1}^q a_j$ belongs to $\mathbb{Z}^*$, hence, by
Lemma 6.18, we have $|a_j| \le 1 \forall j = 2, \dots, q$:

$$ (1) \qquad \left| \prod_{j=1}^{q} a_j \right|_p \ge \frac{1}{\prod_{j=1}^{q} a_j |\infty|} $$

Consequently,

$$ (2) \qquad |a|_p \ge \left| \prod_{j=1}^{q} a_j \right|_p \ge \frac{1}{\prod_{j=1}^{q} |a_j|_\infty} $$

Now, $|\prod_{j=1}^{q} a_j|_\infty = \prod_{j=1}^{q} |a_j|_\infty \le (|a|)^q$. Thus, by (2), we obtain

$$ |a|_p \ge \frac{1}{(|a|)^q}. \qquad \square $$

**Corollary 9.11.** Let $a \in \Omega^*$ be of degree $q$ over $\mathbb{Q}$ and let $t = \text{den}(a)$. Then, $|a|_p \ge \frac{1}{t^q (|a|)^q}$.

**Corollary 9.12.** Let $a \in \Omega^*$ be of degree $q$ over $\mathbb{Q}$ and let $t = \text{den}(a)$. Then, $\log(|a|_p) \ge -2q(s(a))$.
# Chapter 10

## Algebras $R(D)$

The idea of considering rational functions with no pole inside a domain $D$, in order to define analytic functions in $D$, is due to Marc Krasner [69]. The behavior of rational functions in $K$ is determined by circular filters which characterize all multiplicative norms on rational functions. We make a general study of the set of multiplicative semi-norms of a normed algebra which is locally compact with respect to the topology of pointwise convergence. Results are first due to B. Guennebaud and G. Garandel [61], [62]. Henceforth, the idea of considering the topologic space of multiplicative semi-norms continuous with respect to the topology of a normed algebra was used in many works on Banach algebra.

**Notation.** In this chapter, we denote by $D$ an infinite subset of $K$ and then $K^D$ is provided with the topology $U_D$ of uniform convergence on $D$.

We denote by $R(D)$ the $K$-algebra of rational functions $h(x) \in K(x)$ with no pole in $D$. Since $D$ is infinite, $R(D)$ is clearly a $K$-subalgebra of $K^D$ and is provided with the topology induced by $U_D$ that makes it a topological subgroup of $K^D$. Algebraically, $R(D)$ is a $K$-subalgebra of $K(x)$ and more precisely is of the form $(S(D))^{-1}K[x]$ with $S(D)$ the multiplicative set of polynomials whose zeros do not belong to $D$.

We denote by $R_b(D)$ the $K$-subalgebra of $R(D)$ consisting of the $f \in R(D)$ which are bounded in $D$. Finally, if $D$ is not bounded, we denote by $R_0(D)$ the $K$-subalgebra of $R(D)$ that consists of the $f \in R(D)$ such that $\lim_{|x| \to +\infty, x \in D} f(x) = 0$.
For every $f \in \mathbb{K}^D$, we set $\|f\|_D = \sup_{x \in D} |f(x)| \in [0, +\infty]$.

Recall that an algebra-semi-norm $\psi$ of a $\mathbb{K}$-algebra $A$ is said to be semi-multiplicative or power multiplicative if it satisfies $\psi(x^n) = (\psi(x))^n \quad \forall x \in A$ and is said to be multiplicative if $\psi(xy) = \psi(x)\psi(y) \quad \forall x, y \in A$.

Lemma 10.1 is then immediate.

**Lemma 10.1.** *R*(D) is a principal ideal ring. Every ideal is of the form *P*(x)*R*(D) with *P* a polynomial whose zeros belong to D.

Lemma 10.2 is an immediate application of general properties of the supremum, once the set $[0, +\infty]$ is provided with the classical extensions of the addition and the multiplication:

$$a + (+\infty) = +\infty \quad \text{for every } a \in [0, +\infty],$$

$$a \cdot (+\infty) = +\infty \quad \text{for every } a \in ]0, +\infty].$$

**Lemma 10.2.** For every $g, h \in R(D)$, we have

(i) $\|h\|_D = 0$ if and only if $h = 0$,

(ii) $\|\lambda h\|_D = |\lambda| \|h\|_D$ for every $\lambda \in \mathbb{K}^*$,

(iii) $\|h + g\|_D \le \max(\|h\|_D, \|g\|_D)$,

(iv) If $(\|f\|_D, \|g\|_D)$ is different from $(0, +\infty)$ and from $(+\infty, 0)$, then $\|hg\|_D \le \|h\|_D \cdot \|g\|_D,$

(v) $\|h^n\|_D = (\|h\|_D)^n$ whenever $n \in \mathbb{N}$.

**Theorem 10.3.** $R_b(D) = R(D)$ if and only if $D$ is closed and bounded. Moreover, if $D$ is closed and bounded, then $\| \cdot \|_D$ is a semi-multiplicative ultrametric norm of $\mathbb{K}$-algebra.

**Proof.** We first suppose $D$ to be bounded. By Lemma 10.2, we just have to show that $\|h\|_D < +\infty$ for every $h \in R(D)$ in order to show that $\| \cdot \|_D$ is a norm of $\mathbb{K}$-algebra such that $\|h^n\|_D = \|h\|_D^n$. Since $D$ is bounded, obviously every polynomial $P$ satisfies $\|P\|_D < +\infty$, hence, by Lemma 10.2 (iv), we just have to check that $\left\|\frac{1}{Q}\right\|_D < +\infty$.

To show this, it is sufficient to prove that $\left\|\frac{1}{x-a}\right\|_D < +\infty$ for every $a \in \mathbb{K} \setminus D$. Since $D$ is closed, the distance $r$ from $a$ to $D$ is not zero, hence $\left\|\frac{1}{x-a}\right\|_D \le \frac{1}{r}$.

Now, if $D$ is not bounded, obviously $\|x\|_D = +\infty$. If $D$ is not closed, there exists at least one point $a \in \mathbb{K} \setminus D$ with a sequence
$(a_n)_{n \in \mathbb{N}}$ in $D$ which converges to $a$, hence $\left\| \frac{1}{x-a} \right\|_D = +\infty$. That ends the proof of Theorem 10.3. $\square$

**Theorem 10.4. (G. Garandel, B. Guennebaud) [61], [62]:** Let $\mathcal{F}$ be a large circular filter on $\mathbb{K}$ of diameter $s > 0$. The following three assertions are equivalent:

(i) $\varphi_{\mathcal{F}}(h) \le \|h\|_D$ whenever $h \in R(D)$.

(ii) $\varphi_{\mathcal{F}}$ is a continuous ultrametric multiplicative norm on $R(D)$ with respect to the topology of uniform convergence.

(iii) $\mathcal{F}$ is secant with $D$.

**Proof.** First, (i) and (ii) are obviously equivalent. Next, (iii) clearly implies (i) because if $\mathcal{F}$ is secant with $D$, then $\lim_{\mathcal{F}}|h(x)| = \lim_{\mathcal{F} \cap D}|h(x)| \le \|h\|_D$.

Hence, we just have to show that (i) implies (iii). For this, we assume (iii) to be false and prove that (i) is false. We first assume $\mathcal{F}$ to have center $a$. There exist annuli $\Gamma(a_i, r'_i, r''_i)$ ($1 \le i \le q$) with $|a_i - a_j| = s$ whenever $i \ne j$ and $r'_i < s < r''_i$ such that the set $B = \bigcap_{i=1}^q \Gamma(a_i, r'_i, r''_i)$ belongs to $\mathcal{F}$ and satisfies $B \cap D = \emptyset$. We put $r' = \max_{1 \le i \le q} r'_i$ and $r'' = \min_{1 \le i \le q} r''_i$. Let $\rho' \in]r', s[$, let $\rho'' \in]s, r''[$, and for every $i = 1, \dots, q$ set $b_i \in \Gamma(a_i, \rho', s)$ and set $b \in \Gamma(a, s, \rho'')$.

We put $h(x) = \left(\prod_{i=1}^q \left(\frac{x-a_i}{x-b_i}\right)\right) \left(\frac{\lambda}{x-b}\right)$ with $|\lambda| = b$. We first note that $\varphi_{\mathcal{F}}(h) = 1$ because $\varphi_{\mathcal{F}}(\frac{x-a_i}{x-b_i}) = 1$ whenever $i = 1, \dots, q$ and $\varphi_{\mathcal{F}}(\frac{\lambda}{x-b}) = 1$. Next, it is easily seen that $\|h\|_D \le \max(\frac{r'}{\rho'}, \frac{\rho''}{r''})$. Indeed, $|h(x)| = |\frac{x}{b_i}| \le \frac{r'}{\rho'}$ when $|x - a_i| \le r'$ and $|h(x)| = |\frac{\lambda}{x-b}| \le \frac{\rho''}{r''}$ when $|x - a| \ge r''$. Hence, we have $\|h\|_D < 1$ while $\varphi_{\mathcal{F}}(h) = 1$ and that contradicts the assertion (i).

We now suppose that $\mathcal{F}$ has no center. Let $d(a, r)$ belong to $\mathcal{F}$ such that $d(a, r) \cap D = \emptyset$ and let $\rho \in]s, r[$. There still exists a disk $d(\alpha, \rho) \in \mathcal{F}$ such that $d(\alpha, \rho) \subset d(a, r)$. Let us take $b \in \Gamma(\alpha, \rho, r)$ and $\lambda \in \mathbb{K}$ such that $|\lambda| = |b|$. We just put $h(x) = \frac{\lambda}{x-b}$ and we have $\varphi_{\mathcal{F}}(h) = 1$ because $|h(x)| = 1$ whenever $x \in d(\alpha, \rho)$, while $|h(x)| \le \frac{|b-\alpha|}{r} < 1$ whenever $|x - \alpha| \ge r$, hence finally $\|h\|_D < 1$. That ends the proof of Theorem 10.4. $\square$

In order to describe properties of the multiplicative semi-norms on $R(D)$ and next on analytic elements, we must recall a classical result on continuous multiplicative semi-norms on a normed $\mathbb{K}$-algebra $A$.
**Notation.** We denote by Mult($A$) the set of $\mathbb{K}$-algebra multiplicative semi-norms of a $\mathbb{K}$-algebra $A$. Given $\psi \in \text{Mult}(A)$, we denote by $\text{Ker}(\psi)$ the set of $x \in A$ such that $\psi(x) = 0$ and $\text{Ker}(\psi)$ is called the kernel of $\psi$.

Suppose now $A$ is a normed $\mathbb{K}$-algebra whose norm is denoted by $\| \cdot \|$. We denote by $\text{Mult}(A, \| \cdot \|)$ the set of $\mathbb{K}$-algebra multiplicative semi-norms of $A$ that are continuous functions on $A$ with respect to the norm $\| \cdot \|$ of $A$. Similarly, we denote by $\text{Mult}_m(A, \| \cdot \|)$ the set of $\mathbb{K}$-algebra multiplicative semi-norms of $A$ whose kernel is a maximal ideal that are continuous functions on $A$ and by $\text{Mult}_1(A, \| \cdot \|)$ the set of $\mathbb{K}$-algebra multiplicative semi-norms of $A$ whose kernel is a maximal ideal of codimension 1 that are continuous functions on $A$.

**Lemma 10.5.** Let $A$ be a $\mathbb{K}$-algebra provided with a $\mathbb{K}$-algebra norm $\| \cdot \|$ and let $\varphi \in \text{Mult}(A)$. Then, $\varphi$ belongs to $\text{Mult}(A, \| \cdot \|)$ if and only if $\varphi(x) \le \| x \|$ whenever $x \in A$. Moreover, if $A$ has a unity $u$ and if $\varphi$ is not identically 0, then $\varphi(\lambda u) = |\lambda|$ whenever $\lambda \in \mathbb{K}$. Further, if $\varphi$ belongs to $\text{Mult}(A)$, $\text{Ker}(\varphi)$ is a prime ideal, and if $\text{Mult}(A, \| \cdot \|)$, then $\text{Ker}(\varphi)$ is a closed prime ideal.

**Proof.** Suppose that for some $x \in A$ we have $\varphi(x) > \| x \|$. Since the valuation group of $\mathbb{K}$ is dense, it contains a subgroup of the form $a\mathbb{Z}$, with $a > 0$. Let $q \in \mathbb{N}$ be such that $q(\log(\varphi(x)) - \log(\|x\|)) > a$. Then there clearly exists $\lambda \in \mathbb{K}$ satisfying $\| x \|^{q} < |\lambda| < \varphi(x)^{q}$. So, much the more, we have $\| x^{q} \| < |\lambda| < \varphi(x^{q})$. Let $t = x^{q}$. Then, $\lim_{n \to \infty} \left(\frac{t}{\lambda}\right)^{n} = 0$ but $\lim_{n \to \infty} \varphi\left(\left(\frac{t}{\lambda}\right)^{n}\right) = +\infty$, and then $\varphi$ is not continuous.

Now, let $u$ be a unity in $A$. Either $\varphi(u) = 0$ and then $\varphi(x) = 0$ whenever $x \in A$, or $\varphi(u) = 1$ and then we have $\varphi(\lambda u) = |\lambda|\varphi(u) = |\lambda|$ whenever $\lambda \in L$. The last statement is immediate. This ends the proof of Lemma 10.5. $\square$

Theorem 10.6 is well known and may be found in [62] and in Theorems 6.9 and 6.19 of [45] (see also [53]).

**Theorem 10.6.** Let $A$ be a $\mathbb{K}$-algebra provided with a $\mathbb{K}$-algebra norm $\| \cdot \|$. For every $x \in A$, the sequence $(\| x^n \|^{\frac{1}{n}})_{n \in \mathbb{N}}$ has a limit denoted by $\| x \|_{si}$, satisfying $\|x\|_{si} \le \|x\| \quad \forall x \in A$ and $\|x\|_{si} = \sup\{\phi(x) \mid \phi \in \text{Mult}(A, \| \cdot \|)\}$. Moreover, $\|x^n\|_{si} = (\|x\|_{si})^n \quad \forall f \in A, \forall n \in \mathbb{N}$.
**Theorem 10.7.** Let $A$ be a $\mathbb{K}$-algebra provided with a $\mathbb{K}$-algebra norm $\|\cdot\|$. Then, $\mathrm{Mult}(A, \|\cdot\|)$ is compact with respect to the topology of pointwise convergence.

**Proof.** Let $B$ be the unit ball of $A$. By Lemma 10.5, each $\varphi \in \mathrm{Mult}(A, \|\cdot\|)$ has a restriction $\hat{\varphi}$ to $B$ which satisfies $\hat{\varphi}(B) \subset [0, 1]$. Hence, $\mathrm{Mult}(B, \|\cdot\|)$ is a closed subset of $[0, 1]^B$ provided with the topology of pointwise convergence on $B$. But by Tykhonov's theorem, $[0, 1]$ is compact for this topology and then so is $\mathrm{Mult}(B, \|\cdot\|)$. Moreover, the mapping $\varphi \to \hat{\varphi}$ from $\mathrm{Mult}(A, \|\cdot\|)$ into $\mathrm{Mult}(B, \|\cdot\|)$ is a bijection. Indeed, it is clearly injective and it is surjective because given $\psi \in \mathrm{Mult}(B, \|\cdot\|)$, we may extend $\psi$ to $A$ by putting $\overline{\psi}(x) = |\lambda|\psi(\frac{x}{\lambda})$ with $\lambda \in \mathbb{K}$, $|\lambda| \ge \|x\|$. Finally, this bijection is bicontinuous with respect to the pointwise convergence on both $\mathrm{Mult}(A, \|\cdot\|)$ and $\mathrm{Mult}(B, \|\cdot\|)$ and this ends the proof of Theorem 10.7. $\square$

Theorem 10.8 is given in several works. This proof mainly is given in [53].

**Theorem 10.8.** Let $\mathbb{B}$ be a field extension of $\mathbb{K}$ provided with a non-zero semi-norm of $\mathbb{K}$-algebra $\|\cdot\|$. Then, $\|\cdot\|$ is a norm of $\mathbb{K}$-algebra, and there exists an absolute value $\varphi$ on $\mathbb{B}$ extending that of $\mathbb{K}$ such that $\varphi(x) \le \|x\|$ whenever $x \in A$.

**Proof.** Let $\mathrm{SM}(A, \|\cdot\|)$ denote the set of continuous semi-norms $\phi$ of $A$ satisfying $\phi(f^n) = (\phi(f))^n \quad \forall f \in A, \forall n \in \mathbb{N}$. It is seen that $\|\cdot\|$ is a norm because $\mathrm{Ker}\|\cdot\| = \{0\}$. In the same way, so is the spectral semi-norm $\|\cdot\|_{\mathrm{si}}$ associated with $\|\cdot\|$. Now, $\mathrm{SM}(A, \|\cdot\|_{\mathrm{si}})$ is easily checked to be inductive with respect to the order $\ge$, i.e. given a totally ordered subset $W$ of $\mathrm{SM}(A, \|\cdot\|_{\mathrm{si}})$, the mapping $\psi$ defined in $A$ by $\psi(x) = \inf\{\theta(x)|\theta \in W\}$ belongs to $\mathrm{SM}(A, \|\cdot\|_{\mathrm{si}})$. Then by Zorn's lemma, $\mathrm{SM}(A, \|\cdot\|_{\mathrm{si}})$ admits a minimal element $\varphi$. As we just saw, $\varphi$ is a norm of $\mathbb{K}$-algebra and we have $\varphi(x) \le \|x\|_{\mathrm{si}}$ whenever $x \in A$. We will prove that $\varphi(ab) = \varphi(a)\varphi(b)$ whenever $a, b \in A$. Let $a \in A\setminus\{0\}$. For every $x \in A$, we put $u_n(x) = \frac{\varphi(a^n x)}{\varphi(a)^n}$. The sequence $(u_n(x))_{n \in \mathbb{N}}$ is seen to be decreasing. We put $\sigma(x) = \lim_{n \to +\infty} \frac{\varphi(a^n x)}{\varphi(a)^n}$ whenever $x \in A$.

First, we check that $\sigma$ is a norm of $\mathbb{K}$-algebra. Obviously, it is seen that for every $n \in \mathbb{N}$, $u_n$ is a norm of $\mathbb{K}$-vector space, hence so is $\sigma$.
Next, we have

$$u_n(x)u_n(y) = \frac{\phi(a^n x)\phi(a^n y)}{\phi(a^n)\phi(a^n)} = \frac{\phi(a^n x)\phi(a^n y)}{\phi(a^{2n})} \ge \frac{\phi(a^{2n}xy)}{\phi(a^{2n})} \ge \sigma(xy),$$

whenever $x, y \in A$, hence $\sigma(x)\sigma(y) \ge \sigma(xy)$. So, $\sigma$ is a norm of $\mathbb{K}$-algebra. Now, we check that $\sigma$ is semi-multiplicative, because

$$\lim_{n \to +\infty} \left( \frac{\varphi(a^n x^q)}{\varphi(a^n)} \right) = \lim_{n \to +\infty} \left( \frac{\varphi(a^{qn} x^q)}{\varphi(a^{qn})} \right) = \lim_{n \to +\infty} \left( \frac{\varphi(a^n x)}{\varphi(a^n)} \right)^q = \sigma(x)^q.$$

Then, since $\sigma$ satisfies $\sigma(x) \le \varphi(x) \le \|x\|_{si}$ whenever $x \in A$, it clearly belongs to $SM(A, \| \cdot \|_{si})$. But since $\varphi$ is minimal in $SM(A, \| \cdot \|_{si})$, actually $\varphi$ is equal to $\sigma$.

Now, as the sequence $(u_n)_{n \in \mathbb{N}}$ is decreasing, we have $\sigma(x) \le \frac{\varphi(ax)}{\varphi(a)}$, hence $\varphi(x)\varphi(a) \le \frac{\varphi(ax)}{\varphi(a)}$ and hence $\varphi(a)\varphi(x) \le \varphi(ax)$. But since $\varphi$ is a norm of $\mathbb{K}$-algebra, $\varphi(ax) \le \varphi(a)\varphi(x)$, which proves that $\varphi$ is a multiplicative norm. $\square$

Theorem 10.9 is immediate.

**Theorem 10.9.** Let $A$ be a Banach $\mathbb{K}$-algebra. For every maximal ideal $\mathcal{M}$ of $A$, there exists $\varphi \in \mathrm{Mult}(A, \| \cdot \|)$ such that $\mathrm{Ker}(\varphi) = \mathcal{M}$. If $\mathcal{M}$ is of codimension 1, the mapping $\tau$ from $A$ onto $\mathbb{K}$ admitting $\mathcal{M}$ for kernel satisfies $|\tau(f)| \le \|f\| \quad \forall f \in A$.

**Proof.** Let $\mathcal{M}$ be a maximal ideal of $A$ and let $\mathbb{B}$ be the field $\mathbb{A}/\mathcal{M}$. Since $A$ is complete, $\mathcal{M}$ is closed, therefore $\mathbb{B}$ is provided with the quotient norm. By Theorem 10.9, $\mathbb{B}$ admits an absolute value $| \cdot |$ which extends that of $\mathbb{K}$. Let $\psi$ be the canonical surjection from $A$ to $\mathbb{B}$. On $A$ we put $\varphi(x) = |\psi(x)|$. Then, $\varphi$ is an element of $\mathrm{Mult}(A, \| \cdot \|)$ such that $\mathrm{Ker}(\varphi) = \mathcal{M}$.

Suppose now $\mathcal{M}$ is of codimension 1 and suppose $|\tau(f)| > \|f\|$. Then the series $\sum_{n=0}^{+\infty} \left(\frac{f}{\tau(f)}\right)^n$ converges and shows that $f - \tau(f)$ is invertible in $A$, a contradiction since $\tau(f - \tau(f)) = 0$. $\square$

**Corollary 10.10.** Let $A$ be a Banach $\mathbb{K}$-algebra. Every $\mathbb{K}$-algebra homomorphism from $A$ to $\mathbb{K}$ is continuous.

The characterization of the continuous multiplicative norms of $(R(D), \| \cdot \|_D)$ by means of the large circular filters secant with $D$
suggests us extending this characterization to the multiplicative semi-norms of $R(D)$.

**Theorem 10.11.** Let $D$ be a closed bounded subset of $K$. The mapping $\Xi$ from $\Phi(D)$ into $\text{Mult}(R(D))$ defined as $\Xi(\mathcal{F}) = \varphi_{\mathcal{F}}$ is a bijection from $\Phi(D)$ onto $\text{Mult}(R(D), U_D)$. Moreover, $\varphi_{\mathcal{F}}$ is an absolute value if and only if $\mathcal{F}$ does not converge in $D$. Further, $\text{Mult}(R(D, \|\cdot\|_D))$ is provided with the topology of pointwise convergence for which it is compact.

**Proof.** On the one hand, by Theorems 4.10 and 10.4, it is clearly seen that the mapping defined on $\Phi^\circ(D)$ by $\mathcal{F} \to \varphi_{\mathcal{F}}$ is a bijection from this set onto the set of continuous multiplicative norms on $R(D)$.

On the other hand, every $a \in D$ defines a multiplicative semi-norm $\psi$ by $\psi(h) = |h(a)|$, the kernel of which is the maximal ideal $(x-a)R(D)$. Thus, we have a mapping from the set of convergent circular filters on $D$ into the set of multiplicative semi-norms which are not norms: this mapping is obviously injective.

Finally, let $\psi$ be a multiplicative semi-norm whose kernel is not zero. Then, $\text{Ker}(\psi)$ is a prime ideal hence a maximal ideal of $R(D)$, and therefore, it is of the form $(x-a)R(D)$ with $a \in D$. Then, $\psi(x-a) = 0$, hence $\psi(x-b) = |a-b|$ whenever $b \in K$, and therefore, $\psi$ is of the form $\varphi_{\mathcal{F}_a}$ with $\mathcal{F}_a$ the filter of the neighborhoods of $a$ in $D$. Thus, $\Xi$ is a bijection from $\Phi(D)$ onto $\text{Mult}(R(D), U_D)$ and $\varphi_{\mathcal{F}}$ is a norm if and only if $\mathcal{F}$ is not convergent. Finally, by Theorem 10.7, $\text{Mult}(R(D, \|\cdot\|_D))$ is compact with respect to the topology of pointwise convergence. That ends the proof of Theorem 10.11. $\square$

**Corollary 10.12.** $\text{Mult}(\mathbb{K}[X])$ is provided with the topology of pointwise convergence for which it is locally compact.
# Chapter 11

## Analytic Elements

Due to the fact that any disk $d(a, r)$ is exactly the same as $d(b, r)$ for every $b \in d(a, r)$, it is easily seen that a power series $\sum_{n=0}^{\infty} a_n (x-a)^n$ which admits the disk $d(a, r)$ for disk of convergence may not be extended outside its convergence disk as it is done in complex analysis, by means of a change of origin.

However, by Runge's theorem, we remember that a holomorphic function in a compact subset $D$ of $\mathbb{C}$ is equal to the limit of a sequence of rational functions with respect to the uniform convergence on $D$. This is why Marc Krasner introduced analytic elements on a subset $D$ of $\mathbb{K}$ directly by considering limits of sequences of rational functions with respect to the uniform convergence on $D$ [69].

Actually, Marc Krasner constructed a theory of analytic functions $f$ defined on a quasi-connected set $D$ equal to the union of a chained family of quasi-connected sets $(D_i)_{i \in I}$ such that the restriction of $f$ to each $D_i$ is an analytic element on $D_i$ [69] (this construction was widened to the analytic infraconnected sets by Philippe Robba) [83].

Another theory was defined by John Tate, consisting (in one variable) of using infraconnected affinoid sets. Here, we only describe some basic properties of analytic elements on infraconnected sets in order to apply them to power series and various Laurent series that are used for studying meromorphic functions. A comparison between Krasner's theory and Tate's theory was made in [37]. Here, we aim at studying meromorphic functions in the field $\mathbb{K}$ and applications to problems of value distribution. This is why we do not repeat the study of Krasner-Tate algebras [37]. However, interesting properties
of injective analytic elements can be proved with the use of Krasner–Tate algebra. In Chapter 42, we recall these properties letting us prove the famous relation satisfied by many kind of injective analytic functions in $K$.

We examine algebras of analytic elements, particularly Banach algebras of bounded analytic elements. We see the characterization of sets $D$ such that the space of analytic elements on $D$ is a $K$-algebra. We examine some basic properties of analytic elements such as poles when the set $D$ is not closed and we see that analytic elements on a closed bounded set $D$ are uniformly continuous. If $D$ has finitely many infraconnected components, for each one, its characteristic function is an analytic element on $D$.

**Definition and notation.** Let $D$ be an infinite subset of $K$. We denote by $H(D)$ the completion of $R(D)$ for the topology $U_D$ of uniform convergence on $D$. The elements of $H(D)$ are called the analytic elements on $D$ [35], [44], [69].

The set $H(D)$ is then provided with the topology of uniform convergence on $D$ for which it is complete and every $f \in H(D)$ defines a function on $D$ which is the uniform limit (on $D$) of a sequence $(h_n)_{n \in \mathbb{N}}$ in $R(D)$. Thus, given two infinite sets $D, D'$ such that $D \subset D'$, the restriction to $D$ of elements of $H(D')$ enables us to consider that $H(D')$ is included in $H(D)$.

Next, $H(D)$ is a $K$-vectorial space and a complete topological group with respect to the topology $U_D$. The question of whether the product of two analytic elements on $D$ is an analytic element on $D$ is studied later. However, it is easily seen that given $f \in H(D)$, the function $f^n$ also belongs to $H(D)$.

**Lemma 11.1.** For every $f, g \in H(D)$, we have

(i) $\|f\|_D = 0$ if and only if $f = 0$,

(ii) $\|\lambda f\|_D = |\lambda| \|f\|_D$ whenever $\lambda \in K^*$,

(iii) $\|f + g\|_D \le \max(\|f\|_D, \|g\|_D)$,

(iv) if $(\|f\|_D, \|g\|_D)$ is different from $(0, +\infty)$ and from $(+\infty, 0)$, then the function $fg$ satisfies $\|fg\|_D \le \|f\|_D \|g\|_D$,

(v) $\|f^n\|_D = \|f\|_D^n$ whenever $n \in \mathbb{N}^*$.

**Notation.** We denote by $H_b(D)$ the set of elements $f \in H(D)$ bounded on $D$. Then, $H_b(D)$ is clearly a $K$-vectorial subspace of
$H(D)$ and is closed in $H(D)$. Moreover, $\| \cdot \|_D$ is a norm on $H_b(D)$ that makes it a Banach $\mathbb{K}$-algebra. If $D$ is unbounded, we denote by $H_0(D)$ the set of the $f \in H(D)$ such that $\lim_{\substack{|x| \to +\infty \\ x \in D}} f(x) = 0$.

Theorem 11.2 is an immediate consequence of Theorem 10.3.

**Theorem 11.2.** *Let $H_b(D)$ be a Banach $\mathbb{K}$-subalgebra of $\mathbb{K}^D$. The following three conditions are equivalent:*

(i) $H_b(D) = H(D)$,

(ii) $H(D)$ is topological $\mathbb{K}$-vector space,

(iii) $(H(D), \| \cdot \|_D)$ is a Banach $\mathbb{K}$-algebra,

(iv) $D$ is closed and bounded.

*If these conditions are satisfied, then $\| \cdot \|_D$ is a semi-multiplicative norm.*

**Definition.** Let $f \in H(D)$ have no zero in $D$. The element $f$ is said to be *invertible in $H(D)$* if the function $\frac{1}{f}$ (also denoted by $f^{-1}$) belongs to $H(D)$. This definition holds even if $H(D)$ is not a ring.

The following lemma is classical.

**Lemma 11.3.** Let $f \in H(D)$ be such that $\inf_{x \in D} |f(x)| > 0$. Then, $f^{-1}$ belongs to $H(D)$. Moreover, if $D$ is closed and bounded, $f^{-1}$ belongs to $H(D)$ if and only if $\inf_{x \in D} |f(x)| > 0$.

Let $g \in H(D)$ satisfy $|g(x)| = 1$ for all $x \in D$ and $\|f - 1\|_D > \|g - 1\|_D$. Then, we have $\|fg - 1\|_D = \|f - 1\|_D$.

**Proof.** We suppose $\inf_{x \in D} |f(x)| = \lambda > 0$. Let $(h_n)_{n \in \mathbb{N}}$ be a sequence in $R(D)$ such that $\lim_{n \to \infty} \|h_n - f\|_D = 0$. For $n$ big enough, we have $|h_n(x)| \ge \lambda$ whenever $x \in D$, hence $\left|\frac{1}{h_n(x)} - \frac{1}{f(x)}\right| = \left|\frac{f(x)-h_n(x)}{h_n(x)f(x)}\right| \le \frac{\|f-h_n\|_D}{\lambda^2}$, hence the sequence $\frac{1}{h_n}$ converges to $\frac{1}{f}$.

Conversely, if $D$ is closed and bounded and if $\frac{1}{f} \in H(D)$, $\frac{1}{f}$ has to be bounded by some $M \in \mathbb{R}_+$, hence $|f(x)| \ge \frac{1}{M}$ whenever $x \in D$.

Now, let $g \in H(D)$ satisfy $|g(x)| = 1$ for all $x \in D$ and $\|f - 1\|_D > \|g - 1\|_D$. For every $x \in D$, we have $|f(x) - 1| \|g(x)| > |g(x) - 1|$, and therefore, $|f(x)g(x) - 1| = |f(x) - 1| \|g(x)| = |f(x) - 1|$. This finishes proving Lemma 11.3. $\square$

**Theorem 11.4.** Let $f \in H(D)$ and let $h \in R_b(D)$. Then, $fh$ belongs to $H(D)$.
**Proof.** Let $\varepsilon$ be > 0 and let $g \in R(D)$ satisfy $\|f - g\|_D < \varepsilon$. Then, we have $\|hf - hg\|_D < \varepsilon\|h\|_D$ and this clearly shows that $fh \in H(D)$. $\square$

When $D$ is not closed or is not bounded, we show how to split an element $f \in H(D)$.

**Theorem 11.5.** The vector space $H(D)$ is equal to the direct sum $R_0(\mathbb{K}\setminus(\overline{D}\setminus D)) \oplus H(\overline{D})$. Moreover, if $D$ is not bounded, then $H(\overline{D})$ is equal to the direct sum $\mathbb{K}[x] \oplus H_0(\overline{D})$.

**Proof.** Let $(f_n)_{n \in \mathbb{N}}$ be a sequence in $R(D)$ such that $\lim_{n \to \infty} \|f_n - f\|_D = 0$. In particular, there exists $N \in \mathbb{N}$ such that $f_n - f_N$ is bounded when $n \ge N$. We put $g_n := f_n - f_N$. The sequence $g_n$ converges in $H(D)$ to $f - f_N$. Let $g := f - f_N$. On the other hand, since each $f_n - f_N$ belongs to $R(D)$ and is bounded in $D$, $f_n - f_N$ belongs to $R(\overline{D})$. Obviously, $\|f_n - f_N\|_D = \|f_n - f_N\|_{\overline{D}}$, hence finally, $g$ belongs to $H_b(\overline{D})$. Now, we may obviously split $f_N$ in the form $E(x) + h_1(x) + h_2(x)$ with $E(x) \in \mathbb{K}[x], h_1 \in R_0(\overline{D}), h_2 \in R_0(\mathbb{K}\setminus(\overline{D}\setminus D))$. We put $f^* = h_2$ and $\overline{f} = E + h_1 + g$. We have clearly split $f$ in the form $f^* + \overline{f}$ with $f^* \in R_0(\mathbb{K}\setminus(\overline{D}\setminus D)), \overline{f} \in H(\overline{D})$. Hence, we have proven that $H(D) = R_0(\mathbb{K}\setminus(\overline{D}\setminus D)) + H(\overline{D})$.

This sum is easily seen to be direct. Indeed, suppose that we have $h \in R_0(\mathbb{K}\setminus(\overline{D}\setminus D))$ and $g \in H(\overline{D})$ such that $h + g = 0$, with $h \ne 0$. Then, $h$ has a pole $\alpha \in \overline{D}\setminus D$ and it may be written in the form $\sum_{i=1}^q \frac{\lambda_i}{(x-\alpha)^i} + h_\alpha$ with $\lambda_i \ne 0$ ($1 \le i \le q$) and $h_\alpha \in R_0(\mathbb{K}\setminus(\overline{D}\setminus(D \cup \{\alpha\})))$. But $g$ is obviously bounded around $\alpha$, hence $h$ has to be bounded when $x$ approaches $\alpha$, hence finally $\alpha$ does not exist. This shows that the sum is direct.

Now, we suppose $D$ is unbounded. First, we prove that every element $f \in H_b(D)$ admits a limit when $|x|$ tends to $+\infty$. Let $\epsilon \in \mathbb{R}_+^*$ and let $h \in R(D)$ satisfy $\|f - h\|_D < \epsilon$. Since $f$ is bounded in $D$, so is $h$. But then $h$ is of the form $\frac{P}{Q}$, with $\deg(P) = \deg(Q)$, and therefore, $h$ has a limit $\lambda$ when $|x|$ tends to $+\infty$. Let $\epsilon > 0$ be such that $|h(x) - \lambda| < \epsilon$ whenever $x \in D\setminus d(0, r)$. Clearly, we have $|f(x) - \lambda| < \epsilon$ whenever $x \in D\setminus d(0, r)$. This proves that $f$ does converge along the filter $\mathcal{F}$ which admits as a basis the family of sets $D\setminus d(0, r)$ ($r > 0$).

Let $f \in H(\overline{D})$ be unbounded. Let $(x_n)_{n \in \mathbb{N}}$ be a sequence such that $\lim_{n \to \infty} |f(x_n)| = +\infty$. Suppose the sequence $(x_n)_{n \in \mathbb{N}}$ does not tend
to $+\infty$. Then there exists a bounded subsequence $(x_{n_q})_{q \in \mathbb{N}}$ such that $\lim_{q \to \infty} |f(x_{n_q})| = +\infty$, but this is impossible due to Theorem 11.2 because such a sequence lies in a closed bounded set $D'$ included in $D$. Now, there exists $h \in R(\bar{D})$ such that $f - h$ is bounded, and therefore, we have $\lim_{n \to \infty} |h(x_n)| = +\infty$. Let $h(x) = P(x) + u(x)$ with $P \in \mathbb{K}[x]$ and $u \in R_0(\bar{D})$. Since $f - h$ is bounded, clearly, $f - P$ belongs to $H_b(\bar{D})$, hence we have proven that $H(\bar{D}) = \mathbb{K}[x] + H_b(\bar{D})$. Moreover, since $u$ has a limit when $|x|$ tends to $+\infty$ in $D$, we have $H_b(\bar{D}) = H_0(\bar{D}) + \mathbb{K}$, and therefore, $H(\bar{D}) = \mathbb{K}[x] + H_0(\bar{D})$. Finally, by considering elements when $|x|$ tends to $+\infty$, this sum is easily seen to be direct and this ends the proof. $\square$

**Definition.** Let $\alpha \in \bar{D} \setminus D$ and $f \in H(D)$, and let $f = f^* + \bar{f}$, with $f^* \in R_0(\mathbb{K}\setminus(\bar{D} \setminus D))$ and $\bar{f} \in H(\bar{D})$. Let $\alpha \in \bar{D} \setminus D$ be a pole of $f^*$ and let $f^*(x) = \sum_{j=1}^q \frac{\lambda_j}{(x-\alpha)^j} + u(x)$, with $u \in R_0(\mathbb{K}\setminus(\bar{D} \setminus (D \cup \{\alpha\})))$. The pole $\alpha$ of order $q$ of $f^*$ is called a *pole of order $q$ of $f$* and $\lambda_1$ is called the *residue of $f$ at $\alpha$* and is denoted by $\text{res}(f, \alpha)$.

Let $a_i$, $1 \le i \le n$, be the poles of $f$ and for each $i$ let $q_i$ be the order of $a_i$. The polynomial $\prod_{i=1}^n (x - a_i)^{q_i}$ is named the polynomial of poles of $f$ in $\bar{D} \setminus D$.

**Corollary 11.6.** Let $\alpha \in \bar{D} \setminus D$ and $f \in H(D)$ be such that $|f(x)|$ is bounded in $d(\alpha, r) \cap D$ with $r > 0$. Then $f \in H(D \cup \{\alpha\})$.

**Proof.** Indeed, as $\bar{f}$ is obviously bounded in $D \cap d(\alpha, r)$, so is $f^*$, and therefore, $f^*$ has no pole at $\alpha$. $\square$

**Corollary 11.7.** $H(D) = H_b(\bar{D}) + R(\mathbb{K}\setminus(\bar{D} \setminus D))$, $H_b(D) = H_b(\bar{D}) \subset H(\bar{D})$ and $H_b(D) = H_b(\bar{D})$. If $D$ is bounded, then $H_b(D) = H(\bar{D})$. If $D$ is not bounded, for every unbounded $f \in H(D)$, there exists a unique $q \in \mathbb{N}^*$ such that $x^{-q}f(x)$ has a finite non-zero limit when $|x|$ tends to $+\infty$, $x \in D$. Let $d(a, r^{-})$ be a hole of $D$. If $f$ belongs to $H(\bar{D})$ and if $x^{-q}f(x)$ has a finite non-zero limit, then $\frac{f(x)}{x^{-a}}$ belongs to $H_b(D)$.

**Corollary 11.8.** If $\bar{D} = \mathbb{K}$, then $H(D) = R(D)$.

Corollary 11.9 comes from the definition of the poles and from Theorem 11.5.

**Corollary 11.9.** Let $f \in H(D)$ and let $\alpha \in \bar{D} \setminus D$. Then $\alpha$ is a pole of order $n > 0$ for $f$ if and only if $(x - \alpha)^n f(x)$ has a finite non-zero
limit at $\alpha$. If there exists no $r \in \mathbb{R}_+^*$ such that $|f(x)|$ is bounded in $d(\alpha, r)\setminus\{\alpha\}$, then $\alpha$ is a pole of order $n \ge 1$ for $f$ and $(x-\alpha)^n f(x)$ has a finite non-zero limit at $\alpha$.

**Theorem 11.10.** Let $f \in H(D)$ and let $\alpha \in \bar{D} \setminus D$. Either $f$ belongs to $H(D \cup \{\alpha\})$ or $\alpha$ is a pole for $f$.

**Proof.** If $f$ does not belong to $H(D \cup \{\alpha\})$, by Corollary 11.6, $f$ is unbounded in any disk $d(\alpha, r)$ whenever $r > 0$. Hence, by means of the notation of Theorem 11.5, $\alpha$ is clearly a pole of $f^*$ and therefore is a pole of $f$. $\square$

We must note Theorem 11.11:

**Theorem 11.11.** Let $D$ be closed and bounded and let $f \in H(D)$. Then, $f$ is uniformly continuous in $D$.

**Proof.** The claim is immediate when $f$ is a polynomial. Suppose now that $f(x) = \frac{P(x)}{Q(x)} \in R(D)$. Since $D$ is closed and bounded, there exists $m > 0$ such that $|Q(x)| \ge m, \forall x \in D$. Consequently, $\frac{P(x)}{Q(x)}$ also is uniformly continuous. Then, when $f \in H(D)$, since $f$ is a uniform limit of rational functions, $f$ is also uniformly continuous. $\square$

**Notation.** We denote by Alg the family of sets $E$ such that $H(E)$ is a $\mathbb{K}$-subalgebra of $\mathbb{K}^E$.

**Theorem 11.12.** Let $f \in H(D)$. There exists $W \in R_b(D)$, whose zeros lie in $\bar{D} \setminus D$ and $h \in H(\bar{D})$ such that $f = \frac{h}{W}$. Further, if $D$ is bounded or if $D \in \text{Alg}$, then there exists $g \in H(\bar{D})$ such that $f = \frac{g}{Q}$ with $Q$ the polynomial of poles of $f$ in $\bar{D} \setminus D$.

**Proof.** We may summarize Theorem 11.5 in this way: $f$ is of the form $\tilde{f}(x) + \hat{f}(x)$ with $\tilde{f} \in R(\mathbb{K}\setminus(\bar{D}\setminus D))$ and $\hat{f} \in H_b(\bar{D})$. Indeed, if $D$ is bounded, we just take $\hat{f} = \bar{f}$, and if $D$ is not bounded, $\hat{f}$ is the one defined in Theorem 11.5. Thus, $\tilde{f}(x)$ can be written in the form $\frac{P(x)}{Q(x)}$ with $Q(x) = \prod_{i=1}^n (x - a_i)^{q_i}$ i.e. the polynomial of the poles of $f$ in $\bar{D}\setminus D$, and $P(x) \in \mathbb{K}[x]$. Let $q = \sum_{i=1}^n q_i$. Theorem 11.12 is obviously trivial if $D$ has no hole, hence we may assume $D$ to have at least one hole $T = d(a, r^-)$. Let $W(x) = \frac{Q(x)}{(x-a)^q}$. We know that $W \in R_b(\bar{D})$ hence $W\hat{f} \in H_b(\bar{D})$. On the other hand, we see that
$W\tilde{f} \in R(\bar{D})$, hence $Wf \in H(\bar{D})$. We just put $h = Wf$ and have the factorization $f = \frac{h}{W}$.

If $D$ is bounded, we see that both $Q, h$ are bounded in $D$, hence $Q\hat{f}$ belongs to $H_b(\bar{D})$ and then $Qf$ belongs to $H(\bar{D})$. In the same way, if $H(D)$ is supposed to be a ring, then $Q\hat{f}$ belongs to $H(\bar{D})$ and then $Qf = P + Q\hat{f}$ belongs to $H(\bar{D})$. This ends the proof. $\square$

**Corollary 11.13.** Let $S(D)$ be the set of polynomials whose zeros belong to $\bar{D} \setminus D$. If $D \in \text{Alg}$, then $H(D) = S(D)^{-1}H(\bar{D})$.

**Theorem 11.14.** Let $D$ be closed. Let $g \in H(D)$ and let $P \in \mathbb{K}[x]$ be such that $Pg$ belongs to $H(D)$. For every $Q \in \mathbb{K}[x]$ such that $\deg(Q) \le \deg(P)$, $Qg$ also belongs to $H(D)$.

**Proof.** Theorem 11.14 is clearly trivial when $D$ belongs to $\text{Alg}$. Now, suppose that $D$ does not belong to $\text{Alg}$, hence $D$ is unbounded. If $D$ has no hole, then $D = \mathbb{K}$, hence by Corollary 11.8, $H(D)$ is equal to $\mathbb{K}[x]$. Thus, we may assume that $D$ admits at least one hole and then, without loss of generality, we can assume that this hole is $d(0, r^{-})$. Let $q = \deg(P)$. Let $P_q = P$ and let $P_{q-1} = \frac{P(x)-P(0)}{x}$. Then, $P_{q-1}$ is a polynomial of degree $q-1$. We see that $(P(x)-P(a))g(x)$ belongs to $H(D)$. But $\left\|\frac{1}{x}\right\|_D$ is bounded and then by Theorem 11.4, $P_{q-1}(x)g(x)$ belongs to $H(D)$. Hence, by induction, it is seen that for each $j = 1, \dots, q$ there exists a polynomial $P_j$ of degree $j$ such that $P_j g$ belongs to $H(D)$ and this clearly completes the proof. $\square$

Now, when $D$ is not infraconnected, we have to note an easy result on characteristic functions that shows how rich the algebra $H(D)$ is.

**Proposition 11.15.** Let $D$ have an empty annulus $\Lambda$. Let $w_1, w_2$ be the functions defined on $D$ by $w_1(x) = 1, w_2(x) = 0$ if $x \in \mathcal{I}(\Lambda)$ and $w_1(x) = 0, w_2(x) = 1$ if $x \in \mathcal{E}(\Lambda)$. Then, $w_1$ and $w_2$ belong to $H(D)$.

**Proof.** Let $\Lambda = \Gamma(a, r_1, r_2)$, with $a \in D$. With no loss of generality, we may obviously assume $a = 0$. Let $\alpha \in \Lambda$ be such that $r_1 < |\alpha| < r_2$ and for each $n \in \mathbb{N}^*$, let $u_n = \frac{1}{1-(\frac{x}{\alpha})^n}$. Then, $\|1-(\frac{x}{\alpha})^n-1\|_{\mathcal{I}(\Lambda)} \le (\frac{r_1}{|\alpha|})^n$, while $|1-(\frac{x}{\alpha})^n| \ge (\frac{r_2}{|\alpha|})^n$ for every $x \in \mathcal{E}(\Lambda)$, hence finally, $\|u_n - w_1\|_D \le \max((\frac{r_1}{|\alpha|})^n, (\frac{|\alpha|}{r_2})^n)$. Thus, we see that $w_1 = \lim_{n \to \infty} u_n \in H(D)$ and $w_2 = 1-w_1 \in H(D)$. $\square$
**Theorem 11.16.** Let E have finitely many infraconnected components $E_1, \dots, E_q$. For each $i = 1, \dots, q$, the characteristic function of $E_i$ belongs to $H(E)$.

**Proof.** Let A be one of the infraconnected components of E. By Theorem 2.21, there exist empty annuli $(\Lambda_j)_{0 \le j \le n}$ such that A is either of the form

$$ (\alpha) \mathcal{I}_E(\Lambda_0) \cap \left( \bigcap_{j=1}^n \mathcal{E}_E(\Lambda_j) \right) \quad \text{or of the form} \quad (\beta) \bigcap_{j=1}^n \mathcal{E}_E(\Lambda_j). $$

But by Proposition 11.15, the characteristic function $u_j$ of $\mathcal{E}_E(\Lambda_j)$ belongs to $H_b(E)$ ($1 \le j \le n$) and so does the characteristic function $u_0$ of $\mathcal{I}_E(\Lambda_0)$. Since all the $u_j$ belong to $H_b(E)$, we see that the products $u = \prod_{j=0}^n u_j$ and $w = \prod_{j=1}^n u_j$ belong to $H(E)$. Then, when A is of the form $(\alpha)$ (respectively $(\beta)$), its characteristic function is equal to $u$ (respectively $w$) and therefore belongs to $H(E)$. $\square$
# Chapter 12

## Composition of Analytic Elements

Given $A$ and $B \subset \mathbb{K}$, $f \in H(A)$ such that $f(A) \subset B$ and $g \in H(B)$, a basic question is whether $g \circ f \in H(A)$. There is an immediate application to the study of homomorphisms from an algebra $H(D)$ to another $H(D')$.

**Lemma 12.1.** Let $A$ and $B$ be subsets of $\mathbb{K}$ and let $f \in H(A)$ be such that $f(A) \subset B$. For every $\lambda \in \mathbb{K} \setminus \bar{B}$, $f - \lambda$ is invertible in $H(A)$. Moreover, if for every $h \in R(B)$, $h \circ f$ belongs to $H(A)$, then for every $g \in H(B)$, $g \circ f$ belongs to $H(A)$ and for every $\lambda \in \bar{B} \setminus B$, $f - \lambda$ is invertible in $H(A)$.

**Proof.** Let $r = \delta(\lambda, B)$. If $\lambda \notin \bar{B}$, we have $r > 0$ and then $|f(x) - \lambda| \ge r$ whenever $x \in A$. Hence, by Lemma 11.3, $f - \lambda$ is invertible in $H(A)$.

Now, we assume that for every $h \in R(B)$, $h \circ f$ belongs to $H(A)$. Let $g \in H(B)$, let $\epsilon > 0$, and let $h \in R(B)$ satisfy $\|g - h\|_B \le \epsilon$. It is seen that $\|g \circ f - h \circ f\|_A \le \epsilon$. Since $h \circ f$ belongs to $H(A)$, then so does $g \circ f$.

Finally, let $\lambda \in \bar{B} \setminus B$ and let $h(u) = \frac{1}{u-\lambda}$. Since $h \circ f$ belongs to $H(A)$, $f - \lambda$ is invertible in $H(A)$. $\square$

**Theorem 12.2.** Let $A, B$ be subsets of $\mathbb{K}$ and let $f \in H(A)$ satisfy $f(A) \subset B$:

(i) If $f \in R(A)$, then $g \circ f \in H(A)$ whenever $g \in H(B)$.

(ii) If $A \in \text{Alg}$, then $g \circ f \in H(A)$ for all $g \in H(B)$ if and only if $f - \lambda$ is invertible in $H(A)$ for all $\lambda \in \bar{B} \setminus B$.
**Proof.** By Lemma 12.1, we just have to show that for every $g \in R(B)$, $g \circ f$ belongs to $H(A)$ in each one of these two hypotheses:

(H1) $f \in R(A)$.

(H2) $f - \lambda$ is invertible in $H(A)$ for all $\lambda \in \bar{B} \setminus B$.

So, we take $g(u) = \frac{P(u)}{Q(u)} \in R(B)$ and show that $g \circ f \in H(A)$ in each hypothesis. Let $\lambda_1, \dots, \lambda_q$ be the poles of $g$ in $K \setminus B$.

(H1) For every $j = 1, \dots, q$, $f - \lambda_j$ is invertible in $R(A)$ because $f - \lambda_j$ has no zero in $A$. Hence, $Q \circ f$ is invertible in $R(A)$ and then $g \circ f$ belongs to $R(A)$.

(H2) For each $j = 1, \dots, q$, either $\lambda_j$ belongs to $\bar{B} \setminus B$ or it belongs to $K \setminus \bar{B}$. In both cases, by Lemma 12.1, each $f - \lambda_j$ is invertible in $H(A)$. Since $A$ belongs to Alg, $Q \circ f$ is clearly invertible in $H(A)$ and $P \circ f$ belongs to $H(A)$. Hence, so does $g \circ f$. $\square$

**Corollary 12.3.** Let $A \in \text{Alg}$ and let $B$ be a closed subset of $K$. Let $f \in H(A)$ satisfy $f(A) \subset B$ and let $g \in H(B)$. Then, $g \circ f$ belongs to $H(A)$.

**Example.** Let $r, s \in \mathbb{R}_+^*$, let $f \in H(d(0, r))$ be such that $f(d(0, r)) \subset d(0, s)$, and let $g \in H(d(0, s))$. Then, $g \circ f$ belongs to $H(d(0, r))$.

**Lemma 12.4.** Let $h \in R(D)$ and let $D' = h(D)$. Let $f \in H(D')$. If $f$ is invertible in $H(D')$, then $f \circ h$ is invertible in $H(D)$. If $h$ is a Moebius function, $f$ is invertible in $H(D')$ if and only if $f \circ h$ is invertible in $H(D)$.

**Proof.** First, we suppose $f$ invertible in $H(D')$. Let $g = \frac{1}{f}$. Then, by Theorem 12.2, $g \circ h$ belongs to $H(D)$ and is clearly equal to $\frac{1}{f \circ h}$. Now, we assume that $h$ is a Moebius function and we put $\ell = h^{-1}$. If $f \circ h$ is invertible in $H(D)$, $(f \circ h) \circ \ell$ is invertible in $H(D')$ and this ends the proof of Lemma 12.4. $\square$

We are now going to study the $K$-algebra homomorphisms from $H(D)$ into $H(D')$. First, we consider homomorphisms from $R(D)$ into $R(D')$.

**Proposition 12.5.** Let $D, D'$ be subsets of $K$ and let $\gamma \in R(D')$ satisfy $\gamma(D') \subset D$. Let $\phi_\gamma$ be the mapping from $R(D)$ into $R(D')$ defined as $\phi_\gamma(f) = f \circ \gamma (f \in R(D))$. Then $\phi_\gamma$ is a homomorphism
from $R(D)$ into $R(D')$ and this homomorphism is injective if and only if $\gamma$ is not a constant. Every $\mathbb{K}$-algebra homomorphism is of this form and the mapping $\gamma \to \phi_\gamma$ is a bijection from the set of $\gamma \in R(D')$ such that $\gamma(D') \subset D$ onto the set of $\mathbb{K}$-algebra homomorphisms from $R(D)$ into $R(D')$.

**Proof.** Let $\gamma \in R(D')$ satisfy $\gamma(D') \subset D$. Then it is seen that $\phi_\gamma$ takes values in $R(D')$, is a $\mathbb{K}$-algebra homomorphism, and is obviously injective if and only if $\gamma$ is not a constant.

Conversely, let $\psi$ be a $\mathbb{K}$-algebra homomorphism from $R(D)$ into $R(D')$ and let $\gamma = \psi(I_{D'})$ with $I_{D'}$ the identical mapping in $D'$. Then, we have $\psi(P) = P \circ \gamma$ for every polynomial $P$. On the other hand, if $\alpha \notin D$, then $(x - \alpha)$ is invertible in $R(D)$ and $\psi(\frac{1}{x-\alpha}) = (\psi(x-\alpha))^{-1} = (\gamma-\alpha)^{-1}$. Therefore, $\psi(h) = h \circ \gamma$ whenever $h \in R(D)$. The mapping $\gamma \to \phi_\gamma$ is obviously injective and hence is a bijection. $\square$

**Proposition 12.6.** Let $D, D'$ be sets in $\mathbb{K}$ and let $\gamma \in H(D')$ satisfy $\gamma(D') \subset D$ and $f \circ \gamma \in H(D')$ for all $f \in H(D)$. Let $\phi_\gamma$ be the mapping from $H(D)$ into $H(D')$ defined as $\phi_\gamma(f) = f \circ \gamma$. Then $\phi_\gamma$ is a linear mapping from $H(D)$ into $H(D')$ continuous with respect to the topology of uniform convergence on $D$ for $H(D)$ and on $D'$ for $H(D')$. Moreover, given $f, g \in H(D)$ such that $fg \in H(D)$ we have $\phi_\gamma(fg) = \phi_\gamma(f)\phi_\gamma(g)$. The restriction of $\phi_\gamma$ to $H_b(D)$ is a Banach $\mathbb{K}$-algebra homomorphism from $H_b(D)$ into $H_b(D')$.

If $\gamma$ is a bijection from $D'$ onto $D$ and if $\gamma^{-1} \in H(D)$, then $\phi_\gamma$ is a $\mathbb{K}$-vector space isomorphism from $H(D)$ onto $H(D')$ bicon- continuous with respect to the topology of uniform convergence on $D$ for $H(D)$ and on $H(D')$ for $H(D')$, satisfying $(\phi_\gamma)^{-1} = \phi_{\gamma^{-1}}$ and the restriction of $\phi_\gamma$ to $H_b(D)$ is a Banach $\mathbb{K}$-algebra isomorphism from $H_b(D)$ onto $H_b(D')$. Further, if $\gamma(D') = D$, then the equality $\|\phi_\gamma(f)\|_{D'} = \|f\|_D$ is true for every $f \in H(D)$ and the restriction of $\phi_\gamma$ to $H_b(D)$ is an isometric Banach $\mathbb{K}$-algebra isomorphism from $H_b(D)$ onto $H_b(D')$.

**Proof.** It is easily seen that $\phi_\gamma$ is linear and satisfies $\phi_\gamma(fg) = \phi_\gamma(f)\phi_\gamma(g)$ when $fg \in H(D)$. Next, $\phi_\gamma$ is clearly continuous because $\|\phi_\gamma(f)\|_{D'} = \|f \circ \gamma\|_D = \sup_{x \in D'} |f(\gamma(x))| \le \sup_{u \in D} |f(u)| = \|f\|_D$. In particular, we note that if $\gamma(D') = D$, we have $\|\phi_\gamma(f)\|_{D'} = \|f \circ \gamma\|_D = \sup_{x \in D'} |f(\gamma(x))| = \sup_{u \in D} |f(u)| = \|f\|_D$.
If $f \in H_b(D)$, obviously, $f \circ \gamma \in H_b(D')$. Now, let $\gamma$ be a bijection from $D'$ onto $D$ such that $\gamma^{-1} \in H(D)$. It is seen that $(\phi_{\gamma^{-1}}) \circ \phi_{\gamma} = I_{H(D)}$ while $\phi_{\gamma} \circ (\phi_{\gamma^{-1}}) = I_{H(D')}$, hence $\phi_{\gamma}$ is an isomorphism such that $\phi_{\gamma^{-1}} = (\phi_{\gamma})^{-1}$. $\square$

We study the $K$-algebra homomorphisms from $H(D)$ into $H(D')$.

**Notation.** Given subsets $D$ and $D'$ of $K$, we denote by $\Xi(D', D)$ the set of the $\gamma \in H(D')$ such that $\gamma(D') \subset D$ and such that for every $\lambda \in \bar{D} \setminus D$, $\gamma - \lambda$ is invertible in $H(D')$.

Given two $K$-algebras $A$ and $B$, we denote by $\operatorname{Hom}(A, B)$ the set of $K$-algebra homomorphisms from $A$ into $B$.

**Remark.** In particular, $\Xi(D', D)$ contains the set of the $h \in R(D')$ such that $h(D') \subset D$.

**Theorem 12.7.** Let $D, D' \in \operatorname{Alg}$ and let $\gamma \in \Xi(D', D)$. The mapping $\phi_\gamma$ defined in $H(D)$ by $\phi_\gamma(f) = f \circ \gamma$ has values in $H(D')$ and is a $K$-algebra homomorphism from $H(D)$ into $H(D')$. Conversely, every $K$-algebra homomorphism from $H(D)$ into $H(D')$ is continuous and of this form. Further, the mapping $\gamma \to \phi_\gamma$ from $\Xi(D', D)$ onto $\operatorname{Hom}(H(D), H(D'))$ is a bijection.

Let $D'' \in \operatorname{Alg}$ and let $\tau \in \Xi(D'', D')$. Then, $\gamma \circ \tau \in \Xi(D'', D)$ and $\phi_{\gamma \circ \tau} = \phi_\gamma \circ \phi_\tau$.

Further, a homomorphism $\phi_\gamma$ from $H(D)$ into $H(D')$ is an isomorphism if and only if $\gamma$ is a bijection from $D'$ onto $D$ such that $\gamma^{-1} \in H(D)$ and then, when it is satisfied, we have $(\phi_\gamma)^{-1} = \phi_{\gamma^{-1}}$.

**Proof.** By Corollary 12.3, $f \circ \gamma$ belongs to $H(D')$ whenever $f \in H(D)$, and then by Proposition 12.6, $\phi_\gamma$ is a $K$-algebra homomorphism from $H(D)$ into $H(D')$. Let $\psi \in \operatorname{Hom}(H(D), H(D'))$ and first let us show that $\psi$ satisfies $\|\psi(f)\|_{D'} \le \|f\|_D$ whenever $f \in H(D)$. Indeed, suppose that for some $f \in H(D)$ we have $\|\psi(f)\|_{D'} > \|f\|_{D'}$. Let $g = \psi(f)$. There exists $\alpha \in D'$ such that $|g(\alpha)| > \|f\|_{D'}$. Let $\lambda = g(\alpha)$. The series $\frac{f}{\lambda} \sum_{n=0}^{\infty} (\frac{f}{\lambda})^n$ does converge in $H(D)$ to $(\lambda - f)^{-1}$. Thus, $\lambda - f$ is invertible in $H(D)$ and then $\lambda - g = \psi(\lambda - f)$ is invertible in $H(D')$. But by hypothesis, $\alpha$ is a zero for $\lambda - g$, hence $\lambda - g$ is not invertible in $H(D')$ and this shows that $\|\psi(f)\|_{D'} \le \|f\|_D$. Now, let $\gamma = \psi(I_D) \in H(D')$ and let us show that $\gamma \in \Xi(D', D)$. Let $\alpha \in K \setminus D$. Since $\psi \in \operatorname{Hom}(H(D), H(D'))$, $\psi$ must
satisfy $\psi(\frac{1}{x-\alpha}) = \frac{1}{\psi(x)-\alpha} = \frac{1}{\gamma-\alpha}$, hence $\gamma - \alpha$ has to be invertible in $H(D')$ for every $\alpha \in \mathbb{K}\backslash D$. But this just means that $\gamma \in \Xi(D', D)$.

In the same way, we see that for every $h \in R(D)$, we have $\psi(h) = h(\psi(x)) = h \circ \gamma$. Finally, since $\psi$ is continuous, the equality $\psi(f) = f \circ \gamma$ holds in all $H(D)$. Obviously, given $\gamma, \tau \in \Xi(D', D)$, if $\phi_\gamma = \phi_\tau$, then $\phi_\gamma(I_D) = \phi_\tau(I_D)$, hence $\gamma = \tau$. The mapping $\gamma \to \phi_\gamma$ is then a bijection from $\Xi(D', D)$ onto $\text{Hom}(H(D), H(D'))$.

Now, let $D'' \in \text{Alg}$ and $\tau \in \Xi(D'', D')$. It is seen that $\gamma \circ \tau \in \Xi(D'', D)$ and $\phi_{\gamma \circ \tau}(I_D) = \gamma \circ \tau = \phi_\tau(\gamma) = \phi_\tau(\phi_\gamma(I_D)) = \phi_\tau \circ \phi_\gamma(I_D)$, hence $\phi_{\gamma \circ \tau} = \phi_\tau \circ \phi_\gamma$.

By Proposition 12.6, if $\gamma$ is a bijection from $D'$ onto $D$ and such that $\gamma^{-1} \in H(D)$, then $\phi_\gamma$ is an isomorphism of $\mathbb{K}$-vector space, hence it is an isomorphism of $\mathbb{K}$-algebra, and then, by Proposition 12.6, we have $(\phi_\gamma)^{-1} = \phi_{\gamma^{-1}}$. Conversely, if $\phi_\gamma$ is an isomorphism, then $(\phi_\gamma)^{-1}$ is in the form $\phi_\tau$, with $\tau \in \Xi(D, D')$ and $\phi_\tau \circ \phi_\gamma(I_D) = \gamma \circ \tau(I_D) = I_D$ and $\phi_\gamma \circ \phi_\tau(I_{D'}) = \tau \circ \gamma(I_{D'}) = I_{D'}$. Hence, $\gamma$ is a bijection from $D'$ onto $D$ such that $\gamma^{-1} = \tau \in H(D)$. That finishes showing Theorem 12.7. $\square$

**Definition.** Let $A, B$ be subsets of $\mathbb{K}$. If there exists a bijection $f \in H(A)$ from $A$ onto $B$ such that $f^{-1}$ belongs to $H(B)$, then $f$ is named a *bianalytic element from A onto B*.

Propositions 12.8 and 12.9 are often useful to transform unbounded domains into bounded domains.

**Proposition 12.8.** Let $D \in \text{Alg}$ and let $h \in R(D)$ be a Moebius function. Let $D' = h(D)$. Then, $D'$ belongs to Alg and $H(D')$ is isomorphic to $H(D)$ with respect to the mapping $\psi$ defined in $H(D')$ as $\psi(f) = f \circ h$.

**Proof.** By Theorem 12.2, for every $f \in H(D')$, $f \circ h$ belongs to $H(D)$, and by Proposition 12.6, this mapping is a $\mathbb{K}$-vector space isomorphism which satisfies $\psi(fg) = \psi(f)\psi(g)$ whenever $f, g \in H(D')$. Hence, the space $H(D') = \psi^{-1}(H(D))$ is a $\mathbb{K}$-algebra isomorphic to $H(D)$. In particular, $D'$ belongs to Alg and $\psi$ is a $\mathbb{K}$-algebra isomorphism. $\square$

**Proposition 12.9.** Let $D$ be a set with a hole $T = d(a, r^-)$, let $\gamma(x) = \frac{1}{x-a}$, and let $D' = \gamma(D)$. Then, $D' \in \text{Alg}$ and $H(D')$ is isomorphic to $H_b(D)$.
**Proof.** Without loss of generality, we may clearly assume $D$ to be closed because by Corollary 11.7, $H_b(D)$ is equal to $H_b(\bar{D})$. For every $f \in H(D)$, let $\psi(f) = f \circ \gamma \in H(D')$. Then, $\psi(H_b(D))$ is a $\mathbb{K}$-algebra included in $H(D')$. If $D$ is bounded, $D'$ is bounded and closed like $D$, hence by Proposition 12.8, $\psi$ is an isomorphism from $H(D)$ onto $H(D')$. Now, we suppose $D$ unbounded. Then, $D'$ is bounded and $\psi(H_b(D))$ is obviously included in $H_b(D')$ which, by Corollary 11.7, is just equal to $H(\bar{D}')$. On the other hand, $\gamma$ clearly maps $R_b(D)$ onto $R(\bar{D}')$, hence $\psi(H_b(D)) = H(\bar{D}')$. $\square$

**Theorem 12.10.** Let $T = d(a, r^-)$ be a hole of $D$ and let $\gamma(x) = \frac{1}{x-a}$. Let $D' = \gamma(D)$. The mapping $\psi$ from $H_b(D')$ into $H_b(D)$ defined as $\psi(f) = f \circ \gamma$ is a $\mathbb{K}$-algebra isomorphism.

**Proof.** $D'$ is bounded, hence by Corollary 11.7, $H_b(D')$ is equal to the Banach $\mathbb{K}$-algebra $H(\bar{D}')$. Now, by Theorem 12.2, we see that $\gamma \in \Xi(D, D')$ and $\gamma^{-1} \in \Xi(D', D)$. Hence, $\psi$ is clearly a $\mathbb{K}$-Banach space isomorphism from $H_b(D')$ onto $H_b(D)$. Now, $\psi$ satisfies $\psi(fg) = \psi(f)\psi(g)$ whenever $f, g \in H(D)$ such that $fg \in H(D')$. But both $H_b(D)$, $H_b(D')$ are Banach $\mathbb{K}$-algebras, hence $\psi$ is a Banach $\mathbb{K}$-algebra isomorphism. $\square$

**Theorem 12.11.** Let $A$ be a point in $D$ which is not isolated. Let $\gamma(x) = \frac{1}{x-a}$ and let $D' = \gamma(D \setminus \{a\})$. Then given $f \in H(D')$, $f \circ \gamma$ belongs to $H(D)$ if and only if $f(x)$ has a limit when $|x|$ tends to $+\infty$.

**Proof.** If $f \circ \gamma$ belongs to $H(D)$, then we just have

$$ \lim_{|x| \to +\infty} f(x) = \lim_{x \to a} f \circ \gamma(x) = f \circ \gamma(a). $$

Conversely, if $f$ has a limit $l$ when $|x|$ tends to $+\infty$, then $f \circ \gamma$ is bounded in certain disks $d(a, r) \setminus \{a\}$. Therefore, by Corollary 11.6, $f \circ \gamma$ belongs to $H(D)$. $\square$

**Corollary 12.12.** Let $D \in \text{Alg}$, and let $A$ be a point in $D$ which is not isolated such that $(D \setminus \{a\})$ belongs to $\text{Alg}$. Let $\gamma(x) = \frac{1}{x-a}$ and let $D' = \gamma(D \setminus \{a\})$. Then, $H(D)$ is isomorphic to the subalgebra of $H(D')$ which consists of $f$ such that $|f(x)|$ is bounded when $|x|$ approaches $+\infty$.
# Chapter 13

## Multiplicative Spectrum of $H(D)$

In Chapter 10, we studied and characterized the multiplicative semi-norms on a $\mathbb{K}$-algebra $R(D)$ of rational functions. We apply these properties to the completion $H(D)$ of $R(D)$ by considering multiplicative semi-norms that are continuous with respect to the topology of $H(D)$. On $H(D)$ as on $R(D)$, the role of circular filters is obviously crucial: each continuous multiplicative semi-norm of $H(D)$ is defined by a circular filter secant with $D$ exactly as it was explained for rational functions. However, circular filters that are not secant with $D$ play no role with regards to $H(D)$.

**Notation.** Throughout this chapter, $D$ is an infraconnected subset of $\mathbb{K}$. We denote by $\text{Mult}(H(D), U_D)$ the set of continuous multiplicative semi-norms $\psi$ of the $\mathbb{K}$-vector space $H(D)$ that satisfy $\psi(fg) = \psi(f)\psi(g)$ whenever $f, g \in H(D)$ such that $fg \in H(D)$.

**Remark.** This notation does not require $H(D)$ to be a $\mathbb{K}$-algebra, though it coincides with the notation already introduced for any topological algebra when $H(D)$ is a normed $\mathbb{K}$-algebra. Multiplicative semi-norms appeared to be the main tool for studying analytic elements [36], [44], [61]. They also are at the basis of Berkovich theory [9].

**Theorem 13.1** ((G. Garandel) [61]). For every $\mathcal{F} \in \Phi(D)$, the multiplicative semi-norm $\varphi_{\mathcal{F}}$ defined on $R(D)$ extends by continuity to $H(D)$ to a continuous semi-norm of $\mathbb{K}$-vector space$_{D}\varphi_{\mathcal{F}}$ of $H(D)$ that satisfies$_{D}\varphi_{\mathcal{F}}(f.g) = _{D}\varphi_{\mathcal{F}}(f) _{D}\varphi_{\mathcal{F}}(g)$ whenever $f, g \in H(D)$
such that $fg \in H(D)$. Moreover, the mapping: $\mathcal{F} \to {}_D\varphi_{\mathcal{F}}$, from $\Phi(D)$ into $\text{Mult}(H(D), \mathcal{U}_D)$, is a bijection.

**Proof.** We may obviously extend $\varphi_{\mathcal{F}}$ by continuity to ${}_D\varphi_{\mathcal{F}}$ satisfying ${}_D\varphi_{\mathcal{F}}(fg) = {}_D\varphi_{\mathcal{F}}(f){}_D\varphi_{\mathcal{F}}(g)$ whenever $f, g \in H(D)$ such that $fg \in H(D)$. We now check that the mapping $\mathcal{F} \to {}_D\varphi_{\mathcal{F}}$ from $\Phi(D)$ into $\text{Mult}(H(D), \mathcal{U}_D)$ is a bijection. It is obviously injective because if ${}_D\varphi_{\mathcal{F}_1} = {}_D\varphi_{\mathcal{F}_2}$, then $\varphi_{\mathcal{F}_1} = \varphi_{\mathcal{F}_2}$, hence by Corollary 10.10, $\mathcal{F}_1 = \mathcal{F}_2$. Now, let $\psi \in \text{Mult}(H(D), \mathcal{U}_D)$. The restriction of $\psi$ to $R(D)$ is an element $\psi_0$ of $\text{Mult}(R(D), \mathcal{U}_D)$, hence by Corollary 10.10, $\psi_0$ is of the form $\varphi_{\mathcal{F}}$ and then, by continuity, we have $\psi = {}_D\varphi_{\mathcal{F}}$. $\square$

From Theorem 10.4, the following theorem is immediate concerning a space $H(D)$.

**Corollary 13.2.** Let $\mathcal{F}$ be a large circular filter on $\mathbb{K}$ of diameter $s > 0$. The following three assertions are equivalent:

(i) $\varphi_{\mathcal{F}}(h) \le \|h\|_D$ whenever $h \in H(D)$.

(ii) $\varphi_{\mathcal{F}}$ is a continuous ultrametric multiplicative norm on $H(D)$ with respect to the topology of uniform convergence.

(iii) $\mathcal{F}$ is secant with $D$.

By Corollary 10.10, we have Corollary 13.3:

**Corollary 13.3.** Let $D$ be a closed bounded subset of $\mathbb{K}$. Then, $\text{Mult}(H(D), \|\cdot\|_D)$ is compact with respect to the topology of point-wise convergence.

**Remark.** If $\mathcal{F}$ is a large circular filter, we know that $\varphi_{\mathcal{F}}$ is a norm on $R(D)$. But we don't know whether ${}_D\varphi_{\mathcal{F}}$ is a norm on $H(D)$. This not trivial question is linked to the problem of $T$-filters which is a so big question that it would require another book [38], [44], [53].

**Definition and notation.** For convenience, for every $a \in D$, we put $\varphi_a(f) = |f(a)|$ whenever $f \in H(D)$ and so we define the semi-norms $\varphi_a \in \text{Mult}(H(D), \mathcal{U}_D)$.

An element $\psi \in \text{Mult}(H(D), \mathcal{U}_D)$ is said to be *punctual* if it is of the form $\varphi_a$ with $a \in D$, i.e. if its circular filter is punctual.

Let $\mathcal{F}$ be a monotonous filter on $D$. By Proposition 3.14, there exists a unique circular filter $\mathcal{G}$ on $D$ less thin than $\mathcal{F}$. Then, we put ${}_D\varphi_{\mathcal{F}}(f) = {}_D\varphi_{\mathcal{G}}(f)$ for all $f \in H(D)$.
For simplicity, when $\mathcal{G}$ has center 0 and diameter $r$, we set $|f|(r) = {}_D\varphi_{\mathcal{G}}(f)$. So, when $f$ belongs to $R(D)$, this is the definition already given in Chapter 4.

Now, let $D$ be infraconnected. Let $a \in \tilde{D}$ and let $r$ satisfy $\delta(a, D) \le r \le \text{diam}(D)$. The circular filter $\mathcal{F}$ of center $a$ and diameter $r$ is then secant with $D$. We put ${}_D\varphi_{a,r} = {}_D\varphi_{\mathcal{F}}$. Let $A$ be a bounded subset of $\tilde{D}$ and let $\tilde{A} = d(a, r)$. If $\delta(a, D) \le r \le \text{diam}(D)$, we put ${}_D\varphi_A = {}_D\varphi_{a,r}$. In particular, this notation applies to holes of an infraconnected set $D$.

Let $\mathcal{F}$ be a circular filter or a monotonous filter on $D$. We denote by $\mathcal{J}(\mathcal{F})$ the set of the $f \in H(D)$ such that $\lim_{\mathcal{F}} f(x) = 0$. Hence, $\mathcal{J}(\mathcal{F})$ is equal to $\text{Ker}({}_D\varphi_{\mathcal{F}})$, and therefore, if $D \in \text{Alg}$, $\text{Ker}({}_D\varphi_{\mathcal{F}})$ is a closed prime ideal of $H(D)$.

If $\mathcal{F}$ is a monotonous filter on $D$, we denote by $\mathcal{J}_0(\mathcal{F})$ the set of the $f \in H(D)$ such that $f(x) = 0$ whenever $x \in \mathcal{B}(\mathcal{F})$.

Finally, given $a \in D$, we denote by $\mathcal{J}(a)$ the set of the $f \in H(D)$ such that $f(a) = 0$. Then, if $D \in \text{Alg}$, $\mathcal{J}_0(\mathcal{F})$ and $\mathcal{J}(a)$ are closed prime ideals of $H(D)$.

Among many ultrametric properties, we note the following.

**Lemma 13.4.** Let $\mathcal{F}$ be a circular filter or a monotonous filter on $D$ and let $f \in H(D)$. There exists $A \in \mathcal{F}$ such that $|f(x)|$ is bounded in $A$. Moreover, for every sequence $(a_n)_{n \in \mathbb{N}}$ thinner than $\mathcal{F}$, we have $\lim_{n \in \mathbb{N}} \varphi_{a_n} = \varphi_{\mathcal{F}}$. Given $a, b \in D$ and $r \in ]0, \text{diam}(D)[$ such that $|a-b| \le r$, we have ${}_D\varphi_{a,r} = {}_D\varphi_{b,r}$.

**Proof.** Indeed, there does exist $A \in \mathcal{F}$ such that $|f(x)| \le {}_D\varphi_{\mathcal{F}}(f) + 1$ for all $x \in A$. The last statements come from properties seen on $R(D)$. $\square$

**Lemma 13.5.** Let $f \in H(D)$ be invertible in $H(D)$. Then for every $\psi \in \text{Mult}(H(D), \mathcal{U}_D)$, we have $\psi(f) \ne 0$.

**Proof.** Indeed, we have $\psi(f)\psi(\frac{1}{f}) = 1$. $\square$

**Lemma 13.6.** Let ${}_D\varphi_{\mathcal{F}} \in \text{Mult}(H(D), \mathcal{U}_D)$, let $f \in H(D)$, and let $g \in H(D)$ be such that $\|f-g\|_D < {}_D\varphi_{\mathcal{F}}(f)$. Then, ${}_D\varphi_{\mathcal{F}}(f) = {}_D\varphi_{\mathcal{F}}(g)$.

**Proof.** Indeed, we know that ${}_D\varphi_{\mathcal{F}}(f-g) \le \|f-g\|_D$, hence ${}_D\varphi_{\mathcal{F}}(f-g) < {}_D\varphi_{\mathcal{F}}(f)$, and therefore, ${}_D\varphi_{\mathcal{F}}(g) = {}_D\varphi_{\mathcal{F}}(f)$. $\square$
**Lemma 13.7.** Let $D$ be unbounded and let $f \in H_b(D)$. Then, $|f(x)|$ has a limit $\lim_{D\varphi_D}(f)$ when $|x|$ tends to $+\infty$, while $x$ lies in $D$ and $\lim_{D\varphi_D}(f)$ belongs to $\text{Mult}(H_b(D), \|\cdot\|_D)$.

**Proof.** By Corollary 11.7, $f(x)$ admits a limit $\lambda$ when $|x|$ tends to $+\infty$ ($x \in D$). Hence, $\lim_{|x|\to+\infty, x \in D} |f(x)| = |\lambda|$. Thus, the mapping $\varphi_\infty$, defined as $\varphi_\infty(f) = \lim_{|x|\to+\infty} |f(x)|$, belongs to $\text{Mult}(R_b(D), \|\cdot\|_D)$ and obviously has continuation by continuity to an element $\lim_{D\varphi_D} \in \text{Mult}(H_b(D), \|\cdot\|_D)$ which satisfies $\lim_{D\varphi_D}(f) = \lim_{|x|\to+\infty, x \in D} |f(x)|$. $\square$

**Notation.** When there is no risk of confusion about the set $D$, we just write $\varphi_F$ (respectively $\varphi_{a,r}$, respectively $\varphi_D$, and respectively $\varphi_A$), instead of $\lim_{D\varphi_D}$ (respectively $\lim_{D\varphi_{a,r}}$, respectively $\lim_{D\varphi_D}$, and respectively $\lim_{D\varphi_A}$). Next, when $D$ is unbounded, $\lim_{D\varphi_D}$ is also denoted by $\lim_{D\varphi_\infty}$.

**Theorem 13.8.** Let $\psi \in \text{Mult}(H_b(D), \|\cdot\|_D) \setminus \text{Mult}(H(D), \mathcal{U}_D)$. If $D$ is bounded, $\psi$ is of the form $\varphi_a$ with $a \in \bar{D} \setminus D$. If $D$ is not bounded, $\psi$ is either of the form $\varphi_a$, with $a \in \bar{D} \setminus D$ or of the form $\lim_{D\varphi_\infty}$.

**Proof.** First, we suppose $D$ bounded. By Corollary 11.7, we have $H_b(D) = H(\bar{D})$. Hence, $\psi$ is equal to some $\lim_{D\varphi_F}$, with $F$ a circular filter on $\bar{D}$. If $F$ is large, it is a large circular filter secant with $D$ and then $\psi$ belongs to $\text{Mult}(H(D), \mathcal{U}_D)$. If $F$ is not large, it is the filter of neighborhoods of a point $a \in \bar{D}$. But if $a \in D$, obviously $\varphi_a$ belongs to $\text{Mult}(H(D), \mathcal{U}_D)$. Hence, $a \in \bar{D} \setminus D$.

Now, we suppose $D$ unbounded. If $D$ has no hole, we just have $H_b(D) = R_b(D) = \mathbb{K}$, hence $\text{Mult}(H_b(D), \|\cdot\|_D) = \text{Mult}(H(D), \mathcal{U}_D)$. Thus, we may assume $D$ to have a hole $T = d(a, r^-)$. Without loss of generality, we may assume $a = 0$. Let $\gamma(x) = \frac{1}{x}$ and let $D' = \gamma(D)$. Then, $D'$ is bounded and, by Proposition 12.9, we know that the algebra $H_b(D)$ is isomorphic to $H(D')$. By Proposition 12.8, the mapping $f \to f \circ \gamma$ defines a $\mathbb{K}$-vector space isomorphism from $H(D)$ onto $H(D')$ and a $\mathbb{K}$-algebra isomorphism from $H_b(D)$ onto $H_b(D') = H(\bar{D}')$. Hence, we may define $\psi' \in \text{Mult}(H(\bar{D}'))$ by $\psi'(f \circ \gamma) = \psi(f)$ whenever $f \in H_b(D)$. If $\psi'$ belonged to $\text{Mult}(H(D'), \mathcal{U}_{D'})$, then we would have $\psi(f) = \psi'(f \circ \gamma)$ whenever $f \in H(D)$, and therefore, $\psi \in \text{Mult}(H(D), \mathcal{U}_D)$. Hence, $\psi'$ does not belong to $\text{Mult}(H(D'), \mathcal{U}_{D'})$ and then $\psi'$ is of the form
Multiplicative Spectrum of H(D)

107

$\varphi_b$ with $b \in \overline{D'\setminus D}$. If $b \neq 0$, then $\psi = \varphi_{\frac{1}{b}}$. If $b = 0$, then $\psi = {}_D\varphi_D$ and this ends the proof. $\square$

**Theorem 13.9.** Let $(\alpha_n)_{n \in \mathbb{N}}$ be a bounded sequence in $D$ such that no subsequence converges to any point of $\overline{D} \setminus D$. There exists a subsequence $(\alpha_{n_s})_{s \in \mathbb{N}}$ such that the sequence $(\varphi_{\alpha_{n_s}})_{s \in \mathbb{N}}$ converges in $\text{Mult}(H(D), \mathcal{U}_D)$.

**Proof.** By Theorem 3.1, we may extract either a convergent subsequence, or a monotonous distances subsequence, or an equal distances subsequence from the sequence $(\alpha_n)_{n \in \mathbb{N}}$. Let $(\alpha_{n_s})_{s \in \mathbb{N}}$ be such a subsequence. If this subsequence converges to a point $\alpha \in \mathbb{K}$, then by hypothesis, $\alpha$ lies in $D$, hence $\lim_{s \to +\infty} \varphi_{\alpha_{n_s}} = \varphi_{\alpha}$. If this subsequence is a monotonous distances subsequence, or an equal distances subsequence, then by Proposition 3.18, on $D$ there exists a large circular filter $\mathcal{F}$ less thin than the sequence $(\alpha_{n_s})$ and then we see that for every $f \in H(D)$ we have $\lim_{s \to +\infty} |f(\alpha_{n_s})| = {}_D\varphi_{\mathcal{F}}(f)$, hence $\lim_{s \to +\infty} \varphi_{\alpha_{n_s}}(f) = \varphi_{\mathcal{F}}(f)$. Thus, in every case, we have proven that the subsequence $\varphi_{\alpha_{n_s}}$ converges in $\text{Mult}(H(D), \mathcal{U}_D)$. $\square$
# Chapter 14

## Power and Laurent Series

A power series on a *p*-adic field admits a disk of convergence whose radius is defined in the same way as on $\mathbb{C}$. The difference in behavior between power series in $\mathbb{C}$ and in a field such as $\mathbb{K}$ concerns what happens when $|x|$ is equal to the radius of convergence. We show that the norm of uniform convergence in a disk $d(a, s) \subset d(0, R^{-})$ is multiplicative and satisfies $\| \sum_{n=0}^{+\infty} a_n x^n \|_{d(0,s)} = \sup_{n \in \mathbb{N}} |a_n| s^n$. As a consequence, the product of two power series converging in $d(0, R^{-})$ is bounded if and only if both are bounded. We show that the algebra of power series with a radius of convergence equal to $R$ is equal to the intersection of algebras of analytic elements $H(d(0, s))$ when $s < R$. We show that all analytic elements in $d(0, R^{-})$ are power series converging in $d(0, R^{-})$. The converse is false. However, we see that the analytic elements in $d(0, R)$ are exactly the power series converging in this disk.

**Definitions.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ be a power series with coefficients in $\mathbb{K}$.

As usual, when $\limsup_{n \to \infty} \sqrt[n]{|a_n|} \neq 0$, we call *radius of convergence* of $f$ the number $r = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}$ (with $r = 0$ when $\limsup_{n \to \infty} \sqrt[n]{|a_n|} = +\infty$).

When $\limsup_{n \to \infty} \sqrt[n]{|a_n|} = 0$, we define the radius of convergence of $f$ as $+\infty$.

**Example.** Let $f(x) = \sum_{n=1}^{\infty} nx^n$. The radius of convergence of this series is 1. This function obviously defines the rational function $\frac{x}{(1-x)^2}$ in $d(1, 1^{-})$.
**Remark.** If a sequence of positive numbers $(u_n)_{n \in \mathbb{N}}$ is such that the sequence $(\frac{u_{n+1}}{u_n})_{n \in \mathbb{N}}$ converges to a limit $l \ge 0$, then so does the sequence $(\sqrt[n]{u_n})_{n \in \mathbb{N}}$. On the field $\mathbb{K}$, as in Archimedean analysis, it is a way to compute easily many radii of convergence.

Lemma 14.1 is immediate.

**Lemma 14.1.** Let $f = \sum_{n=0}^{\infty} a_n x^n$ be a power series with coefficients in $\mathbb{K}$. The series converges if and only if $\lim_{n \to \infty} |a_n x^n| = 0$. Let $r$ be its radius of convergence. If $|x| < r$, then the series converges. If $|x| > r$, then the series diverges.

**Definition and notation.** Power series whose radius of convergence is $\infty$ are called *entire functions on $\mathbb{K}$* and the set of entire functions is denoted by $\mathcal{A}(\mathbb{K})$.

For every $a \in \mathbb{K}$, $r \in \mathbb{R}_+^*$, similarly we denote by $\mathcal{A}(d(a, r^-))$ the set of power series in $x-a$ whose radius of convergence is superior or equal to $r$ and by $\mathcal{A}_b(d(a, r^-))$ the set of functions $f \in \mathcal{A}(d(a, r^-))$ that are bounded in $d(a, r^-)$. The set $\mathcal{A}(d(a, r^-)) \setminus \mathcal{A}_b(d(a, r^-))$ is denoted by $\mathcal{A}_u(d(a, r^-))$.

Similarly, we denote by $\mathcal{A}(\mathbb{K}\setminus d(a, r))$ the set of Laurent series converging whenever $|x-a| > r$, by $\mathcal{A}_b(\mathbb{K}\setminus d(a, r))$ the set of bounded Laurent series converging whenever $|x-a| > r$, and by $\mathcal{A}_u(\mathbb{K}\setminus d(a, r))$ the set of unbounded Laurent series converging whenever $|x-a| > r$.

Finally, given $r', r''$ such that $0 < r' < r''$, we denote by $\mathcal{A}(\Gamma(a, r', r''))$ the set of Laurent series converging whenever $r' < |x-a| < r''$. And we denote by $\mathcal{A}_b(\Gamma(a, r', r''))$ the set of functions $f \in \mathcal{A}(\Gamma(a, r', r''))$ that are bounded in $\Gamma(a, r', r'')$.

From Lemma 14.1, we can derive Corollary 14.2:

**Corollary 14.2.** Let $\sum_{n=-\infty}^{+\infty} a_n x^n$ be a Laurent series with coefficients in $\mathbb{K}$, let $r'' = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}$ with $r'' = 0$ whenever $\limsup_{n \to \infty} \sqrt[n]{|a_n|} = +\infty$, and let $r' = \frac{1}{\limsup_{n \to -\infty} \sqrt[n]{-|a_n|}}$ with $r' = 0$ whenever $\limsup_{n \to -\infty} \sqrt[n]{-|a_n|} = +\infty$. If $r' < |x| < r''$, the series converges. If $|x| > r''$ or if $|x| < r'$, the series diverges.

**Corollary 14.3.** Let $r', r'' \in \mathbb{R}_+$ satisfy $0 < r' < r''$. Then, $\mathcal{A}(\Gamma(0, r', r''))$ is the set of Laurent series $\sum_{n=-\infty}^{+\infty} a_n x^n$ such that $r' \le \frac{1}{\limsup_{n \to -\infty} \sqrt[n]{|a_n|}} \le r''$.
**Corollary 14.4.** Let $r', r'' \in \mathbb{R}_+$ be such that $r' < r''$ and let $f(x) = \sum_{n=-\infty}^{+\infty} a_n x^n \in \mathcal{A}(\Gamma(a, r', r''))$. For each $r \in ]r', r''[$, one has

$$ \lim_{n \to +\infty} |a_n|r^n = \lim_{n \to -\infty} |a_n|r^n = 0. $$

Lemma 14.5 is useful in certain further problems:

**Lemma 14.5.** Let $q \in \mathbb{N}^*$ and let $\mathbb{L}$ be a complete algebraically closed extension of $\mathbb{K}$. Let $f \in \mathcal{A}(d(a, R^{-}))$ and suppose that there exists a power series $g$ with coefficients in $\mathbb{L}$, with radius of convergence $\ge R$ such that $(g(x))^q = f(x)$, $\forall x \in d(a, R^{-})$. Then $g$ has all coefficients in $\mathbb{K}$ and belongs to $\mathcal{A}(d(a, R^{-}))$.

**Proof.** Without loss of generality, we can obviously suppose $a = 0$. Let $f(x) = \sum_{n=0}^{+\infty} b_n x^n$ (with $b_n \in \mathbb{K}$) and let $g(x) = \sum_{n=0}^{+\infty} a_n x^n$. Then, $(a_0)^q = b_0$, hence $a_0 \in \mathbb{K}$ because $\mathbb{K}$ is algebraically closed. Now, suppose we have proven that $a_n \in \mathbb{K}$, $\forall n \le t - 1$. We can see that $b_t$ is of the form $a_t(a_0)^{q-t} + h$, where $h$ is a polynomial in $a_0, a_1, \dots, a_{t-1}$. Therefore, $a_t$ also belongs to $\mathbb{K}$. Consequently, $g$ has all coefficients in $\mathbb{K}$, which ends the proof. $\square$

**Notation.** Let $r \in \mathbb{R}_+^*$ and let $\sum_{n=-\infty}^{+\infty} a_n x^n \in H(C(0, r))$. By hypothesis, we have $\lim_{n \to +\infty} |a_n|r^n = \lim_{n \to -\infty} |a_n|r^n = 0$. Generalizing the notation already introduced for rational functions, we denote by $\nu^+(f, \log r)$ the highest of the integers $m \in \mathbb{Z}$ such that $|a_m|r^m = \sup_{n \in \mathbb{Z}} |a_n|r^n$ and by $\nu^-(f, \log r)$ the lowest of the integers $m \in \mathbb{Z}$ such that $|a_m|r^m = \sup_{n \in \mathbb{Z}} |a_n|r^n$. Next, when $\nu^+(f, \log r) = \nu^-(f, \log r)$, we just set $\nu(f, \log r)$.

Recall that we have, given a circular filter $\mathcal{F}$ of center 0 and diameter $r$, for every element of $H(d(0, r))$ and particularly for every analytic function $f \in \mathcal{A}(\mathbb{K})$, put $|f|(r) = \lim_{\mathcal{F}} |f(x)|$.

**Theorem 14.6.** Let $r \in \mathbb{R}_+^*$, let $\mathcal{F}$ be the circular filter of center 0 and diameter $r$ on $\mathbb{K}$, and let $E = d(0, r)$. Then, $H(E)$ is the set of power series $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ such that $\lim_{n \to +\infty} |a_n|r^n = 0$ and we have $\|f\|_E = |f|(r) = \max_{n \in \mathbb{N}} |a_n|r^n = \mathcal{E} \varphi_{\mathcal{F}}(f) = \|f\|_{C(0,r)}$.

For every $\alpha \in E$, $H(E)$ is also equal to the set of series $f(x) = \sum_{n=0}^{+\infty} b_n(x - \alpha)^n$ such that $\lim_{n \to +\infty} |b_n|r^n = 0$.

Let $B = \mathbb{K} \backslash d(0, r^-)$. Then, $H(B)$ is the set of Laurent series $f(x) = \sum_{n=0}^{+\infty} \frac{a_n}{x^n}$ such that $\lim_{n \to +\infty} |a_n|r^{-n} = 0$ and we have $\|f\|_B = \max_{n \in \mathbb{N}} |a_n|r^{-n} = \mathcal{B} \varphi_{\mathcal{F}}(f) = \|f\|_{C(0,r)}$.
For every $\alpha \in d(0, r^{-})$, $H(B)$ is also equal to the set of series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{(x-\alpha)^n}$ such that $\lim_{n \to \infty} |b_n|r^{-n} = 0$.

Let $r' \ge r$ and let $D = \Delta(0, r, r')$. Then $H(D)$ is the set of Laurent series $f(x) = \sum_{-\infty}^{\infty} a_n x^n$ such that $\lim_{n \to -\infty} |a_n|r^n = 0$ and $\lim_{n \to +\infty} |a_n|(r')^n = 0$ and we have $\|f\|_D = \max(\max_{n<0} |a_n|r^n, \max_{n \ge 0} |a_n|(r')^n)$. Moreover, for every $\alpha \in d(0, r^{-})$, $H(D)$ is also equal to the set of power series $f(x) = \sum_{-\infty}^{\infty} b_n (x-\alpha)^n$ such that $\lim_{n \to -\infty} |b_n|r^n = 0$ and $\lim_{n \to +\infty} |b_n|(r')^n = 0$.

**Proof.** Let $S(r)$ be the set of power series $f(x) = \sum_{n=0}^{\infty} a_n x^n$ such that $\lim_{n \to \infty} |a_n|r^n = 0$. Such a power series obviously is a uniform limit of polynomials because $|f(x) - \sum_{n=0}^{q} a_n x^n| \le \sup_{n \ge q} |a_n|r^n$ and hence it belongs to $H(E)$. Moreover, $E$ is closed and bounded, hence by Theorem 11.2, $H(E)$ is a $\mathbb{K}$-Banach algebra with respect to the norm of uniform convergence on $E$. By Lemma 4.7, on $\mathbb{K}[x]$ the norm $\| \cdot \|_E$ is $\varphi_{\mathcal{F}}$, and by Theorem 13.1, that equality has continuation to $H(E)$.

Now, for a polynomial $P(x) = \sum_{n=0}^{q} a_n x^n$, by Lemma 4.7, we have $\|P\|_E = \sup_{0 \le n \le q} |a_n|r^n$, hence this equality also has continuation to $f$. Consequently, $\|f\|_E = \lim_{\mathcal{F}} |f(x)| = \varphi_{\mathcal{F}}(f)$. Particularly, $S(r)$ is a subset of $H(E)$.

In order to show that $S(r) = H(E)$, we first show that $S(r)$ is closed in $H(E)$. Since $\mathcal{F}$ is secant with $C(0, r)$, we have $\|f\|_{C(0,r)} = {}_{E}\varphi_{\mathcal{F}}(f)$. But we know that $\varphi_{\mathcal{F}}(P_n) \le \max_{0 \le i \le n} |a_i|r^i$. Since $\varphi_{\mathcal{F}}$ extends continuously to ${}_{E}\varphi_{\mathcal{F}} \in \text{Mult}(H(E), \| \cdot \|_E)$, for $n$ big enough, we have ${}_{E}\varphi_{\mathcal{F}}(f) = {}_{E}\varphi_{\mathcal{F}}(P_n) = |a_j|r^j$ with $j < n$ and $|a_n|r^m < |a_j|r^j$ whenever $m < j$, hence finally ${}_{E}\varphi_{\mathcal{F}}(f) = |a_j|r^j = \max_{n \in \mathbb{N}} |a_n|r^n$. Consequently, we have $\|f\|_{C(0,r)} = \|f\|_E \le |a_j|r^j = \max_{0 \le n \le q} |a_n|r^n$, and therefore, $\|f\|_{C(0,r)} = \|f\|_E = \max_{0 \le n \le q} |a_n|r^n$. This finishes showing that $S(r)$ is a closed subset of $H(E)$.

Now, we show that $R(E)$ is included in $S(r)$. For this, we just have to show that, given any $\beta \in \mathbb{K} \setminus E$, $(\frac{1}{x-\beta})^q = (-\frac{1}{\beta} \sum_{n=0}^{\infty} (\frac{x}{\beta})^n)^q$ belongs to $F$. When developing $(\sum_{n=0}^{\infty} (\frac{x}{\beta})^n)^q$, we see that for every fixed $q \in \mathbb{N}$, the coefficient $A_q$ of $x^q$ is a sum of terms of the form $\frac{s}{\beta^q}$, with $s \in \mathbb{N}$, hence finally, $|A_q| \le \frac{1}{|\beta|^q}$, and therefore, $|A_q|r^q \le (\frac{r}{|\beta|})^q$. Since $|\beta| > r$, this shows that $(\frac{1}{x-\beta})^q \in S(r)$. So, we have proven
the inclusion $R(E) \subset S(r) \subset H(E)$. Since $S(r)$ is closed, we have $S(r) = H(E)$.

Now, let $\alpha \in E$. Since $d(\alpha, r) = d(0, r)$, after the change of variable $x = \alpha + u$, the same reasoning shows that a series $f(x) = \sum_{n=0}^{\infty} a_n x^n \in H(E)$ is also of the form $\sum_{n=0}^{\infty} b_n(\alpha)(x - \alpha)^n$ with $\lim_{n \to \infty} |b_n(\alpha)|r^n = 0$. Conversely, $H(E)$ is clearly equal to the set of series $\sum_{n=0}^{\infty} b_n(x + \alpha)^n$ such that $\lim_{n \to \infty} |b_n|r^n = 0$ because any series $g$ of the form $\sum_{n=0}^{\infty} b_n u^n$, with $\lim_{n \to \infty} |b_n|r^n = 0$, can be written as $\sum_{n=0}^{\infty} a_n(u + \alpha)^n$.

The statements about $H(B)$ are an obvious consequence of those about $H(E)$ after the change of variable $y = \frac{1}{x}$ and more generally, $y = \frac{1}{x-\alpha}$. So, are the statements about $H(D)$. $\square$

We can easily check the following corollaries:

**Corollary 14.7.** Let $f \in A(\mathbb{K})$. The following three statements are equivalent:

(i) $\lim_{r \to +\infty} \frac{|f(r)|}{r^q} = +\infty, \forall q \in \mathbb{N}$.

(ii) There exists no $q \in \mathbb{N}$ such that $\lim_{r \to +\infty} \frac{|f(r)|}{r^q} = 0$.

(iii) $f$ is not a polynomial.

**Corollary 14.8.** Let $f, g \in A(\mathbb{K})$. Then $f.g$ is a polynomial if and only if both $f, g$ are polynomials.

**Corollary 14.9.** Let $r \in \mathbb{R}_+^*$ and let $D = d(0, r)$. Then, $H(D)$ is the set of power series $f(x) = \sum_{n=0}^{\infty} a_n x^n$ such that $\lim_{|n|_{\infty} \to \infty} |a_n|r^n = 0$ and we have

$$ \|f\|_D = \max_{n \in \mathbb{N}} |a_n|r^n = {}_D\varphi_{\mathcal{F}}(f). $$

Moreover, the norms $\| \cdot \|_{C(0,r)}$, $| \cdot |(r)$ and $\| \cdot \|_{d(0,r)}$ are equal and are multiplicative.

**Corollary 14.10.** Let $r \in \mathbb{R}_+^*$ and let $D = d(0, r)$ (respectively $D = d(0, r^{-})$). Then the norm $\| \cdot \|_D$ on $H(D)$ is multiplicative.

**Corollary 14.11.** Let $\alpha \in D$ and $r \in \mathbb{R}_+^*$ be such that $d(\alpha, r) \subset D$. Let $f \in H(D)$. In $d(\alpha, r)$, $f(x)$ is equal to a power series of the form $\sum_{n=0}^{\infty} a_n(x - \alpha)^n$ such that $\lim_{n \to \infty} |a_n|r^n = 0$. If $f(\alpha) = 0$ and if $f(x)$ is not identically zero in $d(\alpha, r)$, then there exists a unique
integer $q \in \mathbb{N}^*$ such that $a_n = 0$ for every $n < q$ and $a_q \neq 0$ and $\alpha$ is an isolated zero of $f$ in $d(\alpha, r)$.

**Proposition 14.12.** Let $r \in \mathbb{R}_+^*$ and let $f(x) = \sum_{n=0}^{\infty} a_n x^n$. The following statements are equivalent:

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

(b) $f \in \bigcap_{s<r} H(d(0, s))$.

(c) The series $f$ is convergent in all of $d(0, r^{-})$.

**Proof.** (b) and (c) are clearly equivalent to the condition $\lim_{n \to \infty} |a_n|s^n = 0$ whenever $s < r$ and, in the same way as in archimedean analysis, it is shortly checked that this is also equivalent to $\limsup_{n \to \infty} \sqrt[n]{|a_n|} \le \frac{1}{r}$. $\square$

**Remark.** If $f$ is convergent for some $\alpha \in C(0, r)$, then $\lim_{n \to \infty} |a_n|r^n = 0$, hence $f$ belongs to $H(d(0, r))$.

**Corollary 14.13.** Let $r \in \mathbb{R}_+^*$ and let $f(x) = \sum_{n=-\infty}^{0} a_n x^n$. The following statements are equivalent:

(a) $f \in \mathcal{A}(\mathbb{K}\setminus d(0, r))$.

(b) $f \in \bigcap_{s>r} H_b(\mathbb{K}\setminus d(0, s))$.

(c) The series $f$ is convergent in all of $\mathbb{K}\setminus d(0, r)$.

**Corollary 14.14.** Let $r_1, r_2 \in \mathbb{R}_+^*$ with $r_1 < r_2$ and let $f(x) = \sum_{n=-\infty}^{+\infty} a_n x^n$. The following statements are equivalent:

(a) $f \in \mathcal{A}(\Gamma(0, r_1, r_2))$.

(b) $f \in \bigcap_{r_1<s_1<s_2<r_2} H(\Delta(0, s_1, s_2))$.

(c) The series $f$ is convergent in all of $\Gamma(0, r_1, r_2)$.

**Corollary 14.15.** Let $f \in \mathcal{A}(d(0, r^{-}))$ be not identically zero. For every $\alpha \in d(0, r^{-})$, $f(x)$ is equal to a power series $\sum_{n=0}^{\infty} b_n(\alpha)(x - \alpha)^n$. If $f$ is not identically zero and if $\alpha$ is a zero of $f$ in $d(0, r^{-})$, $\alpha$ is an isolated zero and $f$ factorizes in $\mathcal{A}(d(0, r^{-}))$ in the form $(x - \alpha)^q g(x)$, with $q \in \mathbb{N}^*$ and $\alpha \neq 0$.

**Definition.** Let $f \in H(D)$ and let $\alpha \in \mathring{D}$, let $r > 0$ be such that $d(\alpha, r) \subset D$ and suppose $f(x) = \sum_{n=q}^{\infty} b_n(x - \alpha)^n$ whenever $x \in d(\alpha, r)$, with $b_q(\alpha) \neq 0$ and $q > 0$. Then, $\alpha$ is called a zero of
multiplicity order $q$, or more simply, a zero of order $q$. In the same way, $q$ is named the multiplicity order of $\alpha$.

**Remark.** In particular, these definitions apply to functions $f \in \mathcal{A}(d(a, r^{-}))$, at any point $\alpha \in d(a, r^{-})$.

**Corollary 14.16.** Let $a \in \mathbb{K}$, $R \in \mathbb{R}_+^*$ and let $f \in \mathcal{A}(d(a, R^{-}))$ (respectively $f \in \mathcal{A}(\mathbb{K}))$. Let $a_1, \dots, a_q$ be zeros of $f$ of respective order $s_j$ and let $P(x) = \prod_{j=1}^q (x - aj)^{s_j}$. Then, $f$ factorizes in the form $P(x)u(x)$ with $u \in \mathcal{A}(d(a, R^{-}))$ (respectively $u \in \mathcal{A}(\mathbb{K})$).

**Corollary 14.17.** Let $a \in \mathbb{K}$, $R$, $R' \in \mathbb{R}_+^*$ with $R < R'$ and let $\Lambda = d(a, R^{-})$ (respectively $\Lambda = \mathbb{K}\backslash d(a, R)$, respectively $\Lambda = \Gamma(a, R, R')$) and let $C(b, r)$ be a circle included in $\Lambda$. Then, $\Lambda\varphi_{b,r}$ applies to $\mathcal{A}(\Lambda)$.

**Corollary 14.18.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(\mathbb{K})$. If $f$ is not a constant, then $\lim_{r \to +\infty} |f(r)| = +\infty$.

**Notation.** Let $R \in \mathbb{R}_+^*$ and let $f \in \mathcal{A}(d(a, R^{-}))$. Given $r \in ]0, R[$, by Proposition 14.12, $f$ belongs to $H(d(a, r))$, hence for every circular filter $\mathcal{F}$ secant with $d(a, r)$, $\varphi_{\mathcal{F}}(f)$ is defined. Particularly, if $a = 0$, $|f|(r)$ is defined.

**Theorem 14.19.** Let $R \in \mathbb{R}_+^*$ and let $f \in \mathcal{A}(d(a, R^{-}))$. Then, $f$ is invertible in $\mathcal{A}(d(a, R^{-}))$ if and only if $f$ has no zero in $d(a, R^{-})$.

**Proof.** Suppose that $f$ has no zero in $d(a, R^{-})$. For each $r \in ]0, R[$, $f$ belongs to $H(d(a, r))$, and hence by Lemma 11.3, it is invertible in $H(d(a, r))$. Consequently, the function defined in $d(a, r)$ as $g(x) = \frac{1}{f(x)}$ belongs to $H(d(a, r))$. This is true for all $r \in ]0, R[$ and shows that $f^{-1}$ belongs to $\mathcal{A}(d(a, R^{-}))$. The converse is obvious. $\square$

**Theorem 14.20.** Let $R \in \mathbb{R}_+^*$. The $\mathbb{K}$-subalgebra $\mathcal{A}_b(d(0, R^{-}))$ of $\mathcal{A}(d(0, R^{-}))$ is a Banach $\mathbb{K}$-algebra with respect to the norm $\| \cdot \|_{d(0,R^{-})}$. Further, this norm is multiplicative and satisfies $\|f\|_{d(0,R^{-})} = \lim_{r \to R} |f|(r) = \sup_{n \in \mathbb{N}} |a_n|R^n$.

Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, R^{-}))$. Then, $f$ is bounded in $d(0, R^{-})$ if and only if so is the sequence $(|a_n|R^n)_{n \in \mathbb{N}}$. Moreover, if $f$ is bounded, then $\|f\|_{d(0,R^{-})} = \sup_{n \in \mathbb{N}} |a_n|R^n$.

**Proof.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}_b(d(0, R^{-}))$. By Theorem 14.6, we have $\|f\|_{d(0,R^{-})} = \sup_{n \in \mathbb{N}} |a_n|R^n$. The norm $\| \cdot \|_{d(0,R^{-})}$ is a norm
of $\mathbb{K}$-algebra, hence $\|f g\|_{d(0,R^{-})} \le \|f\|_{d(0,R^{-})} \|g\|_{d(0,R^{-})}$. On the other hand, by Theorem 14.6, the norm $\| \cdot \|_{d(0,s)}$ is multiplicative on $H(d(0,s))$ for every $s < R$, hence $\|fg\|_{d(0,R^{-})} \ge \|fg\|_{d(0,s)} = \|f\|_{d(0,s)} \|g\|_{d(0,s)}$ whenever $s < R$, and therefore, $\| \cdot \|_{d(0,R^{-})}$ is multiplicative on $A_b(d(0, R^{-}))$. Now, let $(f_m)_{m \in \mathbb{N}}$ be a Cauchy sequence in $A_b(d(0, R^{-}))$. We put $f_m(x) = \sum_{n=0}^{\infty} a_{n,m} x^n$. By hypothesis, for every $\epsilon > 0$, we have an integer $N(\epsilon)$ such that $|a_{n,m} - a_{n,q}|R^n \le \epsilon$ for every $n \in \mathbb{N}$, whenever $m, q \ge N(\epsilon)$. Thus, it is easily seen that each sequence $(a_{n,m})_{m \in \mathbb{N}}$ converges in $\mathbb{K}$ to a limit $a_n$ that satisfies $|a_n - a_{n,m}|R^n \le \epsilon$ whenever $m \ge N(\epsilon)$ and then the series $f(x) = \sum_{n=0}^{\infty} a_n x^n$ satisfies $\|f - f_m\|_{d(0,R^{-})} \le \epsilon$. Obviously, $f$ belongs to $H(d(0,s))$ for all $s < R$ and then the sequence $(f_m)$ is proven to converge in $A_b(d(0, R^{-}))$.

For every $s \in ]0, R[$, we have

$$|f|(s) = \|f\|_{d(0,s)} = \sup_{n \in \mathbb{N}} |a_n| s^n \le \sup_{n \in \mathbb{N}} |a_n| R^n = \|f\|_{d(0,R^{-})},$$

and hence we can check that the real increasing bounded function $h$ defined in $]0, R[$ as $h(s) = \sup_{n \in \mathbb{N}} |a_n| s^n$ is obviously continuous at $R$. Consequently, $\|f\|_{d(0,R^{-})} = \lim_{r \to R} |f|(r) = \sup_n |a_n| R^n$. Therefore, obviously, $f$ is bounded in $d(0, R^{-})$ if and only if so is the sequence $(|a_n| R^n)_{n \in \mathbb{N}}$. $\square$

**Corollary 14.21.** Let $R \in \mathbb{R}_+^*$ and let $f, g \in A(d(a, R^{-}))$. Then $fg$ belongs to $A_b(d(a, R^{-}))$ if and only if so do both $f, g$ and $A_b(d(a, R^{-}))$ is $\mathbb{K}$-subalgebra of $A(d(a, R^{-}))$.

**Theorem 14.22.** Suppose that $\mathbb{K}$ has characteristic different from 2. Let $f, g \in A(\mathbb{K}) \setminus \mathbb{K}$ (respectively $f, g \in A_u(d(0, r^{-}))$) be distinct. Then, $f^2 - g^2$ belongs to $A(\mathbb{K}) \setminus \mathbb{K}$ (respectively $f^2 - g^2 \in A_u(d(0, r^{-}))$).

**Proof.** Indeed, $f^2 - g^2 = (f - g)(f + g)$. Suppose that $|f + g|(r)$ is bounded when $r$ tends to $+\infty$ (respectively to $R$). Since the characteristic of $\mathbb{K}$ is different from 2, $|f - g|(r)$ is obviously unbounded when $r$ tends to $+\infty$ (respectively to $R$). Consequently, since the norm $| \cdot |(r)$ is multiplicative, $|f^2 - g^2|(r)$ cannot be bounded when $r$ tends to $+\infty$ (respectively to $R$). Therefore, $f^2 - g^2$ belongs to $A(\mathbb{K}) \setminus \mathbb{K}$ (respectively to $A_u(d(0, r^{-}))$). Similarly, if $|f - g|(r)$ is bounded when $r$ tends to $+\infty$ (respectively to $R$), we have the symmetric proof. $\square$
**Theorem 14.23.** For every $r \in \mathbb{R}_+^*$, $H(d(0, r^-))$ is included in $\mathcal{A}_b(d(0, r^-))$.

**Proof.** Since $\mathcal{A}_b(d(0, r^-))$ is complete with respect to the norm $\| \cdot \|_{d(0, r^-)}$, we just have to show that $R(d(0, r^-)) \subset \mathcal{A}_b(d(0, r^-))$, hence finally, we just have to show that given $\alpha \in \mathbb{K} \setminus d(0, r^-)$, $\frac{1}{x-\alpha} \in \mathcal{A}_b(d(0, r^-))$. But we have $\frac{1}{x-\alpha} = -\frac{1}{\alpha(1-\frac{x}{\alpha})} = -\frac{1}{\alpha} \sum_{n=0}^{\infty} \left(\frac{x}{\alpha}\right)^n$ for all $x \in d(0, r^-)$ because $|\frac{x}{\alpha}| < 1$, hence $\frac{1}{x-\alpha} \in \mathcal{A}_b(d(0, r^-))$ and that finishes proving Theorem 14.23. $\square$

**Remarks.** We see later that $H(d(0, r^-))$ is much smaller than $\mathcal{A}_b(d(0, r^-))$. In particular, we see that $\sqrt[4]{1+x}$ belongs to $\mathcal{A}_b(d(0, 1^-))$ but does not belong to $H(d(0, 1^-))$.

Let $\sum_{n=0}^{+\infty} a_n x^n$ be a power series whose radius of convergence is $r$. Suppose first that $r \in |\mathbb{K}|$. If there is at least one point $\alpha \in C(0, r)$ such that the series converges at $\alpha$, then this implies that $\lim_{n \to +\infty} |a_n|r^n = 0$ and hence the series converges in all $C(0, r)$ and defines an element of $H(d(0, r))$. If $r$ does not belong to $|\mathbb{K}|$, the power series converging in $d(0, r)$ are just the power series converging in $d(0, r^-)$. This is why we don't have to consider *analytic functions* inside a disk $d(a, r)$.

**Theorem 14.24.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, R^-))$ and suppose that $a_n \in \mathbb{Q}_p, \forall n \in \mathbb{N}$. Then, for every $x \in d(0, R^-)$, if $x$ is algebraic over $\mathbb{Q}_p$, so is $f(x)$.

**Proof.** Suppose $x$ is algebraic, of degree $q$ over $\mathbb{Q}_p$ and let $\mathbb{E} = \mathbb{Q}_p[x]$. For every $m \in \mathbb{N}$, $\sum_{n=0}^m a_n x^n$ belongs to $\mathbb{E}$. But since $\mathbb{E}$ is a finite extension of $\mathbb{Q}_p$, it is complete, hence $f(x)$ also belongs to $\mathbb{E}$. $\square$
# Chapter 15

## Krasner Mittag-Leffler Theorem

The wonderful Mittag-Leffler theorem for analytic elements is due to Marc Krasner who showed it on quasi-connected sets [69]. The same proof holds on infraconnected sets as it was shown by Philippe Robba [83]. The theorem shows that a Banach space $H_b(D)$ is a direct topological sum of elementary subspaces and is indispensable to have a clear image of the space $H(D)$. Further, it appears necessary when studying meromorphic functions as we see later.

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

We remember that if $D$ is unbounded, $H_0(D)$ denotes the set of the $f \in H(D)$ such that $\lim_{|x| \to +\infty} f(x) = 0$.

**Theorem 15.1** ((M. Krasner) [69], [83]). Let $D$ be closed and bounded (respectively unbounded) and let $f \in H_b(D)$. There exists a unique sequence of holes $(T_n)_{n \in \mathbb{N}^*}$ of $D$ and a unique sequence $(f_n)_{n \in \mathbb{N}}$ in $H(D)$ such that $f_0 \in H(\tilde{D})$ (respectively $f_0 \in \mathbb{K}$), $f_n \in H_0(\mathbb{K} \setminus T_n)$ ($n > 0$), $\lim_{n \to \infty} f_n = 0$ satisfying further:

$$ (1) \quad f = \sum_{n=0}^{\infty} f_n \quad \text{and} \quad \|f\|_D = \sup_{n \in \mathbb{N}} \|f_n\|_D. $$

For every hole $T_n = d(a_n, r_n^{-})$, we have

$$ (2) \quad \|f_n\|_D = \|f_n\|_{K \setminus T_n} = {}_D\varphi_{a_n, r_n}(f_n) \le {}_D\varphi_{a_n, r_n}(f) \le \|f\|_D. $$
If $D$ is bounded and if $\tilde{D} = d(a, r)$, we have

$$ (3) \quad \|f_0\|_D = \|f_0\|_{\tilde{D}} = {}_D\varphi_{a,r}(f_0) \le {}_D\varphi_{a,r}(f) \le \|f\|_D. $$

If $D$ is not bounded, then $|f_0| = \lim_{|x| \to \infty} |f(x)| \le \|f\|_D$.

Let $D' = \tilde{D} \setminus (\cup_{n=1}^\infty T_n)$. Then, $f$ belongs to $H(D')$ (respectively $H_b(D')$) and its decomposition in $H(D')$ is given again by (1) and $f$ satisfies $\|f\|_{D'} = \|f\|_D$.

**Proof.** Since $f \in H_b(D)$, by Corollary 11.7, we know that $f \in H(\tilde{D})$. Hence, without loss of generality, we may assume that $D$ is closed. Obviously, we may also assume $0 \in \tilde{D}$.

First, we suppose $f \in R(D)$. Then, $f$ has decomposition in the form $E(x) + \sum_{j=1}^t \frac{\lambda_j}{(x-\alpha_j)^{q_j}}$ with $E(x) \in \mathbb{K}[x]$ and $\alpha_j \in \mathbb{K} \setminus D$. Now, for each $j$, either $\alpha_j$ belongs to a hole $T$ or $\alpha_j$ belongs to $\mathbb{K} \setminus \tilde{D}$. Let $T_1, \dots, T_s$ be the holes that contain some $\alpha_j$. Then, $\sum_{j=1}^t \frac{\lambda_j}{(x-\alpha_j)^{q_j}}$ is of the form $\sum_{n=1}^s f_n + h_0$ with $f_i \in H_0(\mathbb{K} \setminus T_i)$ and $h_0 \in H_b(\tilde{D})$. Finally, we put $f_0 = E(x) + h_0$ and we have the announced decomposition: $f = \sum_{n=0}^s f_n$ with $f_i \in H_0(\mathbb{K} \setminus T_i)$ and $f_0 \in H_b(\tilde{D})$. In the case when $D$ is unbounded, $f_0$ is just a constant.

For each $i = 1, \dots, s$, $f - f_i$ clearly belongs to $H_b(D \cup T_i)$ and obviously $f$ belongs to $H_b(\tilde{D} \setminus (\cup_{i=1}^s T_i))$.

First, we show that for any $n \in \mathbb{N}^*$, we have $\|f_n\|_D = \|f_n\|_{\mathbb{K} \setminus T_n}$. Let $\mathcal{F}_n$ be the circular filter on $\mathbb{K}$ of center $\alpha_n$ and diameter $r_n$. By Theorem 14.6, we have

$$ (4) \quad \|f_n\|_{\mathbb{K} \setminus T_n} = \lim_{\mathcal{F}_n \cap (\mathbb{K} \setminus T_n)} |f_n(x)|. $$

But by Proposition 3.17, $\mathcal{F}_n$ is secant with $D$, hence

$$ (5) \quad \lim_{\mathcal{F}_n \cap (\mathbb{K} \setminus T_n)} |f_n(x)| = \lim_{\mathcal{F}_n \cap D} |f_n(x)| $$

and obviously,

$$ (6) \quad \lim_{\mathcal{F}_n \cap D} |f_n(x)| \le \|f_n\|_D \le \|f_n\|_{\mathbb{K} \setminus T_n}. $$

Finally, by (4), (5), and (6), we obtain

$$ (7) \quad \|f_n\|_{\mathbb{K} \setminus T_n} = \|f_n\|_D = {}_D\varphi_{\mathcal{F}_n}(f_n). $$
In the same way, when $D$ is bounded, say $\tilde{D} = d(0, r)$, we consider the circular filter $\mathcal{F}_0$ of center 0 and diameter $r$, in order to prove that

$$ (8) \quad \|f_0\|_D = \|f_0\|_{\tilde{D}} = {}_D\varphi_{\mathcal{F}_0}(f). $$

Now, let us show that $\|f\|_D \ge \|f_n\|_D$ for any $n \in \mathbb{N}^*$. Since $f \in R(D)$, there exists an annulus $\Gamma(a_n, r_n, r'_n)$ such that $f$ has neither any zero nor any pole inside $\Gamma(a_n, r_n, r'_n)$. We put $I = ]\log(r_n), \log(r'_n)]$. By hypothesis, $f_n$ has no pole in $K\setminus d(0, r_n)$. Hence, since $\lim_{|x|\to\infty} f_n(x) = 0$, by Corollary 3.17, we see that $\frac{d\Psi_{a_n}}{d\mu}(f_n, \mu) < 0$ whenever $\mu \in I$. Let $g_n = f - f_n \in R(D \cup T_n)$. Since $g_n$ has no pole inside $T_n$, by Corollary 4.17, we see that $\frac{d_r\Psi_{a_n}}{d\mu}(g_n, \mu) \ge 0$ whenever $\mu < \log(r_n)$.

Therefore, the equation $\Psi_{a_n}(f_n, \mu) = \Psi_{a_n}(g_n, \mu)$ has at most one solution in $I$ and then $\Psi_{a_n}(f, \mu)$ is equal to $\max(\Psi_{a_n}(f_n, \mu), \Psi_{a_n}(g_n, \mu))$ whenever $\mu \in I$, hence $\Psi_{a_n}(f, \mu) \ge \Psi_{a_n}(f_n, \mu)$ whenever $\mu \in I$. It follows that the multiplicative semi-norm $\varphi_{\mathcal{F}_n}$ defined on $R(D)$ satisfies $\log(\varphi_{\mathcal{F}_n}(f_n)) = \Psi_{a_n}(f_n, \log(r_n)) \le \Psi_{a_n}(f, \log(r_n)) = \log(\varphi_{\mathcal{F}_n})(f)$, hence

$$ (9) \quad \varphi_{\mathcal{F}_n}(f_n) \le \varphi_{\mathcal{F}_n}(f). $$

But $\varphi_{\mathcal{F}_n}(f_n) = \|f_n\|_D$ and $\varphi_{\mathcal{F}_n}(f) \le \|f\|_D$, hence by (9), we have $\|f\|_D \ge \|f_n\|_D$. Finally, by (7), we see that (2) is clearly proven.

When $D$ is bounded, we put $\tilde{D} = d(0, r)$ and we prove (3) in the same way as above when proving (2) by considering an annulus $\Gamma(0, r, r')$ such that $f$ has neither any zero nor any pole inside $\Gamma(0, r, r')$. Then the element $g_0 = f - f_0$ is of the form $\frac{P}{Q}$ with $P, Q \in \mathbb{K}[x]$, all the zeros of $Q$ in $\tilde{D}$ and $\deg(P) < \deg(Q)$ because $g_0 \in R_0(\mathbb{K}\setminus\tilde{D})$. Hence, we have $\frac{d\Psi}{d\mu}(g_0, \mu) < 0$, while $\frac{d\Psi}{d\mu}(f_0, \mu) \ge 0$ whenever $\mu \in ]\log r, \log r']$, so we have ${}_D\varphi_{\mathcal{F}_0}(f_0) \le {}_D\varphi_{\mathcal{F}_0}(f)$, and hence, by (8), we obtain (3).

When $D$ is not bounded, the inequality $|f_0| \le \|f\|_D$ is obvious because $\lim_{|x|\to\infty} f(x) - f_0 = 0$. This finishes proving the Mittag-Leffler theorem when $f \in R(D)$.

Now, let $f \in H(D)$ and let $(h_m)_{m \in \mathbb{N}}$ be a sequence in $R(D)$ that converges to $f$ in $H(D)$. The set of holes of $D$ that contain at least one pole of some $h_m$ is clearly countable. Hence, there
exists a sequence of holes $(T_n)_{n \in \mathbb{N}^*}$ such that, denoting by $D'$ the set $\tilde{D} \setminus (\bigcup_{n \in \mathbb{N}^*}^{\infty} T_n)$, then $h_m$ belongs to $H(D')$ whenever $m \in \mathbb{N}$. For each $m \in \mathbb{N}$, $h_m$ splits in $H(D')$ in the form $h_m = \sum_{n=0}^{\infty} h_{m,n}$ with $h_{m,0} \in H(\tilde{D})$, $h_{m,n} \in H_0(\mathbb{K}\setminus T_n)$. In particular, for each fixed $n \in \mathbb{N}$, we have $\|h_{m,n} - h_{q,n}\|_D \le \|h_m - h_q\|_D$. Thus, we see that the sequence $(h_{m,n})_{m \in \mathbb{N}}$ converges in $H(\mathbb{K}\setminus T_n)$ for $n > 0$ (respectively in $H(\tilde{D})$ for $n = 0$) to a limit $f_n$ and then we have $f = \sum_{n=0}^{\infty} f_n$ in $H(D')$. Obviously, $\|f\|_D = \sup_{n \in \mathbb{N}} \|f_n\|_D$, whereas $\|f_n\|_D = \|f_n\|_{D'}$ whenever $n \in \mathbb{N}$ and so, $\|f\|_D = \|f\|_{D'}$. This ends the proof of Theorem 15.1. $\square$

**Corollary 15.2.** Let $(T_i)_{i \in I}$ be the family of holes of $D$. Let $J$ be a subset of $I$ and let $S = I \setminus J$. Let $E = D \cup (\bigcup_{i \in J} T_i)$ and let $F = D \cup (\bigcup_{i \in S} T_i)$. Then we have $H(D) = H_0(E) \oplus H(F)$ and for each $g \in H_0(E)$, $h \in H(F)$, we have $\|g+h\|_D = \max(\|g\|_E, \|h\|_F)$.

The Mittag-Leffler theorem suggests some new definitions.

**Definition and notation.** Let $f \in H_b(D)$. We consider the series $\sum_{n=0}^{\infty} f_n$ obtained in Theorem 15.1, whose sum is equal to $f$ in $H(D)$, with $f_0 \in H(\tilde{D})$, $f_n \in H(\mathbb{K}\setminus T_n) \setminus \{0\}$ and with the $T_n$ holes of $D$. Each $T_n$ is called an *f-hole* and $f_n$ is called the *Mittag-Leffler term of f associated with $T_n$*, whereas $f_0$ is called the *principal term of f*. For each *f-hole* $T$ of $D$, the Mittag-Leffler term of $f$ associated with $T$ is denoted by $\overline{f}_T$, whereas the principal term of $f$ is denoted by $\overline{f}_0$. The series $\sum_{n=0}^{\infty} f_n$ is called the *Mittag-Leffler series of f on the infraconnected set D*. More generally, let $E$ be an infraconnected set and let $f \in H(E)$. According to Theorem 11.5, $f$ is of the form $g+h$ with $g \in R(\mathbb{K}\setminus(E\setminus E))$ and $h \in H_b(E)$ whereas such a decomposition is unique, up to an additive constant. For every hole $T$ of $E$, we denote by $\overline{f}_T$ the Mittag-Leffler term of $h$ associated with $T$ and $\overline{f}_T$ is still named *the Mittag-Leffler term of f associated with $T$*.

**Corollary 15.3.** Let $f \in H_b(D)$, let $(T_n)_{n \in \mathbb{N}^*}$ be the sequence of the *f-holes*, with $T_n = d(a_n, \rho_n^{-})$, and let $f_0 = \overline{f}_0$ and $f_n = \overline{f}_{T_n}$ for every $n \in \mathbb{N}^*$. Let $\tilde{D} = d(a, s)$ (respectively $\tilde{D} = \mathbb{K}$). There exists $q \in \mathbb{N}$ such that $\|f\|_D = \|f_q\|_D$. If $q \ge 1$, then $\|f\|_D = {}_D\varphi_{a_q, r_q}(f) = {}_D\varphi_{a_q, r_q}(f_q)$. If $q = 0$ and if $D$ is bounded (respectively is not bounded),
Krasner Mittag-Leffler Theorem

123

then $\|f\|_D = {}_D\varphi_{a,s}(f) = {}_D\varphi_{a,s}(f_0)$ (respectively $\|f\|_D = |f_0|$). Further, given a hole $T$ of $D$, if $f$ belongs to $H_b(D)$ and if $g$ belongs to $H_0(\mathbb{K}\setminus T)$ and satisfies $f - g \in H(D \cup T)$, then $f_T$ is equal to $g$.

**Definition.** Let $D$ be bounded, of center $\alpha$ and diameter $r$. A circular filter $\mathcal{F}$ on $D$ is called the specific filter of a hole $T = d(a, r^{-})$ if it is the circular filter on $D$ of diameter $r$. If $D$ is bounded and $\tilde{D} = d(a, R)$, the circular filter of center $a$ and diameter $R$ is called the specific filter of $\tilde{D}$. In general, a specific filter of a hole of $D$ or of $\tilde{D}$ is called a specific filter of $D$.

**Corollary 15.4.** Let $f \in H_b(D)$. There exists a large circular filter $\mathcal{F}$ with center $\alpha \in \tilde{D}$ secant with $D$ such that ${}_D\varphi_{\mathcal{F}}(f) = \|f\|_D$. If $D$ is bounded, there exists a specific filter $\mathcal{F}$ of $D$ such that ${}_D\varphi_{\mathcal{F}}(f) = \|f\|_D$.

**Corollary 15.5.** Let $f \in H(d(0, 1^{-}))$ and let $(d(\alpha_m, 1^{-}))_{m \in \mathbb{N}^*}$ be the family of the $f$-holes. Then $f$ is of the form

$$ (1) \quad \sum_{n=0}^{\infty} a_{n,0}x^n + \sum_{m,n \in \mathbb{N}^*} \frac{a_{n,m}}{(x - \alpha_m)^n} $$

with $\lim_{n \to \infty} a_n = 0$, $\lim_{n \to \infty} |a_{n,m}| = 0$ whenever $m \in \mathbb{N}^*$ and $\lim_{m \to \infty} (\sup_{n \in \mathbb{N}^*} |a_{n,m}|) = 0$. On the other hand, $f$ satisfies $\|f\|_{d(0,1^{-})} = \sup_{m,n \in \mathbb{N}^*} |a_{n,m}|$.

Conversely, every function of the form (1), with the $\alpha_m$ satisfying $|\alpha_m| = |\alpha_j - \alpha_m| = 1$ whenever $m \neq j$, belongs to $H(d(0, 1^{-}))$. The norm $\| \cdot \|_{d(0,1^{-})}$ is multiplicative and equal to $\|d(0,1^{-})\varphi_{0,1}\|$.

**Corollary 15.6.** Let $r_1, r_2 \in \mathbb{R}_+$ satisfy $0 < r_1 < r_2$. Then, $H(\Delta(0, r_1, r_2))$ is equal to the set of the Laurent series $\sum_{-\infty}^{+\infty} a_n x^n$ with $\lim_{n \to -\infty} |a_n| r_1^n = \lim_{n \to -\infty} |a_n| r_2^n = 0$ and we have $\| \sum_{-\infty}^{+\infty} a_n x^n \|_{\Delta(0, r_1, r_2)} = \max(\sup_{n \ge 0} |a_n| r_1^n, \sup_{n < 0} |a_n| r_2^n)$.

**Proof.** On Theorem 14.6, we saw that $H(\Delta(0, r_1, r_2))$ is equal to the set of the Laurent series $\sum_{-\infty}^{+\infty} a_n x^n$ with $\lim_{n \to -\infty} |a_n| r_1^n = \lim_{n \to -\infty} |a_n| r_2^n = 0$. Then, the conclusions on the norm come from Theorem 15.1. $\square$

**Theorem 15.7.** Let $r \in \mathbb{R}_+$. Then, $H(C(0, r))$ is equal to the set of the Laurent series $\sum_{-\infty}^{+\infty} a_n x^n$ with $\lim_{|n|_\infty \to \infty} |a_n| r^n = 0$ and we have
$\| \sum a_n x^n \|_{C(0,r)} = \sup_{n \in \mathbb{Z}} |a_n| r^n$. Next, the norm $\| \cdot \|_{C(0,r)}$ is multiplicative and equal to $C(0,r) \varphi_{0,r}$.

**Proof.** We put $\Lambda = C(0, r)$. We may apply Theorem 14.6 by taking $r_1 = r_2 = r$ and we obtain all the conclusions but the fact that $\| \cdot \|_{\Lambda}$ is multiplicative. Let us show this. Let $h \in R(\Lambda)$. Hence, $h$ is of the form $\frac{P}{Q}$, with $P, Q \in \mathbb{K}[x]$ and $Q(x)$ has no zero in $\Lambda$. Let $\Theta$ be a class of $\Lambda$. By Lemma 4.9, we have $|Q(x)| = \varphi_{0,r}(Q)$ whenever $x \in \Theta$ and $|P(x)| \le \varphi_{0,r}(P)$ whenever $x \in \Theta$. Hence, we see that $\|\frac{P}{Q}\|_{\Lambda} \le \varphi_{0,r}(\frac{P}{Q})$, and therefore, $\|h\|_{\Lambda} = \varphi_{0,r}(h)$ whenever $h \in R(\Lambda)$. Consequently, we have $\|f\|_{\Lambda} = \varphi_{0,r}(f)$ whenever $f \in H(\Lambda)$. $\square$

**Proposition 15.8.** Let $r_1, r_2 \in \mathbb{R}_+^*$, with $r_1 < r_2$:

(i) $\mathcal{A}(\Gamma(0, r_1, r_2)) = A(d(0, r_2^-)) \oplus \mathcal{A}_0(\mathbb{K}\setminus d(0, r_1))$ and $\mathcal{A}_b(\Gamma(0, r_1, r_2)) = \mathcal{A}_b(d(0, r_2^-)) \oplus \mathcal{A}_{0,b}(\mathbb{K}\setminus d(0, r_1))$.

(ii) Let $f(x) = \sum_{n=-\infty}^{+\infty} a_n x^n \in \mathcal{A}(\Gamma(0, r_1, r_2))$. Then, $f \in \mathcal{A}_b(\Gamma(0, r_1, r_2))$ if and only if $\max(\sup_{n \ge 0} |a_n| r_2^n, \sup_{n < 0} |a_n| r_1^n) < +\infty$. Moreover, if $f \in \mathcal{A}_b(\Gamma(0, r_1, r_2))$, then

$$ \|f\|_{\Gamma(0,r_1,r_2)} = \max \left( \sup_{n \ge 0} |a_n| r_2^n, \sup_{n < 0} |a_n| r_1^n \right). $$

(iii) $\mathcal{A}_b(\Gamma(0, r_1, r_2))$ is a Banach $\mathbb{K}$-algebra that contains $H(\Gamma(0, r_1, r_2))$.

**Proof.** (i) is obvious. We show (ii). Let $f \in \mathcal{A}_b(\Gamma(0, r_1, r_2))$ and let $f = f_1 + f_2$ with $f_2 \in \mathcal{A}_b(d(0, r_2^-))$ and $f_1 \in \mathcal{A}_{0,b}(\mathbb{K}\setminus d(0, r_1))$. We put $\Lambda = \Gamma(0, r_1, r_2)$. It is obviously seen that $\|f\|_{\Lambda} \le \max(\|f_1\|_{\Lambda}, \|f_2\|_{\Lambda}) \le \max(\sup_{n \ge 0} |a_n| r_2^n, \sup_{n < 0} |a_n| r_1^n)$. Now, for every $s_1, s_2$ such that $r_1 < s_1 < s_2 < r_2$, we know that $f$ belongs to $H(\Delta(0, s_1, s_2))$ because so do both $f_1, f_2$. Then, by Theorem 15.1, we have

$$ \|f\|_{\Delta(0,s_1,s_2)} = \max (\|f_2\|_{d(0,s_2)}, \|f_1\|_{\mathbb{K}\setminus d(0,s_1^-)}). $$

Finally, $\|f_2\|_{d(0,s_2)} = \sup_{n \in \mathbb{N}} |a_n| s_2^n$, while $\|f_1\|_{\mathbb{K}\setminus d(0,s_1^-)} = \sup_{n < 0} |a_n| s_1^n$. Thus, we see that $\|f\|_{\Gamma(0,r_1,r_2)} \ge \|f\|_{\Delta(0,s_1,s_2)} = \max(\sup_{n \ge 0} |a_n| s_2^n, \sup_{n < 0} |a_n| s_1^n)$. This is true for every $s_1, s_2 \in ]r_1, r_2[$, hence finally, $\|f\|_{\Lambda} = \max(\sup_{n \ge 0} |a_n| r_2^n, \sup_{n < 0} |a_n| r_1^n)$. All statements in (ii) are then proven.
We now prove (iii). By (ii), $\mathcal{A}_b(\Lambda)$ is just the Banach $\mathbb{K}$-algebra

$$ \mathcal{A}_b(d(0, r_2^-)) \oplus \mathcal{A}_{0,b}(\mathbb{K}\setminus d(0, r_1)) $$

provided with the norm $\|f_1 + f_2\|_{\Lambda} = \max (\|f_2\|_{d(0,r_2^-)}, \|f_1\|_{\mathbb{K}\setminus d(0,r_1)})$. We saw that $R(d(0, r_2^-)) \subset \mathcal{A}_b(d(0, r_2^-))$, hence $R(d(0, r_2^-)) \subset \mathcal{A}_b(\Lambda)$.

In the same way, we have $R(\mathbb{K}\setminus d(0, r_1)) \subset \mathcal{A}_b(\mathbb{K}\setminus d(0, r_1))$ and then $\mathcal{A}_b(\mathbb{K}\setminus d(0, r_1))$ is obviously included in $\mathcal{A}_b(\Lambda)$. Since $R(\Lambda) = R(d(0, r_2^-)) + R(\mathbb{K}\setminus d(0, r_1))$, $R(\Lambda)$ is included in $\mathcal{A}_b(\Lambda)$ which is complete for the norm $\|\cdot\|_{\Lambda}$, hence $H(\Lambda) \subset \mathcal{A}_b(\Lambda)$. This finishes proving Proposition 15.8. $\square$

**Notation.** Given a subset $A$ of $\hat{\mathbb{K}}$, we denote by $\hat{H}(A)$ the set of the analytic elements on $A$, taking $\hat{\mathbb{K}}$ as a ground field.

$E^*$

Now, we apply the Mittag-Leffler theorem to the analytic extension of analytic elements.

**Theorem 15.9.** For all $f \in H(D)$, $f$ has continuation to a unique element $\hat{f} \in \hat{H}(\hat{D})$. Further, if $f \in H_b(D)$, the Mittag-Leffler series of $\hat{f}$ in $\hat{D}$ is the same as this of $f$ in $D$.

**Proof.** By Theorem 11.5, we may easily assume that $f$ belongs to $H_b(\hat{D})$. The Mittag-Leffler series of $f$ on $D$ obviously converges on $\hat{D}$, to an element of $\hat{H}(\hat{D})$. This is unique because so is the Mittag-Leffler series of $f$ on $D$. $\square$

**Theorem 15.10.** Let $E$ be an infraconnected set such that $D \cap E$ is infraconnected and such that every hole of $D \cap E$ is either a hole of $D$ or a hole of $E$. Let $F \in H(D)$, $G \in H(E)$, satisfying $F(x) = G(x)$ whenever $x \in D \cap E$. Then there exists $h \in H(D \cup E)$ such that $h(x) = F(x)$ whenever $x \in D$, $h(x) = G(x)$ whenever $x \in E$, such that for every $h$-hole $V$ of $D \cup E$, $h_V$ is either of the form $F_S$, when $V$ is a $F$-hole $S$ of $D$, or of the form $G_T$ when $V$ is a $G$-hole $T$ of $E$.

**Proof.** By Theorem 11.5, it is easily seen that we may assume $F \in H_b(D)$, $G \in H_b(E)$ without loss of generality. Let $A = D \cup E$, $B = D \cap E$. Let $h$ be the restriction of $F$ and $G$ to $B$. Let $(V_n)_{n \in \mathbb{N}^*}$ be the sequence of $h$-holes that are holes of $D$ and let $(W_n)_{n \in \mathbb{N}^*}$ be
the sequence of $h$-holes that are holes of $E$ but not of $D$. For each $q \in \mathbb{N}^*$, as $h(x)$ is equal to $F(x)$ in $B$, $\overline{h}_{W_q}$ is an element of $H_0(D)$ of the form $\sum_{m=1}^{\infty} \overline{F}_{S_m^q}$ with $S_m^q$ some $F$-holes of $D$ included in $W_q$. We put $f_{q,m} = \overline{F}_{S_m^q}$ for every $(q,m) \in (\mathbb{N}^{*2})$. In the same way, for each $q \in \mathbb{N}^*$, $\overline{h}_{V_q}$ is an element of $H_0(E)$ of the form $\sum_{m=1}^{\infty} \overline{G}_{T_m^q}$ with $T_m^q$ the $G$-holes of $E$ included in $V_q$. We put $g_{q,m} = \overline{G}_{T_m^q}$ for every $(q,m) \in (\mathbb{N}^{*2})$. Without loss of generality, we may obviously assume $\tilde{D} \subset \tilde{E}$; we put $h_0(x) = \overline{G_0(x)}$. We note that $A$ is clearly included in the set $A' = \tilde{E} \setminus (\bigcup_{(q,m) \in (\mathbb{N}^{*2})} S_m^q) \cup (\bigcup_{(q,m) \in (\mathbb{N}^{*2})} T_m^q)$. Then, it is easily seen that the series $h_0(x) + \sum_{(m,q) \in \mathbb{N}^{*2}} f_{q,m} + \sum_{(m,q) \in \mathbb{N}^{*2}} g_{q,m}$ converges in $H(A')$ because by Corollary 15.2 we have $\|f_{q,m}\|_{A'} = \|\overline{F}_{S_m^q}\|_D$ and $\|g_{q,m}\|_{A'} = \|\overline{G}_{T_m^q}\|_E$, whereas $\lim_{q+m \to +\infty} \|\overline{F}_{S_m^q}\|_E = \lim_{q+m \to +\infty} \|\overline{G}_{T_m^q}\|_E = 0$. Further, by construction, $h(x)$ is equal to $F(x)$ and $G(x)$ in $B$ and is such that for every $h$-hole $V$ of $D \cup E$, $\overline{h}_V$ is either of the form $\overline{F}_S$, when $V$ is a $F$-hole $S$ of $D$, or of the form $\overline{G}_T$ when $V$ is a $G$-hole $T$ of $E$. This clearly ends the proof of Theorem 15.10. $\square$

**Corollary 15.11.** Let $E$ be an infraconnected set such that $D \subset E$ and such that each hole of $D$ contains a unique hole of $E$. Let $f \in H(E)$ and let $f = \overline{f_0} + \sum_{n=1}^{+\infty} \overline{f_{T_n}}$ be the Mittag-Leffler series of $f$ on the infraconnected $E$. For every $n \in \mathbb{N}^*$, let $V_n$ be the hole of $D$ containing $T_n$. Then the Mittag-Leffler series of $f$ on the infraconnected $D$ is of the form $\overline{f_0} + \sum_{n=1}^{+\infty} \overline{f_{V_n}}$ with $\overline{f_{V_n}} = \overline{f_{T_n}}$, $\forall n \in \mathbb{N}^*$.

In the particular case of affinoid subsets, we can be more accurate for Theorem 15.10:

**Theorem 15.12.** Let $D_1, D_2$ be infraconnected affinoid subsets of $\mathbb{K}$ such that $D_1 \cap D_2 \neq \emptyset$ and let $f_j \in H(D_j)$, $j = 1, 2$ be such that $f_1(x) = f_2(x)$, $\forall x \in D_1 \cap D_2$. Then the function $f$ defined in $D_1 \cup D_2$ as $f(x) = f_j(x)$ $\forall x \in D_j$, $j = 1, 2$, belongs to $H(D_1 \cup D_2)$.

**Proof.** Let $D = D_1 \cup D_2$. Without loss of generality, we can assume that $\tilde{D}_1$ contains $\tilde{D}_2$ and hence $\tilde{D} = \tilde{D}_1$. We can also assume that $0 \in D_1 \cap D_2$. Set $A = D_1 \setminus D_2$. Then $A$ is included in a finite union of
holes of $D_1$. Consider such a hole $T = d(a, r^-)$ of $D_1$ (with $r \in |\mathbb{K}|$) such that $T \cap D_2 \neq \emptyset$. Since $D_1 \cap D_2 \neq \emptyset$, both $D_1, D_2$ have points on $C(a, r)$. Moreover, since both are affinoid, $D_1$ contains all classes of $C(a, r)$ except maybe finitely many because $T$ is a hole of $D_1$. On the other hand, $D_2$ also contains all classes of $C(a, r)$ except maybe finitely many because it has points on $C(a, r)$ and inside $d(a, r^-)$. Consequently, $f_1(x)$ and $f_2(x)$ coincide in all classes of $C(a, r)$ except maybe in finitely many $\Lambda_1, \dots, \Lambda_q$.

Let $g$ be the Mittag-Leffler term of $f_1$ relative to $T$. Let $S_k, 1 \le k \le t$, be the holes of $D_2$ included in $T$, and for each $k=1, \dots, t$, let $h_k$ be the Mittag-Leffler term of $f_2$ relative to the hole $S_k$. Consider now the restrictions $f_1$ of $f_1$ and $f_2$ of $f_2$ on the set $D' = D_1 \cap D_2$.

The two functions are equal in $D'$ and of course have the same Mittag-Leffler term relative to the hole $d(a, r^-)$. Concerning $h_2$, this term is $\sum_{k=1}^t h_k$. Consequently, $g = \sum_{k=1}^t h_k$. Since $g$ and the $h_k$ are Laurent series converging in $\mathbb{K}\setminus d(a, r^-)$, $g$ and $\sum_{k=1}^t h_k$ coincide in all this set. Consequently, in the Mittag-Leffler series of $f_1$, we can replace $g$ by $\sum_{k=1}^t h_k$. Thus, $f_1$ becomes an element of $D_1 \cup ((d(a, r^-) \cap D_2)$. We can do the same with each hole of $D_1$ containing points of $D_2$, and hence, after finitely many similar change, we obtain an element $f$ of $H(D)$ such that $f(x) = f_j(x) \forall x \in D_j, j = 1, 2$. $\square$

**Notation.** Let $E$ be a $\mathbb{K}$-Banach space. We denote by $E^*$ the $\mathbb{K}$-Banach space of continuous linear forms of $E$ provided with its usual norm. The dual of a Banach space $H(D)$ was thoroughly studied by Yvette Amice [2].

**Theorem 15.13 (Y. Amice).** Let $r \in \mathbb{R}_+$. Given $h(t) = \sum_{n=0}^{\infty} \frac{b_n}{t^n} \in \mathcal{A}_b(\mathbb{K}\setminus d(0, r))$, there exists a unique $\phi_h \in H(d(0, r))^*$ satisfying $\phi_h(x^q) = b_q$, $(q \in \mathbb{N})$. Moreover, on the space $\mathcal{A}_b(\mathbb{K}\setminus d(0, r))$ provided with the norm $\| \cdot \|_{\mathbb{K}\setminus d(0,r)}$, the mapping $h \to \phi_h$ is an isometric isomorphism from $\mathcal{A}_b(\mathbb{K}\setminus d(0, r))$ onto $H(d(0, r))^*$.

**Proof.** Let $F = \mathbb{K}\setminus d(0, r)$. First, let $h(t) = \sum_{n=0}^{\infty} \frac{b_n}{t^n} \in \mathcal{A}_b(F)$ and let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in H(d(0, r))$. Since the sequence $\frac{|b_n|}{r^n}$ is bounded and $\lim_{n \to \infty} |a_n|r^n = 0$, it is seen that $\lim_{n \to \infty} a_n b_n = 0$ and then the series $\sum_{n=0}^{\infty} a_n b_n$ is convergent. Hence, we may put $\phi_h(f) = \sum_{n=0}^{\infty} a_n b_n$. Thus, we define a linear form $\phi_h$ of $H(d(0, r))$
that satisfies

$$|\phi_h(f)| \le \sup_{n \in \mathbb{N}} |a_n b_n| \le \left( \sup_{n \in \mathbb{N}} |a_n| r^n \right) \left( \sup_{n \in \mathbb{N}} \frac{|b_n|}{r^n} \right) = \|f\|_{d(0,r)} \|h\|_F.$$

Therefore, with respect to the norm $\| \cdot \|_F$ of $H((d(0, r))^*)$, we have $\|\phi_h\| \le \|h\|_F$. Now, we check that the equality is satisfied. Indeed, let $q \in \mathbb{N}$. We have

$$\frac{|\phi_h(x^q)|}{\|x^q\|_{d(0,r)}} = \frac{|\phi_h(x^q)|}{r^q} = \frac{|b_q|}{r^q} \le \sup_{f \ne 0} \frac{|\phi_h(f)|}{\|f\|_{d(0,r)}}$$

for all $q \ge 0$. Hence, we have $\|h\|_F = \sup_{q \in \mathbb{N}} \frac{|b_q|}{r^q} \le \|\phi_h\|$. So we obtain the announced equality. Thus, we have defined an isometric homomorphism from $A_b(F)$ into $H(d(0, r))^*$.

Now, we check that this mapping is surjective. Indeed, let $\psi \in H(d(0, r))^*$, and for each $n \in \mathbb{N}$, let $b_n = \psi(x^n)$. Obviously, we have $\|\psi\| \ge \frac{|b_n|}{r^n}$ for every $n \in \mathbb{N}$, hence the sequence $(|b_n| r^{-n})_{n \in \mathbb{N}}$ is bounded and therefore defines a function $f(t) = \sum_{n=0}^{\infty} \frac{b_n}{t^n} \in A_b(F)$. Thus, $\psi$ is equal to $\phi_h$, and therefore, the mapping $h \to \phi_h$ is surjective. This ends the proof of Theorem 15.13. $\square$

**Remark.** There obviously exists an isometric homomorphism from $H(d(0, 1))$ into $H(d(0, 1))^*$ defined as follows: let $f = \sum_{n=0}^{+\infty} a_n x^n \in H(d(0, 1^{-}))$ and let $f(x) = f(\frac{1}{x}) \in A_b(K \setminus d(0, 1))$. Then we have an element $f^* \in H(d(0, 1))^*$ equal to $\phi_f$. The question of whether this homomorphism is surjective depends on the ground field $K$. If $K$ is spherically complete, this homomorphism is not surjective. If $K$ is not spherically complete, this homomorphism is surjective [86].

**Corollary 15.14.** Let $r \in \mathbb{R}_+$. For each $h(t) = \sum_{n=0}^{\infty} b_n t^n \in A_b(d(0, r^{-}))$, there exists a unique $\phi_h \in H(K \setminus d(0, r^{-}))^*$ satisfying $\phi_h(x^{-q}) = b_q$ ($q \in \mathbb{N}$). Moreover, the space $A_b(d(0, r^{-}))$ being provided with the norm $\| \cdot \|_{d(0, r^{-})}$, the mapping $h \to \phi_h$ is an isometric isomorphism from $A_b(d(0, r^{-}))$ onto $H(K \setminus d(0, r^{-}))^*$.

**Corollary 15.15.** Let $r \in \mathbb{R}_+$. For each $h(t) = \sum_{n=0}^{\infty} b_n t^n \in A_b(d(0, r^{-}))$ such that $h(0) = 0$, there exists a unique $\phi_h \in H_0(K \setminus d(0, r^{-}))^*$ satisfying $\phi_h(x^{-q}) = b_q$ ($q \in \mathbb{N}^*$). Moreover, this mapping $h \to \phi_h$ from the subspace of the $h \in A_b(d(0, r^{-}))$ such that $h(0) = 0$ into $H_0(K \setminus d(0, r^{-}))^*$ is an isometric isomorphism.
Now, applying Theorem 15.13 to $H(\tilde{D})$ and Corollary 15.15 to the spaces $H_0(\mathbb{K}\setminus T_i)$ for each hole $T_i$ of an infraconnected set $D$, we obtain Corollary 15.16.

**Corollary 15.16 (Y. Amice).** Let $D$ be closed bounded infraconnected. Let $(T_i)_{i \in J}$ be the family of its holes and for every $i \in J$, let $a_i \in T_i$. Let $M \in \mathbb{R}_+^*$. Let $h_0 \in A_b(\mathbb{K}\setminus \tilde{D})$ and let $(h_i)_{i \in J}$ be a family such that for each $i \in J$, $h_i$ belongs to $A_b(T_i)$ and satisfies

$$ (1) \quad h_i(a_i) = 0 $$

and

$$ (2) \quad \|h_i\|_{T_i} \le M \text{ for all } i \in J. $$

There exists a unique $\psi \in H(D)^*$ satisfying

$$ \psi(f) = \phi_{h_0}(f) \text{ for every } f \in H(\tilde{D}), $$

$$ \psi(f) = \phi_{h_i}(f) \text{ for every } f \in H_0(\mathbb{K}\setminus T_i), \text{ whenever } i \in J. $$

Further, for every element $\psi$ of $H(D)^*$, there exists a unique $h_0 \in A_b(\mathbb{K}\setminus d(0, r))$ and a unique family $(h_i)_{i \in J}$ satisfying (1) and (2) for some $M \in \mathbb{R}_+^*$ such that $\psi$ is defined as above.

Now, we use the continuous linear forms to define the residue of an element on a hole.

**Theorem 15.17.** Let $f \in H_b(\mathbb{K}\setminus d(a, r^-))$, and for each $\alpha \in d(a, r^-)$, let $f(x) = \sum_{n=0}^{\infty} \frac{b_n(\alpha)}{(x-\alpha)^n}$. Then, $b_1(\alpha)$ does not depend on $\alpha$ in $d(a, r^-)$.

**Proof.** Let $E = \mathbb{K}\setminus d(a, r^-)$. We know that $\frac{|b_1(\alpha)|}{r} \le \|f\|_E$, and therefore, fixing $\alpha$ in $d(a, r^-)$, the linear form $\psi_\alpha$ on $H_b(E)$ defined as $\psi_\alpha(f) = b_1(\alpha)$ is obviously continuous. We show that $\psi_\alpha(f) = \psi_a(f)$. First, for every $q \in \mathbb{N}$, we put $f_q(x) = \frac{1}{(x-\alpha)^q}$. We have $f_q(x) = \frac{1}{(x-a)^q} (\sum_{j=0}^\infty (\frac{\alpha-a}{x-a})^j)^q$. Therefore, for every $q \ge 2$, we have $\psi_a(f_q) = 0$ and that $\psi_a(f_1) = 1$. Hence, $\psi_\alpha(f_q) = \psi_a(f_q)$ for every $q \in \mathbb{N}$. This shows that $\psi_\alpha(f) = \psi_a(f)$ for every $f \in H_b(E)$. $\square$

**Definition and notation.** Let $f \in H_b(D)$, let $T$ be a hole of $D$, and let $a \in T$. Let $f_T(x) = \sum_{n=1}^{\infty} \frac{b_n(a)}{(x-a)^n}$. By Theorem 15.17, $b_1(a)$
actually does not depend on $a$ in $T$. We set $\mathrm{res}(f, T) = b_1(a)$ and this number $\mathrm{res}(f, T)$ is called the residue of $f$ on the hole $T$.

By Theorem 15.1, Theorem 15.18 is obvious:

**Theorem 15.18.** Let $f \in H(D)$ and let $T$ be a hole of $D$ of diameter $r$. Then,

$$|\mathrm{res}(f, T)| \le r \|\overline{f_T}\|_{K\backslash T} \le r \|f\|_D.$$

We can now characterize $K$-algebra homomorphisms among the continuous linear forms.

**Theorem 15.19.** Let $D$ be a closed bounded infraconnected, let $a \in \tilde{D}$, let $(T_i)_{i \in J}$ be the family of holes of $D$ and for every $i \in J$, and let $a_i \in T_i$. Let $M \in \mathbb{R}_+^*$. Let $h_0 \in A_b(\mathbb{K}\backslash \tilde{D})$ and let $(h_i)_{i \in J}$ be a family such that for each $i \in J$, $h_i$ belong to $A_b(T_i)$ and satisfy Conditions (1) and (2):

$$ (1) \quad h_i(a_i) = 0 $$

and

$$ (2) \quad \|h_i\|_{T_i} \le M \text{ for all } i \in J. $$

Let $\psi \in H(D)^*$ satisfy

$$ \psi(f) = \phi_{h_0}(f) \text{ for every } f \in H(\tilde{D}), $$

$$ \psi(f) = \phi_{h_i}(f) \text{ for every } f \in H_0(\mathbb{K}\backslash T_i), \text{ whenever } i \in J. $$

Then, $\psi$ is a homomorphism of $K$-algebra from $H(D)$ onto $K$ if and only if there exists $\alpha \in D$ such that

$$ (3) \quad h_0(t) = \frac{t-a}{t-\alpha}, $$

and for every $i \in J$,

$$ (4) \quad h_i(t) = \frac{t-a_i}{\alpha-t}. $$

Moreover, every $K$-algebra homomorphism from $H(D)$ to $K$ is continuous and is of this form.

**Proof.** First, we suppose that $\psi$ is a $K$-algebra homomorphism from $H(D)$ onto $K$ and we put $\psi(x) = a$. As $h_0$ is of the form
$h_0(t) = \sum_{n=0}^{\infty} \frac{b_n}{(t-a)^n}$, here for every $n \in \mathbb{N}$, we have $b_n = \psi((x-a)^n) = (\alpha - a)^n$, and therefore, $h_0(t) = \sum_{n=0}^{\infty} \left(\frac{\alpha-a}{t-a}\right)^n = \frac{t-a}{t-\alpha}$. Next, we fix $i \in J$. Then, $h_i$ is of the form $h_i(t) = \sum_{n=1}^{\infty} b_{i,n}(t-a_i)^n$. Hence, for every $n \in \mathbb{N}^*$, we have $b_{i,n} = \psi(\frac{1}{(x-a_i)^n}) = \left(\frac{1}{\alpha-a_i}\right)^n$, and therefore, $h_i(t) = \sum_{n=1}^{\infty} \left(\frac{t-a_i}{\alpha-a_i}\right)^n = \frac{t-a_i}{\alpha-t}$.

Conversely, we suppose (3) and (4) are satisfied. Then, it is easily checked that $h_0(t) = \sum_{n=0}^{\infty} \left(\frac{\alpha-a}{t-a}\right)^n$, and therefore, for every $n \in \mathbb{N}$, we have $\psi((x-a)^n) = (\alpha-a)^n$. Hence, for every $f \in H(\tilde{D})$, we have $\psi(f) = f(\alpha)$.

In the same way, we check that, fixing $i \in J$, we have $\psi(\frac{1}{(x-a_i)^n}) = (\frac{1}{\alpha-a_i})^n$, hence $\psi(f) = f(\alpha)$ for every $f \in H_0(\mathbb{K}\setminus T_i)$. This clearly finishes proving that $\psi(f) = f(\alpha)$ for every $f \in H(D)$.

Now, let $\psi$ be a $\mathbb{K}$-algebra homomorphism from $H(D)$ to $\mathbb{K}$. By Corollary 10.10, $\psi$ is continuous and hence belongs to $H(D)^*$. Consequently, it is of the form defined by the theorem. $\square$
# Chapter 16

## Factorization of Analytic Elements

In $\mathbb{C}$, it is well known that when a (not identically zero) holomorphic function admits a zero at a point $\alpha$, this zero has a finite order of multiplicity. Actually, this is a generalization of a property of rational functions. In the non-Archimedean context, we find again that property among analytic elements and it is essential. In this chapter, $D$ is just a subset of $\mathbb{K}$.

**Lemma 16.1.** Let $\alpha \in \mathring{D}$. Let $q \in \mathbb{N}^*$ and let $(g_n)$ be a sequence of $H(D)$ such that the sequence $(x - \alpha)^q g_n$ converges in $H(D)$. Then the sequence $(g_n)_{n \in \mathbb{N}}$ also converges in $H(D)$.

**Proof.** Without loss of generality, we can assume $\alpha = 0$. Set $f_n = x^q g_n$, $n \in \mathbb{N}$. Since 0 lies in $\mathring{D}$, there exists a disk $d(0, r) \subset D$. Let $E = D \setminus d(0, r)$. Clearly, we have $\|f_s - f_n\|_{d(0,r)} = \|x^q\|_{d(0,r)} \|g_s - g_n\|_{d(0,r)} = r^q \|g_s - g_n\|_{d(0,r)}$ and hence $\|g_s - g_n\|_E \le \frac{\|f_s - f_n\|}{r^q}$. Consequently, $\|g_s - g_n\|_D \le \frac{\|f_s - f_n\|}{r^q}$, and therefore, the sequence $(g_n)$ is a Cauchy sequence, which ends the proof. $\square$

**Theorem 16.2.** Let $\alpha$ belong to $D \cap \mathring{D}$ and let $f \in H(D)$ be such that $f(\alpha) = 0$. Then, $f$ has factorization in $H(D)$ in the form $(x - \alpha)g$ with $g \in H(D)$. If there is no neighborhood $V$ of $\alpha$ such that $f(x) = 0$ whenever $x \in V$, then there exists a unique integer $q \in \mathbb{N}$ and $h \in H(D)$ such that $f(x) = (x - \alpha)^q h(x)$ and $h(\alpha) \ne 0$ and then $\alpha$ is a zero of order $q$ of $f$.
**Proof.** First, we prove the main factorization in the form $(x-\alpha)g$. We may obviously assume $\alpha = 0$. By hypothesis, there exists a disk $d(0, s)$ included in $\bar{D}$. And then, by Theorem 11.5, there exists a disk $d(0, r)$ included in $d(0, s)$ such that $f$ has no pole in $d(0, r)$. Consequently, $f$ belongs to $H(D \cup d(0, r))$. So, we can assume that $0$ is interior to $D$ and that $d(0, r) \subset D$ without loss of generality. By Corollary 14.11, the restriction of $f$ to $d(0, r)$ is equal to a power series $\sum_{n=1}^{\infty} a_n x^n$ for all $x \in d(0, r)$.

Now, let $t_n$ be a sequence in $R(D)$ such that $\lim_{n \to \infty} t_n = f$. Clearly, $|t_n(0)| \le \|t_n - f\|_D$ because $f(0) = 0$, so we have $\|t_n - t_n(0) - f_n\|_D \le \|t_n - f\|_D$. We put $h_n = t_n - t_n(0)$ ($n \in \mathbb{N}$). The sequence $(h_n)$ of $R(D)$ approaches $f$ in $H(D)$ and satisfies $h_n(0) = 0$ whenever $n \in \mathbb{N}$, hence $h_n$ has factorization in $R(D)$ in the form $xg_n$. By Lemma 16.1, the sequence $(g_n)_{n \in \mathbb{N}}$ converges in $H(D)$. Let $g$ be its limit. We show that $\lim_{q \to \infty} \|xg_q - xg\|_D = 0$. Let $\varepsilon > 0$ and let $N \in \mathbb{N}$ be such that $\|h_n - f\|_D \le \varepsilon$ whenever $n \ge N$. We fix $q \ge N$. Then, $\|h_n - h_q\|_D \le \varepsilon$, hence $|xg_n(x) - xg_q(x)| \le \varepsilon$ whenever $x \in D$. So, when $n$ tends to $+\infty$, we see that $|xg(x) - xg_q(x)| \le \varepsilon$ whenever $x$ in $D$. Thus, we have $\|xg - xg_q\|_D \le \varepsilon$, and therefore, $\lim_{q \to \infty} \|xg_q - xg\|_D = 0$. But, by hypothesis, we have $\lim_{q \to \infty} \|f - xg_q\|_D = 0$ and then $f = xg$.

Now, we suppose that $f$ is not identically zero in $d(0, r)$. Then at least one of the coefficients $a_n$ of its power series is not zero. By Corollary 14.11, $f$ admits 0 as a zero of order $q$ and then $q$ is the smallest integer such that $a_q \ne 0$. In $d(0, r)$, we have $f(x) = \sum_{n=q}^{\infty} a_n x^n = xg(x)$, hence $g(x) = \sum_{n=q}^{\infty} a_n x^{n-q}$ whenever $x \in d(0, r)$. Suppose that $f$ has been proven to be factorized in the form $x^s g_s$ with $s < q$ and $g_s \in H(D)$. Clearly, $g_s(x) = \sum_{n=q}^{\infty} a_n x^{n-s}$ whenever $x \in d(0, r)$, hence $g_s(0) = 0$, and therefore, $g_s$ has factorization in the form $xg_{s+1}$ with $g_{s+1} \in H(D)$. Thus, by induction, we obtain $f = x^q g_q(x)$ with $g_q(x) = \sum_{n=q}^{\infty} a_n x^{n-q}$ and then $g_q(0) = a_q \ne 0$. That finishes proving Theorem 16.2. $\square$

**Notation.** Let $a \in \mathring{D}$ and let $f \in H(D)$ be such that $f(a) = 0$, $f(x) \ne 0$ in a disk $d(a, r)$. The order of the zero $a$ of $f$ is denoted by $\omega_a(f)$.

**Corollary 16.3.** Let $\bar{D}$ be open, let $f \in H(D)$, and let $\alpha$ be a zero of $f$ in $D$. Either there exists a disk $d(\alpha, r)$ such that $f(x) \ne 0$
whenever $x \in d(\alpha, r)\setminus\{\alpha\}$ or there exists a disk $d(\alpha, r)$ such that
$f(x) = 0$ whenever $x \in d(\alpha, r)$.

**Corollary 16.4.** Let $f \in H(D)$ have a zero of order $q$ at a point $\alpha \in \mathring{D}$. Then for every $s = 1, \dots, q$, $f$ factorizes in the form $(x - \alpha)^s g_s$, with $g_s \in H(D)$ having a zero of order $q - s$ at $\alpha$.

**Corollary 16.5.** Let $\alpha \in \mathring{D}$ and let $f \in H(D)$. Let $\sum_{n=0}^{\infty} a_n (x - \alpha)^n$ be its power series in a disk $d(\alpha, r) \subset D$. Let $P(x) = \sum_{n=0}^{q} a_n (x - \alpha)^n$ and let $g(x) = f(x) - P(x)$. Then $g$ factorizes in the form $(x - \alpha)^q h(x)$, with $h \in H(D)$.

**Definition.** Let $A \subset D$ be an open subset of $K$ and let $f \in H(D)$ have finitely many zeros $a_1, \dots, a_n$ in $A$ of multiplicity order of $q_1, \dots, q_n$, respectively. The polynomial $\prod_{i=1}^n (x - a_i)^{q_i}$ is named the polynomial of zeros of $f$ in $A$.

We are now able to give the following corollary.

**Corollary 16.6.** Let $A$ be a subset of $D$ open in $K$, let $f \in H(D)$ have finitely many zeros in $A$, and let $P$ be the polynomial of its zeros in $A$. Then $f$ has a factorization in the form $f = Pg$, with $g \in H(D)$ and $g(x) \neq 0$ whenever $x \in A$.

**Definitions.** An element $f \in H(D)$ is said to be semi-invertible (respectively quasi-invertible) if it factorizes in the form $P(x) g(x)$, with $g$ invertible in $H(D)$ and with $P$ a polynomial whose zeros belong to $D$ (respectively to $D \cap \mathring{D}$).

An element $f \in H(D)$ is said to be quasi-minorated if for every bounded sequence $(a_n)_{n \in \mathbb{N}}$ of $D$ such that $\lim_{n \to \infty} f(a_n) = 0$ we can extract a subsequence that converges in $K$.

**Remarks.** (1) If a semi-invertible element of $H(D)$ has no zero in $D$, it is invertible in $H(D)$. (2) Let $D$ belong to Alg. If $f_1, f_2$ are semi-invertible (respectively quasi-invertible) elements of $H(D)$, then $f_1 f_2$ is also semi-invertible (respectively quasi-invertible). However, when $D$ does not belong to Alg, counter-examples show that the product of two semi-invertible (respectively quasi-invertible) elements is not
always semi-invertible (respectively quasi-invertible). Such counter-examples are given in a further remark.

**Lemma 16.7.** Let $D \in \text{Alg}$, let $f \in H(D)$ be quasi-invertible (respectively quasi-minorated), and let $h \in R(D)$ be a Moebius function. Let $D' = h(D)$ and let $g = f \circ h^{-1}$. Then, $g$ is a quasi-invertible (respectively quasi-minorated) element of $H(D')$.

**Proof.** Suppose first $f$ is quasi-invertible in $H(D)$. Let $u = h(x)$. So, $f$ is of the form $P(x)\phi(x)$ with $P$ a polynomial whose zeros are interior to $D$ and $\phi$ is an invertible element of $H(D)$. Then, $\phi \circ h^{-1}$ is invertible in $H(D')$ and $P \circ h^{-1}$ belongs to $R(D')$ and is of the form $\frac{Q(u)}{(u-b)^s}$, where $b$ is the unique pole of $h^{-1}$. Consequently, $g$ is of the form $Q(u)\frac{\phi(u)}{(u-b)^s}$. Thus, $\frac{\phi(u)}{(u-b)^s}$ is invertible in $H(D')$ and hence $g$ is quasi-invertible.

Now, suppose $f$ quasi-minorated. Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $D'$ such that $\lim_{n \to \infty} g(a_n) = 0$ and let $b_n = h^{-1}(a_n)$, $(n \in \mathbb{N})$. Then, $\lim_{n \to \infty} f(b_n) = 0$. Since $f$ is quasi-minorated, one can extract a subsequence $(b_{q(m)})_{m \in \mathbb{N}}$ that either converges or satisfies $\lim_{m \to \infty} |b_{q(m)}| = \infty$. But then, the sequence $(a_{q(m)})$ either converges or satisfies $\lim_{m \to \infty} |a_{q(m)}| = \infty$. Hence, $f$ is quasi-minorated. $\square$

**Theorem 16.8.** Let $D$ be bounded, open, closed and let $f \in H(D)$. If $f$ is quasi-minorated, then it is quasi-invertible.

**Proof.** We suppose $f$ is not quasi-invertible and we prove that $f$ is not quasi-minorated either.

First, we suppose $f$ to have finitely many zeros. Since $D$ is open, by Corollary 16.6, $f$ has factorization in the form $P(x)g(x)$, with $P$ a polynomial whose zeros are interior to $D$ and $g$ an element of $H(D)$ which has no zero in $D$ but is not invertible in $H(D)$ since $f$ is not quasi-invertible. Hence, there exists a bounded sequence $(\alpha_n)_{n \in \mathbb{N}}$ in $D$ such that $\lim_{n \to \infty} g(\alpha_n) = 0$. If $f$ were quasi-minorated, we could extract a convergent subsequence from the sequence $(\alpha_n)_{n \in \mathbb{N}}$ whose limit would belong to $D$ and would be a zero of $g$. Hence, $f$ is not quasi-minorated when it has finitely many zeros in $D$.

Now, we suppose that $f$ has a sequence of (distinct) zeros $(\alpha_n)$ in $D$ and that $f$ is quasi-minorated. Hence, we may extract a convergent subsequence of limit $\alpha$. Obviously, $\alpha$ is another zero of $f$, hence, by Corollary 16.3, $f(x)$ is equal to zero inside a disk $d(\alpha, r)$ and then
f is not quasi-minorated, a contradiction. That ends the proof of
Theorem 16.8. □

**Theorem 16.9.** Let $D$ be closed and bounded. Let $f \in H(D)$ be quasi-minorated and have no zero in $D$. Then, $f$ is invertible in $H(D)$.

**Proof.** Assume that $\inf_{x \in D} |f(x)| = 0$ and let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $D$ such that $\lim_{n \to \infty} f(a_n) = 0$. Since $D$ is bounded and since $f$ is quasi-minorated, we can extract a subsequence $(a_n)$ which converges in $\mathbb{K}$ to a point $a \in \bar{D}$. Since $D$ is closed, $a$ belongs to $D$ and satisfies $f(a) = 0$, which contradicts the hypothesis. Thus, there exists $\lambda > 0$ such that $|f(x)| \ge \lambda$ whenever $x \in D$ and then by Lemma 11.3, $f$ is invertible in $H(D)$. □
# Chapter 17

## Algebras $H(D)$

We have seen that $H(D)$ is a Banach $\mathbb{K}$-algebra if and only if $D$ is closed and bounded. But studying analytic elements, analytic functions require to know algebras of analytic elements which are not necessarily bounded. Thus, we have to examine the class Alg of subsets $D$ of $\mathbb{K}$ such that $H(D)$ is a $\mathbb{K}$-algebra with respect to usual laws [35], [37], [44], [50].

**Notation.** Throughout this chapter, $D$ denotes a subset of $\mathbb{K}$. Let $f \in H(D)$. According to Theorem 11.5, $f$ is of the form $f^* + \bar{f}$, with $f^* \in R_0(\mathbb{K}\setminus(\tilde{D}\setminus D))$ and $\bar{f} \in H(\bar{D})$. We keep that notation throughout this chapter.

**Proposition 17.1.** Let $\alpha$ belong to $\bar{D}$ and let $f \in H(D\setminus\{\alpha\})$. For every $q \in \mathbb{N}$, $\frac{f}{(x-\alpha)^q}$ belongs to $H(D\setminus\{\alpha\})$.

**Proof.** Since $\alpha$ belongs to $\bar{D}$, there exists a disk $d(\alpha, s)$ included in $\bar{D}$. On the other hand, there exists $r \in ]0, s[$ such that $f^*$ has no pole in $d(\alpha, r)\setminus\{\alpha\}$. Hence, by Theorem 11.10, $f$ is of the form $\frac{g}{(x-\alpha)^t}$ with $g \in H(D \cup d(\alpha, r))$ and $t \in \mathbb{N}$. Then, $\frac{f}{(x-\alpha)^q} = \frac{g}{(x-\alpha)^{q+t}}$. Thus, without loss of generality, we may assume that $\alpha$ belongs to $D$ and that $f$ belongs to $H(D \cup d(\alpha, r))$.

By Corollary 14.11, in $d(\alpha, r)$, $f(x)$ is equal to a power series $\sum_{n=0}^{\infty} a_n(x-\alpha)^n$. Let $P(x) = \sum_{n=0}^{q} a_n(x-\alpha)^n$. By Theorem 16.2, $f(x) - P(x)$ factorizes in the form $(x-\alpha)^q h$ with $h \in H(D)$.
Hence, we see that $\frac{f(x)}{(x-\alpha)^q} = \frac{P(x)}{(x-\alpha)^q} + h$. Since $\frac{P}{f} \in R(D)$, it is

*clear* that $\frac{f}{(x-\alpha)^q}$ belongs to $H(D)$. $\square$

**Corollary 17.2.** Let $f \in H(D)$ and let $P$ be a polynomial whose zeros are interior to $D$. Let $a_1, \dots, a_n$ be the zeros of $P$. Then, $\frac{f}{P}$ belongs to $H(D\setminus\{a_1, \dots, a_n\})$.

**Proposition 17.3.** If $D$ is bounded and satisfies $\bar{D}\setminus D \subset (\frac{\circ}{\bar{D}})$, then $D \in \text{Alg}$.

**Proof.** Let $f,g \in H(D)$ and let us show that $fg \in H(D)$. By Theorem 11.5, we have $f = \bar{f}^* + \bar{f}$, $g = g^* + \bar{g}$ with $f^*, g^* \in R_0(\mathbb{K}\setminus(\bar{D}\setminus D))$ and $\bar{f}, \bar{g} \in H(\bar{D})$. Since $D$ is bounded, by Corollary 11.7, we have $H(\bar{D}) = H_b(D)$ and then $\bar{f}\bar{g}$ obviously belongs to $H(D)$ while $f^*g^* \in R_0(\mathbb{K}\setminus(\bar{D}\setminus D))$. Finally, by Corollary 17.2, both $f^*\bar{g}$ and $g^*\bar{f}$ belong to $H(D)$ and therefore so does $fg$. $\square$

**Definition.** Let $\mathcal{F}$ be a filter in $D$. An element $f \in H(D)$ is said to be vanishing along $\mathcal{F}$ if $\lim_{\mathcal{F}} f(x) = 0$. Further, $f$ is said to be properly vanishing along $\mathcal{F}$ if $\lim_{\mathcal{F}} f(x) = 0$ and if $\|f\|_A \neq 0$ whenever $A \in \mathcal{F}$.

Proposition 17.4 is a polyvalent result which helps us characterize
the sets $D \in \text{Alg}$ but also find conditions for $H(D)$ not to be a
Noetherian algebra.

**Proposition 17.4.** Let $\mathcal{F}$ be a pierced filter on $D$, let $(T_n)_{n \in \mathbb{N}}$ be a sequence of holes of $D$ that runs $\mathcal{F}$, and let $E = \mathbb{K}\setminus(\bigcup_{n=0}^{\infty} T_n)$. Let $g_1, \dots, g_q \in H_b(E)$ be vanishing along $\mathcal{F}$, with $g_1$ properly vanishing. For every $x \in E$ let $S(x) = \sup_{1 \le i \le q} |g_i(x)|$, let $\mathcal{J}$ be the ideal generated by $g_1, \dots, g_q$ in $H_b(E)$ and let $\overline{\mathcal{J}}$ be its closure in $H_b(D)$.

There exists a sequence $(z_n)_{n \in \mathbb{N}}$ in $D$, thinner than $\mathcal{F}$, such that $g_1(z_n) \neq 0$ and an element $G \in \overline{\mathcal{J}}$ such that $\lim_{n \to \infty} \frac{|G(z_n)|}{S(z_n)} = +\infty$.

**Proof.** Without loss of generality, we may assume $\mathcal{F}$ to be a decreasing filter or a Cauchy filter. Indeed, if $\mathcal{F}$ is an increasing filter of center $\alpha$ and diameter $R$, consider a hole of $D$ $T(b, \rho)$ included in $d(\alpha, R^{-})$, take $\gamma(x) = \frac{1}{x-b}$, and set $D' = \gamma(D)$. Then, by Theorem 3.11, $D'$ admits a decreasing pierced filter $\mathcal{F}'$, the image of $\mathcal{F}$ by $\gamma$. Next, $D'$ is clearly bounded. By Theorem 12.7, the mapping
$\phi$ from $D$ onto $D'$ defined by $\phi(f) = f \circ \gamma^{-1}$ is an isomorphism from $H(\overline{D}')$ onto $H_b(D)$. Hence, $\mathcal{J}$ is isomorphic to the ideal generated by $\{g_j \circ \gamma^{-1} | 1 \le j \le q\}$ in $H(\overline{D}')$. Hence, we assume $\mathcal{F}$ to be a decreasing pierced filter or a Cauchy pierced filter.

Without loss of generality, we may now clearly assume $D = E$. Since $g_j$ are bounded, we may obviously assume $\|g_j\|_D \le 1$ whenever $j = 1, \dots, q$. Let $R = \text{diam}(\mathcal{F})$ and let $(x_m)_{m \in \mathbb{N}}$ be a sequence in $D$ thinner than $\mathcal{F}$, such that $g_1(x_m) \ne 0$ whenever $m \in \mathbb{N}$, with $|x_{m+2} - x_{m+1}| < |x_{m+1} - x_m|$. Since $\mathcal{F}$ is pierced, there exists a subsequence $(x_{m_q})_{q \in \mathbb{N}}$ of the sequence $(x_m)$ together with a sequence of holes $(T_q)_{q \in \mathbb{N}}$ of $D$ such that

$$T_q \subset d(x_{m_{q+1}}, d_{m_q}) \setminus d(x_{m_{q+2}}, d_{m_{q+1}}).$$

Hence, without loss of generality, we may assume that we have a sequence of holes $(T_m)_{m \in \mathbb{N}}$ of $D$ such that $T_m \subset d(x_{m+1}, d_m) \setminus d(x_{m+2}, d_{m+1})$.

We put $D_m = d(x_{m+1}, d_m) \cap D$ and $A_n = D_{2n+1} \setminus D_{2n+3}$. For each $n$, let $u_n \in A_n$ be such that $|g_1(u_n)| \ge \|g_1\|_{A_n} (\frac{n}{n+1})$. For each $j = 1, \dots, t$, let $M_n^j = \|g_j\|_{A_n}$ and let $M_n = \max_{1 \le j \le t} M_n^j$. Since $g_1(x_m) \ne 0$, we have $M_n > 0$ whenever $n \in \mathbb{N}$, and since $\|g_j\|_D \le 1$ for all $j$, we have $M_n \le 1$ whenever $n \in \mathbb{N}$.

We construct a sequence $(U_n)$ in $H_b(D)$ satisfying

$$(1) \quad |U_n(x)| \le \frac{1}{n+1} \quad \text{whenever } x \in D \setminus A_n,$$

$$(2) \quad \sqrt{M_n} \left( \frac{n+1}{n} \right) > \|g_1 U_n\|_{A_n} > \sqrt{M_n}.$$

For every $n \in \mathbb{N}$, set $T_n = d(\beta_n, \rho_n^{-})$, $u_n = x_{2n+2}$, $a_n = \beta_{n+1}$, $b_n = \beta_{n+2}$, $c_n = \beta_{2n+3}$, and $\epsilon_n \in d(0, \frac{1}{n})$. Let us fix $n \in \mathbb{N}$. It is seen that $|u_n - a_n| > |u_n - b_n|$, hence there exists $q_n \in \mathbb{N}$ such that

$$(3) \quad |\epsilon_n| \left| \frac{u_n - a_n}{u_n - b_n} \right|^{q_n} g(u_n) > \sqrt{M_n}$$

and of course there exists $q'_n$ such that

$$(4) \quad \left( \frac{d_{2n+1}}{d_{2n+2}} \right)^{q_n} \left( \frac{d_{2n+3}}{d_{2n+2}} \right)^{q'_n} < 1.$$

$$\text{We put } h_n(x) = \epsilon_n \left( \frac{x-a_n}{x-b_n} \right)^{q_n} \left( \frac{x-c_n}{x-b_n} \right)^{q'_n}.$$
Then, by (4), we see that

when $|x - c_n| > d_{2n+1}$, we have $|h_n(x)| = |\epsilon_n| < \frac{1}{n}$,

when $|x - c_n| \le d_{2n+3}$, we have $|x - a_n| = |a_n - c_n| = d_{2n+1}$ and $|x - b_n| = |b_n - c_n| = d_{2n+2}$, hence $|h_n(x)| \le |\epsilon_n| \left(\frac{d_{2n+1}}{d_{2n+2}}\right)^{q_n} \left(\frac{d_{2n+3}}{d_{2n+2}}\right)^{q'_n} < \frac{1}{n}$.

But now we note that $x$ belongs to $D \setminus A_n$ if and only if it satisfies: either $|x - c_n| > d_{2n+1}$ or $|x - c_n| \le d_{2n+3}$, hence we have proven that $|h_n(x)| < \frac{1}{n}$ whenever $x \in D \setminus A_n$. This shows $h_n$ satisfies (1).

When $x \in A_n$ i.e. when $d_{2n+3} < |x - c_n| < d_{2n+1}$, we see that $\|g_1 h_n\|_{A_n} \ge |g_1(u_n)h_n(u_n)|$, hence, by (3), we have $\|g_1 h_n\|_{A_n} \ge \sqrt{M_n}$. Hence, there trivially exists $\lambda_n \in (0, 1)$ such that $(\frac{n+1}{n})\sqrt{M_n} > |\lambda_n g_1 h_n|_{A_n} > \sqrt{M_n}$.

Now, we put $U_n = \lambda_n h_n$ and we see that $U_n$ satisfies (1) and (2). In particular, we have $\|g_1 U_n\|_D \le \max(\sqrt{M_n}(\frac{n+1}{n}), \frac{\|g_1\|_D}{n+1})$, hence $\lim_{n \to \infty} \|g_1 U_n\|_D = 0$. Let $T = \sum_{n=0}^{\infty} g_1 U_n$. By definition, $T$ belongs to $\bar{F}$ because for every $t \in \mathbb{N}$, $g \sum_{n=0}^{t} U_n$ belongs to $F$.

By (2), there exists a sequence $(z_n)_{n \in \mathbb{N}}$ in $D$ satisfying $z_n \in A_n$ and

$$ (5) \quad \sqrt{M_n} < |g_1(z_n)U(z_n)| < M_n\left(\frac{n+1}{n}\right), $$

hence we have

$$ (6) \quad |U_n(z_n)| > \frac{\sqrt{M_n}}{|g_1(z_n)|} \ge \frac{1}{\sqrt{M_n}} \quad \text{because } |g_1(z_n)| \le M_n. $$

When $j \ne n$, $z_n$ belongs to $D \setminus A_j$, hence by (1) and (6), we have $|U_j(z_n)| < \frac{1}{j+1} < \frac{1}{\sqrt{M_n}} < |U_n(z_n)|$ whenever $j \ne n$. Hence, we see that $|T(z_n)| = |g_1(z_n)U_n(z_n)|$ whenever $n \in \mathbb{N}$. But then, by (5), we see that $\frac{|T(z_n)|}{S(z_n)} = \frac{|T(z_n)|}{M_n} > \frac{1}{\sqrt{M_n}}$. Consequently, $\lim_{n \to \infty} \frac{|T(z_n)|}{S(z_n)} = +\infty$ and this finishes the proof of Lemma 17.4. $\square$

**Corollary 17.5.** Let $a \in D \setminus D^\circ$ and let $\mathcal{F}$ be the pierced filter of the neighborhoods of $a$. There exists a sequence $(z_n)_{n \in \mathbb{N}}$ in $D$ of limit $a$ and an element $G \in H_b(D)$ vanishing along $\mathcal{F}$ such that $\limsup_{n \to \infty} \frac{|G(z_n)|}{|z_n - a|} = +\infty$.
**Lemma 17.6.** Let $D$ have a hole $T = d(a, r^-)$. Let $\gamma(x) = b + \frac{\lambda}{x-a}$ with $\lambda \in \mathbb{K}$ and let $D' = \gamma(D)$. For every $\alpha \in \bar{D}$, $\alpha$ belongs to $(\frac{\circ}{\bar{D}})$ if and only if $\gamma(\alpha)$ belongs to $(\frac{\circ}{\bar{D}'})$. Moreover, if $D$ is not bounded, then $\mathbb{K}\setminus\bar{D}$ is bounded if and only if $b$ belongs to $(\frac{\circ}{\bar{D}'})$.

**Proof.** $\gamma$ is obviously a bicontinuous bijection from $\mathbb{K}\setminus\{a\}$ onto $\mathbb{K}\setminus\{b\}$. Let $\alpha \in (\frac{\circ}{\bar{D}})$. There exists a disk $d(\alpha, r)$ included in $\bar{D}$. Since $a \notin d(\alpha, r)$, $\gamma$ is bounded in $d(\alpha, r)$. Since $\gamma$ is bicontinuous, $\gamma(d(\alpha, r))$ is open in $\mathbb{K}\setminus\{a\}$, hence it is clearly open in $\mathbb{K}$. So, $\gamma(\alpha)$ belongs to $\gamma(\frac{\circ}{\bar{D}})$. But $\gamma(\bar{D}) \subset \bar{\gamma}(\bar{D})$, hence $\gamma(\alpha) \in (\frac{\circ}{\bar{D}'})$. Let $\xi = \gamma^{-1}$. Then, $\xi(u) = a + \frac{\lambda}{u-b}$ and then what is true for $\gamma$ is also true for $\xi$. Hence, conversely, if $\gamma(\alpha) \in (\frac{\circ}{\bar{D}'})$, we see that $\alpha = \xi(\gamma(\alpha)) \in \frac{\circ}{\bar{D}'}$ because $D = \xi(D')$.

We now suppose $D$ unbounded. If $\mathbb{K}\setminus\bar{D}$ is bounded, then $\bar{D}$ contains a set $E$ of the form $\{|x| \mid x - a \mid \ge s\}$ with $s > |a - b|$ whose image $E'$ is $d(a, \frac{|\lambda|}{s})\setminus\{a\}$, hence $\bar{D}'$ contains $d(a, \frac{|\lambda|}{s})$ and so does $\frac{\circ}{\bar{D}'}$.

Finally, we suppose that $\mathbb{K}\setminus\bar{D}$ is not bounded. Then, $\gamma((\mathbb{K}\setminus\{a\})\setminus\bar{D})$ is an open set $E$ in $\mathbb{K}$ whose closure contains $b$. Since $b \in \frac{\circ}{\bar{D}}\setminus D'$, it is easily seen that there is a sequence of holes of $D'$, $(T_n)_{n \in \mathbb{N}}$, that approaches $b$ (each one is obviously included in $E$), hence $b \notin (\frac{\circ}{\bar{D}'})$. This ends the proof of Lemma 17.6. $\square$

**Corollary 17.7.** Let $D$ have a hole $T = d(a, r^-)$, satisfy $\bar{D}\setminus D \subset (\frac{\circ}{\bar{D}})$, and be such that $\mathbb{K}\setminus\bar{D}$ is bounded. Let $\gamma = \frac{1}{x-a}$ and let $D' = \gamma(D)$. Then, $\bar{D}'\setminus D' \subset (\frac{\circ}{\bar{D}'})$.

**Lemma 17.8.** The following two conditions are equivalent:

(A) $\tilde{D}\setminus\bar{D}$ is bounded.

(A') Either $D$ is bounded or $\mathbb{K}\setminus\bar{D}$ is bounded.

**Proof.** If $D$ is bounded, (A) and (A') are clearly satisfied, hence we have nothing to show. Now, we suppose $D$ to be unbounded.
Hence, $\tilde{D} = \mathbb{K}$ and then $\tilde{D} \setminus \overline{D} = \mathbb{K} \setminus \overline{D}$, so $\tilde{D} \setminus \overline{D}$ is bounded if and only if $\mathbb{K} \setminus \overline{D}$ is bounded. Finally, (A) and (A') are equivalent. $\square$

**Theorem 17.9.** *D belongs to Alg if and only if it satisfies the following two conditions:*

(A) $\tilde{D} \setminus \overline{D}$ is bounded.

(B) $\overline{D} \setminus D \subset (\overset{\circ}{\overline{D}})$.

**Proof.** Suppose first that $D$ satisfies A and B. By Lemma 17.8, it satisfies A' and B. Suppose first that $D$ is bounded. Since $D$ satisfies B, by Proposition 17.3, $D \in \text{Alg}$. Suppose now that $\mathbb{K} \setminus \overline{D}$ is bounded. We may obviously assume $D$ to have a hole $T = d(a, r^-)$ because if $D$ has no hole, then, by Corollary 11.8, we have $H(D) = R(D)$. Let $\gamma = \frac{1}{x-a}$ and let $D' = \gamma(D)$. The set $D'$ is then bounded, and, by Corollary 17.7, $D'$ satisfies $\overline{D'} \setminus D' \subset (\overset{\circ}{\overline{D'}})$ and hence $D' \in \text{Alg}$, therefore by Proposition 17.3, $D$ belongs to $\text{Alg}$.

We now suppose that (B) is not satisfied and prove that $D \notin \text{Alg}$.
Indeed, let $\alpha \in (\overline{D} \setminus D) \setminus (\overset{\circ}{\overline{D}})$. By definitions, $D$ has a Cauchy pierced filter $\mathcal{F}$ that converges in $\mathbb{K}$ to $\alpha$. Let $T = d(a, r^-)$ be a hole of $D$ and let $f = \frac{x-\alpha}{x-a}$. Then, $f \in R_b(D)$. By Proposition 17.4, there exists $S \in H_b(D)$ such that $S(\alpha) = 0$ together with a sequence $(z_n)_{n \in \mathbb{N}}$ in $D$ such that $\lim_{n \to \infty} z_n = \alpha$, while $\lim_{n \to \infty} \frac{|S(z_n)|}{|f(z_n)|} = +\infty$.

Let us assume $D \in \text{Alg}$. Then, $\frac{S(x)}{f(x)} \in H(D)$ because $\frac{x-a}{x-\alpha} \in R(D)$. Since $|\frac{S(x)}{f(x)}|$ is not bounded in any neighbourhood of $\alpha$, by Theorem 11.10 and Corollary 11.9, there exists an integer $n \ge 1$ such that $(x - \alpha)^n \frac{S(x)}{f(x)}$ has a non-zero limit $\ell$ at $\alpha$. But when $x \in D \cap d(\alpha, |a|)$, we have $|(x - \alpha)^n \frac{S(x)}{f(x)}| = |(x - \alpha)^{n-1}(x - a)S(x)| = |a| |x - \alpha|^{n-1} |S(x)|$, hence $\ell = 0$, a contradiction. Consequently, $D \notin \text{Alg}$.

Suppose now that $D$ satisfies (B) but does not satisfy (A). Both $D$ and $\mathbb{K} \setminus \overline{D}$ are unbounded. Since $\mathbb{K} \setminus \overline{D}$ is unbounded, $D$ has a hole $T = d(\alpha, r^-)$, and then by Lemma 17.6, the inversion $\gamma(x) = \frac{1}{x-\alpha}$ maps $D$ onto a bounded set $D'$ such that $\alpha \in (\overline{D'} \setminus D') \subset (\overset{\circ}{\overline{D'}})$. Hence, $D'$ does not belong to Alg and neither does $D$. This ends the proof of Theorem 17.9. $\square$
**Corollary 17.10.** If $D$ is bounded and if $\bar{D}$ is open, then $D$ belongs to Alg.

**Notation.** Throughout this book, Conditions (A) and (B) are those given in Theorem 17.9.

**Theorem 17.11.** Let $D$ belong to Alg and have a hole $T = d(a, r^-)$. Let $h(x) = \frac{1}{x-a}$ and let $D' = h(D)$. Then, $D'$ is bounded and belongs to Alg.

**Proof.** Since $D \in \text{Alg}$, by Theorem 17.10, $D$ satisfies Conditions (A) and (B) and hence we can check that $D'$, being obviously bounded, satisfies (A) and (B) too. If $D$ is bounded, this is immediate. If $D$ is not bounded, then 0 is the unique point that might not belong to $\bar{D}' \setminus D' \subset (\bar{D}')^o$. But if 0 is the limit of a sequence of holes of $D'$, then $\bar{D} \setminus \bar{D}$ is not bounded, which contradicts Condition (A). $\square$

**Lemma 17.12.** Let $D$ be open. Then, $D$ satisfies Condition (B) if and only if $\bar{D}$ is open.

**Proof.** Since $D$ is open, $\bar{D}$ is open if and only if for every $\alpha \in \bar{D} \setminus D$, $\alpha$ is interior to $D$. This is just equivalent to Condition (B). $\square$

**Notation.** Let $D \in \text{Alg}$ and let $a \in D$. We denote by $\mathcal{J}(a)$ the ideal of the $f \in H(D)$ such that $f(a) = 0$.

**Theorem 17.13.** Let $D \in \text{Alg}$ and let $a \in D$. If $a$ belongs to $\bar{D}$, then $\mathcal{I}(a) = (x-a)H(D)$. Else, $\mathcal{J}(a)$ is not of finite type.

**Proof.** Suppose first $a \in \bar{D}$. By Theorem 16.2, it is clearly seen that $\mathcal{J}(a) = (x-a)H(D)$. Now, let $a \notin \bar{D}$. Then the filter of the neighborhoods of $a$ is a Cauchy pierced filter. We denote it by $\mathcal{F}$. Suppose that $\mathcal{J}(a)$ is of finite type and let $\{g_1, \dots, g_q\}$ be a system of generators. For each $j = 1, \dots, s$, $q$, let $Q_j$ be the polynomial of poles of $g_j$ in $\bar{D} \setminus D$. Now, let $d(b, r^-)$ be a hole of $D$. By Corollary 11.7, for each $j = 1, \dots, q$, there exists a rational function of the form $\frac{1}{(x-b)^{t_j}}$ such that the function $h_j = \frac{g_j}{Q_j(x-b)^{t_j}}$ belongs to $H_b(\bar{D})$.
Of course, at least one of $g_j$ is properly vanishing along $\mathcal{F}$, other-
wise all the elements of $\mathcal{J}(a)$ would be equal to 0 inside a neighbor-
hood of $a$ and then $\mathcal{J}(a)$ couldn't contain $x-a$. Consequently, we can
assume that $g_1$ is properly vanishing and so is $h_1$. For every $x \in D$, we
put $S(x) = \max_{1 \le j \le q} |h_j(x)|$. We note that $\mathcal{J}(a)$ is obviously closed
in $H(D)$, hence by Proposition 17.4, there exists $f \in \mathcal{J}(a)$ together
with a sequence $(z_n)_{n \in \mathbb{N}}$ in $D$, of limit $a$, such that $S(z_n) \ne 0$ when-
ever $n \in \mathbb{N}$ and that $\lim_{n \to \infty} \frac{|f(z_n)|}{S(z_n)} = +\infty$. This obviously contra-
dicts the fact that $f$ should be of the form $\sum_{j=1}^q f_j g_j$ with the $f_j$ in
$H(D)$. Thus we have shown that $\mathcal{J}(a)$ is not of finite type and this
ends the proof of Theorem 17.13.
$\square$

**Corollary 17.14.** Let $D \in \text{Alg. If } D \setminus \frac{\partial}{D} \neq \emptyset$ then $H(D)$ is not noetherian.
Chapter 18

Derivative of Analytic Elements

Given an infraconnected set, the main question we consider here is whether an element $f$ of $H(D)$ has a derivative that belongs to $H(D)$ and when it does, whether its Mittag-Leffler series is obtained by deriving that of $f$. Another question is whether an analytic element on $D$ whose derivative is identically zero is a constant. Both questions are answered on an infraconnected closed set.

Throughout this chapter, $D$ is a subset of $\mathbb{K}$ and is supposed to be open and infraconnected and we fix $R > 0$.

**Theorem 18.1.** Let $f(x) = \sum_{n=0}^{+\infty} a_n x^n \in H(d(0, R))$. Then, $f$ has a derivative $f'(\alpha)$ at each point $\alpha \in d(0, R)$ and $f'(0) = a_1$. Moreover, the function $f'$ also belongs to $H(d(0, R))$, is equal to $\sum_{n=0}^{+\infty} na_n x^{n-1}$, and satisfies

$$|f'(r)| \le \frac{|f|(r)}{r} \quad \forall r \in ]0, R].$$

Further, $f$ is indefinitely derivable in $d(0, R)$ and

$$f^k(x) = \sum_{n=k}^{\infty} n(n-1)\cdots(n-k+1)a_n x^{n-k}.$$

**Proof.** Without loss of generality, we can suppose that $R \le 1$. Obviously, $\lim_{x \to 0} \frac{f(x)-a_0}{x} = a_1$, hence $f'(0)$ exists and is equal to $a_1$.
More generally, take $\alpha \in d(0, R)\setminus\{0\}$ and consider

$$ \frac{f(x) - f(\alpha)}{x - \alpha} = \frac{\sum_{n=1}^{\infty} a_n (x^n - \alpha^n)}{x - \alpha} \\ = \sum_{n=1}^{\infty} a_n (x^{n-1} + \alpha x^{n-2} + \cdots + \alpha^{n-1}). $$

Then, $|(x^{n-1} + \alpha x^{n-2} + \cdots + \alpha^{n-1}) - n\alpha^{n-1}| \le |x - \alpha|$ ($\max(|\alpha|, |x|)^{n-1}$. Particularly, when $x$ is close enough to $\alpha$, since $\alpha \ne 0$, we have $|x| = |\alpha|$, hence

$$ |a_n|(x^{n-1} + \alpha x^{n-2} + \cdots + \alpha^{n-1}) - n\alpha^{n-1} \le |a_n||\alpha|^{n-1}|x - \alpha| $$

That proves that $(a_n(x^{n-1} + \alpha x^{n-2} + \cdots + \alpha^{n-1})) - na_n\alpha^{n-1}$ converges to 0 uniformly with respect to $n$ and uniformly with respect to $x$ inside a disk of center $\alpha$. Consequently, $f'(\alpha) = \sum_{n=1}^{\infty} na_n(x - \alpha)^n \forall \alpha \in d(0, R)$. Then,

$$ |f'(r)| = \sup_{n \in \mathbb{N}} (|na_n|r^{n-1}) \le \sup_{n \in \mathbb{N}} (|a_n|r^{n-1}) = \frac{1}{r}|f|(r). $$

The last statement concerning $f^{(k)}$ is then immediate. □

More generally, we can derive the following:

**Theorem 18.2.** Let $f \in H(\Delta(0, R, R'))$. Then, $f^{(k)}$ also belongs to $H(\Delta(0, R, R'))$ for every $k \in \mathbb{N}^*$ and satisfies $|f'(r)| \le \frac{|f|(r)}{r} \forall r \in ]R, R']$. Moreover, if the residue characteristic does not divide $\nu^+(f, \log r)$ or $\nu^-(f, \log r)$, then $|f'(r)| = \frac{|f|(r)}{r}$.

**Proof.** $f(x)$ is equal to a Laurent series $\sum_{n=-\infty}^{+\infty} a_n x^n$ with $\lim_{n \to +\infty} |a_n|R^{n'} = 0$ and $\lim_{n \to -\infty} |a_n|R^n = 0$, hence obviously, $\lim_{n \to -\infty} |na_n|R^{n-1} = 0$ and $\lim_{n \to +\infty} |na_n|R^{n-1} = 0$. Consequently, $f'(x)$ belongs to $H(\Delta(0, R, R'))$. Then,

$$ |f'(r)| = \sup_{n \in \mathbb{Z}} (|na_n|r^{n-1}) \le \sup_{n \in \mathbb{Z}} (|a_n|r^{n-1}) = \frac{1}{r}|f|(r). $$

Suppose now that the residue characteristic $p$ does not divide $\nu^+(f, \log r)$ or $\nu^-(f, \log r)$. If $\nu^+(f, \log r) = \nu^-(f, \log r)$ is an integer $q$, it is obvious that $|f|(r) = |a_q|r^q$ and $|qa_q| = |a_q|$, hence $|f'(r)| = \frac{|f|(r)}{r}$, provided $q \ne 0$. Next, the property has continuation by continuity to the points $\mu$ such that $\nu^+(f, \mu) \ne \nu^-(f, \mu)$. □
**Corollary 18.3.** Let $R, R' \in ]0, +\infty[$ ($R < R'$) and let $f \in \mathcal{A}(d(0, R^-))$ (respectively $f \in \mathcal{A}(\Gamma((0, R, R'))$. Then, $|f'|(r) \le \frac{|f|(r)}{r} \quad \forall r \in ]0, R[$ (respectively $\forall r \in ]R, R']$).

**Corollary 18.4.** Suppose $\mathbb{K}$ has characteristic zero and $f$ belongs to $H(d(0, R))$. Then, $a_n = \frac{f^{(n)}(0)}{n!}$ for every $n \in \mathbb{N}$, and if $\alpha$ is a zero of multiplicity order $q$ of $f$, then we have $f^{(j)}(\alpha) = 0$ for every $j < q$ and $f^{(q)}(\alpha) \neq 0$.

**Corollary 18.5.** Let $h \in \mathbb{K}(x)$. Then, for all $r > 0$, we have $|h'|(r) \le \frac{|h|(r)}{r}$.

**Theorem 18.6.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, r^{-}))$. The power series $\sum_{n=0}^{\infty} n a_n x^{n-1}$ also belongs to $\mathcal{A}(d(0, r^{-}))$ and is equal to the derivative of $f$ in $d(0, r^{-})$. The radius of convergence of $f'$ is superior or equal to the one of $f$. Further, if $\mathbb{K}$ has characteristic 0, the radius of convergence of $f'$ is the same as that of $f$.

**Proof.** By Theorem 18.1, the first statement is clear. Now, we suppose that $\mathbb{K}$ has characteristic zero. If $\mathbb{K}$ has residue characteristic zero, then $|n| = 1$ for all $n \in \mathbb{N}^*$, and therefore, the last statement is clear. Now, we assume that $\mathbb{K}$ has residue characteristic $p \neq 0$. By Lemma 6.17, we have $\frac{1}{n} \le |n| \le 1$ for every $n \in \mathbb{N}^*$, and therefore, $\lim_{n \to +\infty} \sqrt[n]{|n|} = 1$. Consequently, $\limsup_{n \to +\infty} \sqrt[n]{|a_n|} = \limsup_{n \to +\infty} \sqrt[n-1]{|na_n|}$, and finally, $f'$ has the same radius of convergence as $f$. $\square$

**Corollary 18.7.** Suppose $\mathbb{K}$ has characteristic 0. Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(d(0, r^{-}))$. The power series $\sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1}$ also belongs to $\mathcal{A}(d(0, r^{-}))$ and has the same radius of convergence as that of $f$ and is a primitive of $f$ in $d(a, r^{-})$.

**Remark.** Unlike in Archimedean analysis, when the characteristic $p$ of $\mathbb{K}$ is not zero, there do exist power series $f$ whose derivatives have a radius of convergence bigger than that of $f$. For example, let $f(x) = \sum_{n=0}^{\infty} x^{p^n}$: the radius of convergence of $f$ is 1, while this of $f'$ is $+\infty$.

**Theorem 18.8.** Let $a \in \mathbb{K}$ and let $R \in \mathbb{R}_+$. Let $(f_m)_{m \in \mathbb{N}^*}$ be a sequence of $H(d(a, R))$ converging uniformly to a function $f$. Then
the sequence $(f'_m)_{m \in \mathbb{N}^*}$ converges uniformly to $f'$ in $H(d(a, R))$ and we have $\|f'_m - f'\|_{d(a,R)} \le \frac{\|f_m - f\|_{d(a,R)}}{R} \quad \forall m \in \mathbb{N}$.

**Proof.** For each $m \in \mathbb{N}$, set $f_m(x) = \sum_{n=0}^{\infty} a_{n,m}x^n$ and let $f(x) = \sum_{n=0}^{\infty} b_nx^n$. Then for each $m \in \mathbb{N}$, we have $\lim_{n \to +\infty} |a_{n,m}|R^n = 0$, $\lim_{n \to +\infty} |b_n|R^n = 0$. Now, the Banach norm of $f_m - f$ tends to zero when $m$ goes to $\infty$, hence $\lim_{m \to +\infty} (\sup_{n \in \mathbb{N}}(|a_{n,m} - b_n|R^n)) = 0$. Consequently, considering the respective derivatives, we have

$$ \lim_{m \to +\infty} \left( \sup_{n \in \mathbb{N}} (na_{n,m} - nb_n|R^n) \right) = 0, $$

and therefore, by Theorem 18.1, we have $\|f'_m - f'\|_{d(a,R)} \le \frac{\|f_m - f\|_{d(a,R)}}{R}$. We are done. $\square$

**Corollary 18.9.** Let $a \in \mathbb{K}$ and let $R \in \mathbb{R}_+$. Let $(f_m)_{m \in \mathbb{N}^*}$ be a sequence of $H(\mathbb{K}\setminus d(a, R^-))$ converging uniformly to a function $f$. Then the sequence $(f'_m)_{m \in \mathbb{N}^*}$ converges uniformly to $f'$ in $H(\mathbb{K}\setminus d(a, R^-))$ and we have $\|f'_m - f'\|_{\mathbb{K}\setminus d(a, R^-)} \le R\|f_m - f\|_{\mathbb{K}\setminus d(a, R^-)}$ $\forall m \in \mathbb{N}$.

**Theorem 18.10.** Suppose $\mathbb{K}$ has characteristic 0. Let $a \in \mathbb{K}$ and let $R \in \mathbb{R}_+$. Let $(f_m)_{m \in \mathbb{N}^*}$ be a sequence of $A(d(a, R^-))$ such that the sequence $(f'_m)_{m \in \mathbb{N}^*}$ converges uniformly to a function $h$ in $H(d(a, r))$ $\forall r \in ]0, R[$ and such that the sequence $f_m(a)$ converges in $\mathbb{K}$. Then the sequence $(f_m)_{m \in \mathbb{N}^*}$ converges to a function $f \in A(d(a, R^-))$ such that $f' = h$ and the convergence is uniform in $d(a, r)$ for every $r \in ]0, R[$.

**Proof.** Without loss of generality, we can assume $a = 0$. Let us fix $r \in ]0, R[$ and let us show that the sequence $(f_m)_{m \in \mathbb{N}^*}$ converges uniformly to a function $f \in H(d(a, r))$ such that $f' = h$. For every $m \in \mathbb{N}$, let $f'_m = \sum_{n=0}^{+\infty} b_{n,m}x^n$ and let $h(x) = \sum_{n=0}^{+\infty} b_n x^n$. Take $s \in ]r, R[$. By hypothesis, we have

$$ (1) \lim_{m \to +\infty} \sup_{n \in \mathbb{N}} |b_{n,m} - b_n|s^n = 0. $$

But since $|n| \ge \frac{1}{n}$, we have $\lim_{n \to +\infty} \frac{\rho^{n+1}}{|n+1|} = 0$ $\forall \rho < 1$, therefore, by (1), we have $\lim_{m \to +\infty} (\sup_{n \in \mathbb{N}} \left| \frac{b_{n,m} - b_n}{n+1} \right| r^{n+1}) = 0$. Consequently, the sequence $(f_m - f_m(0))_{m \in \mathbb{N}}$ converges uniformly to the
function $g(x) = \sum_{n=0}^{+\infty} \frac{b_n}{n+1} x^{n+1}$. Set $\lim_{m \to +\infty} f_m(0) = b_0$. Then the sequence $(f_m)_{m \in \mathbb{N}}$ converges to $g(x) + b_0$ uniformly in $d(0, r)$ for every $r < R$. $\square$

**Corollary 18.11.** Suppose $\mathbb{K}$ has characteristic 0. Let $\alpha \in \mathbb{K}$ and let $R \in \mathbb{R}_+$ and let $a \in \mathbb{K}\d d(\alpha, R)$. Let $(f_m)_{m \in \mathbb{N}^*}$ be a sequence of $\mathcal{A}(\mathbb{K}\d d(\alpha, R))$ such that the sequence $(f'_m)_{m \in \mathbb{N}^*}$ converges uniformly to a function $h$ in $H(\mathbb{K}\d d(\alpha, r^-)) \forall r > R$ and that the sequence $(f_m(a))_{m \in \mathbb{N}}$ converges in $\mathbb{K}$. Then the sequence $(f_m)_{m \in \mathbb{N}^*}$ converges to a function $f \in \mathcal{A}(\mathbb{K}\d d(\alpha, R))$ such that $f' = h$ and the convergence is uniform in $\mathbb{K}\d d(\alpha, r^-)$ for every $r > R$.

**Theorem 18.12.** Let $r > 0$. If $\mathbb{K}$ has characteristic 0, then an element $f \in H(d(0,r))$ has a derivative identically equal to 0 if and only if it is equal to a constant.

If $\mathbb{K}$ has a characteristic $p \neq 0$, then an element $f \in H(d(0,r))$ has a derivative identically equal to 0 if and only if there exists $g \in H(d(0,r))$ such that $f(x) = (g(x))^p$.

**Proof.** By Theorem 14.6, each element of $H(d(0,r))$ is a convergent power series, hence the statement about the case when $\mathbb{K}$ has characteristic zero is obvious. Now, suppose that $\mathbb{K}$ has a characteristic $p \neq 0$. If there exists $g \in H(d(0,r))$ such that $f(x) = (g(x))^p$, obviously we have $f'(x)$ identically equal to 0.

Now, we suppose $f'(x)$ identically equal to 0. Hence, $f(x)$ is of the form $\sum_{j=0}^{\infty} b_j x^{jp}$, with $\lim_{j \to \infty} |b_j|r^{jp} = 0$. For each $j \in \mathbb{N}$, we can take $c_j \in \mathbb{K}$ such that $(c_j)^p = b_j$. Then, it is seen that $\lim_{j \to \infty} |c_j|r^j = 0$. Now, we can put $g(x) = \sum_{n=0}^{\infty} c_n x^n$, and therefore, $g$ belongs to $H(d(0,r))$. Since $\mathbb{K}$ has characteristic $p$, we have $(g(x))^p = f(x)$. This ends the proof. $\square$

**Corollary 18.13.** Let $r > 0$. If $\mathbb{K}$ has characteristic 0, then an element $f \in H(\mathbb{K}\d d(0, r^-))$ (respectively $f \in H_0(\mathbb{K}\d d(0, r^-))$) has a derivative identically equal to 0 if and only if it is equal to a constant (respectively to 0).

If $\mathbb{K}$ has a characteristic $p \neq 0$, then an element $f \in H(\mathbb{K}\d d(0, r^-))$ (respectively $f \in H_0(\mathbb{K}\d d(0, r^-))$) has a derivative identically equal to 0 if and only if there exists $g \in H(\mathbb{K}\d d(0, r^-))$ (respectively $g \in H_0(\mathbb{K}\d d(0, r^-))$) such that $f(x) = (g(x))^p$.
**Corollary 18.14.** Let $r > 0$. If $K$ has characteristic 0, then a power series $f(x) \in \mathcal{A}(d(0, r^{-}))$ has a derivative identically equal to 0 if and only if it is equal to a constant.

If $K$ has a characteristic $p \neq 0$, then a power series $f(x) \in \mathcal{A}(d(0, r^{-}))$ has a derivative identically equal to 0 if and only if there exists $g \in \mathcal{A}(d(0, r^{-}))$ such that $f(x) = (g(x))^p$.

Theorem 18.15 improves Theorems 18.1 and 18.2 concerning derivatives of order $k > 1$.

**Theorem 18.15.** Let $a \in K$, let $R, R', R'' \in \mathbb{R}_+^*$ with $R' < R''$, and let $f \in H(d(0, r))$ (respectively let $f \in H(K \setminus d(0, R^{-}))$, respectively let $f \in H(\Gamma(0, R', R''))$). Then, for every $k \in \mathbb{N}^*$, for every $r < R$ (respectively $r > R$, respectively $r \in ]R', R''[$), we have

$$|f^{(k)}|(r) \leq |k!| \frac{|f|(r)}{r^k}.$$

**Proof.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n \in H(d(0, R^{-}))$. By Theorem 14.6, we have $|f|(r) = \sup_{n \in \mathbb{N}} |a_n|r^n$ and $|f^{(k)}|(r) = \sup_{n \geq k} |\frac{(n!)}{(n-k)!}a_n|r^{n-1}$. But $\frac{(n!)}{(n-k)!}$ is an integer multiple of $k!$ because the combination $\binom{n}{k}$ belongs to $\mathbb{N}$. Consequently, $|\frac{(n!)}{(n-k)!}| \leq |k!|$, and therefore, we obtain $|f^{(k)}|(r) \leq \sup_{n \geq k} |k!| |a_n|r^{n-k}$. The proof is similar when $f$ belongs to $H(K \setminus d(0, r))$ or to $H(\Gamma(0, r', r''))$. $\square$

**Corollary 18.16.** Let $a \in K$, let $r \in \mathbb{R}_+^*$ and let $f \in H(d(a, r))$. Then,

$$\|f^k\|_{d(a,r)} \leq |k!| \frac{\|f\|_{d(a,r)}}{r^k}.$$

**Theorem 18.17.** Let $f \in H_b(D)$ and let $\rho = \delta(D, (K \setminus \overline{D}))$. If $\rho > 0$, then $f'$ belongs to $H_b(D)$ and satisfies $\|f'\|_D \leq \frac{1}{\rho} \|f\|_D$.

**Proof.** Let $(T_n)_{n \in \mathcal{S}}$ be the sequence of the $f$-holes and let $D' = \tilde{D} \setminus (\bigcup_{n \in \mathcal{S}} T_n)$. By Theorem 15.1, we know that $f \in H_b(D')$ and that (1) $\|f\|_{D'} = \|f\|_D$.

By Theorem 18.1, $f$ has a derivative $f'$ in $D'$ and we first check that the function $f'$ satisfies $\|f'\|_{D'} \leq \frac{1}{\rho} \|f\|$. Let $a \in D'$. The disk $d(a, \rho^{-})$
is obviously included in $D$ because if a point $b \in d(a, \rho^-)$ belonged to $K\backslash D'$, since $D$ is closed, there would be a disk $d(b, r^-) \subset K\backslash D'$ with $r < \rho$. Thus, when $x \in d(a, \rho^-)$, $f(x)$ is of the form $\sum_{n=0}^{\infty} a_n (x-a)^n$ and hence the conclusion comes from Theorem 18.1. $\square$

**Corollary 18.18.** Let $f \in H(D)$ be such that the set of diameters of the $f$-holes has a strictly positive lower bound. Then, $f' \in H(D)$.

**Proof.** By Theorem 11.5, we know that $f$ is in the form $g+h$ with $h \in R(D)$ and $g \in H_b(\bar{D})$. Obviously, $h'$ belongs to $R(D)$, and by Theorem 18.17, we have $g' \in H_b(\bar{D})$. $\square$

**Theorem 18.19.** Let $D \in \text{Alg}$ be closed and open. Let $f \in H_b(D)$, let $(V_n)_{n \in \mathbb{N}^*}$ be the set of $f$-holes, and let $\sum_{n=0}^{\infty} f_n$ be its Mittag-Leffler series on $D$ defined as $f_0 = \overline{f_0}$ and for every $n \in \mathbb{N}^*$, $f_n = \overline{f_{V_n}}$. The following three conditions are equivalent:

(a) $f'$ belongs to $H(D)$.

(b) The series $\sum_{n=0}^{\infty} f'_n$ converges in $H(D)$.

(c) The series $\sum_{n=0}^{\infty} f'_n$ converges to $f'$ in $H_b(D)$.

**Proof.** We first prove the equivalence between (b) and (c). For each $q \in \mathbb{N}$, the sum $\sum_{n=0}^q f'_n$ clearly belongs to $H_b(D)$. Thus, we assume that this series $\sum_{n=0}^{\infty} f'_n$ converges to an element $h \in H(D)$ and we prove that $h = f' \in H_b(D)$. Let $\alpha \in D$. There exists a disk $d(\alpha, r) \subset D$. We show that $h(\alpha) = f'(\alpha)$. For every $\psi \in H(D)$, $\tilde{\psi}$ denotes the restriction of $\psi$ to $d(\alpha, r)$. Since the sequence $(\sum_{j=0}^n f_j)$ converges to $\tilde{f}$, by Theorem 18.17, the sequence of the derivatives $(\sum_{j=0}^n f'_j)$ does converge to $\tilde{f}'$, hence it is clearly seen that $f'(\alpha) = h(\alpha)$, and therefore, $f' = h$. We check that $f'$ is bounded in $D$. The sequence $\|f'_n\|_D$ has limit 0 hence is obviously bounded and therefore its sum $f'$ is bounded in $D$. Thus, (b) and (c) are equivalent.

Since (c) trivially implies (a), we just have to prove that (a) implies (b). Thus, we suppose (a) to be true and prove (b).

First, we suppose $D$ bounded. For each hole $T$ of $D$ that is either an $f$-hole or an $f'$-hole, we denote by $\overline{f}_T$ (respectively $\overline{g}_T$) the Mittag-Leffler term of $f$ (respectively $f'$). Let $S$ be the set of holes
$T$ such that $(\overline{\overline{f_T}})' \neq \overline{\overline{g_T}}$ and let $\mathcal{T}$ be the set of $f$-holes such that $(\overline{\overline{f_T}})' = \overline{\overline{g_T}}$. If we can show that $\mathcal{S} = \emptyset$, then (b) is clearly proven.

Hence, we suppose $\mathcal{S} \neq \emptyset$. All the $\overline{\overline{g_T}}$ are equal to zero except maybe a countable family of them. The series $\sum_{T \in \mathcal{T}} \overline{\overline{g_T}}$ and $\sum_{T \in \mathcal{S}} \overline{\overline{g_T}}$ obviously converge in $H(D)$, and then we have $f' = \sum_{T \in \mathcal{T}} (\overline{\overline{f_T}})' + \sum_{T \in \mathcal{S}} \overline{\overline{g_T}}'$. Since (b) implies (c), the series $\sum_{T \in \mathcal{T}} (\overline{\overline{f_T}})'$ is clearly equal to the derivative of $\sum_{T \in \mathcal{T}} \overline{\overline{f_T}}$. Let $h = \sum_{T \in \mathcal{S}} \overline{\overline{f_T}} = f - \sum_{T \in \mathcal{T}} \overline{\overline{f_T}}$. Then, $h' = f' - \sum_{T \in \mathcal{T}} (\overline{\overline{f_T}})' = \sum_{T \in \mathcal{S}} \overline{\overline{g_T}}'$. Let $\mathcal{D}$ be the family of diameters of the holes $T$ that belong to $\mathcal{S}$ and let $\lambda$ be its lower bound. Suppose $\lambda > 0$. By Theorem 18.17, the series $\sum_{T \in \mathcal{S}} (\overline{\overline{f_T}})'$ converges to $h'$, hence $\sum_{T \in \mathcal{S}} (\overline{\overline{f_T}})'$ is the Mittag-Leffler series of $h'$ on $\mathcal{D}$, hence $(\overline{\overline{f_T}})' = \overline{\overline{g_T}}$ for all $T \in \mathcal{S}$ and that contradicts the definition of $\mathcal{S}$. Hence, $\lambda = 0$.

Now, we prove that there exists a hole $V = d(a, r^-) \in \mathcal{S}$ with an annulus $\Gamma(a, r, s)$ such that the set $\mathcal{U}$ of the diameters $\rho$ of the $f$-holes included in $\Gamma(a, r, s)$ has a strictly positive lower bound. Indeed, suppose such a hole $V$ does not exist. Then, we can easily construct a sequence of $f$-holes $(T_n)_{n \in \mathbb{N}^*}$ of the form $T_n = d(a_n, r_n^-)$ with (1) $r_n \le \frac{1}{n}$ and

$$ (2) \quad |a_{n+1} - a_n| \le \frac{2}{n}. $$

For example, assume the sequence has just been constructed up to the rank $q$, satisfying (1) and (2) for $n \le q$. Since $V$ does not exist, then in $\Gamma(a_q, r_q, \frac{2}{q})$, we can find an $f$-hole $T_{q+1} = d(a_{q+1}, r_{q+1}^-)$ with $r_{q+1} < \frac{2}{q+1}$ and then the sequence is clearly constructed by induction by taking any first $f$-hole $T_1 = d(a_1, r_1^-)$. The sequence $(T_n)_{n \in \mathbb{N}^*}$ clearly converges to a point $w \in \mathcal{D}$ and that contradicts the hypothesis "$\mathcal{D}$ is closed and open". Hence, we have now proven the existence of the $f$-hole $V$ with an annulus $\Gamma(a, r, s)$ and a number $\xi > 0$ such that every $f$-hole $T \subset \Gamma(a, r, s)$ satisfies

$$ (3) \quad \mathrm{diam}(T) \ge \xi. $$

Let $\mathcal{L}$ be this family of $f$-holes included in $\Gamma(a, r, s)$. Let $l = \sum_{T \in \mathcal{L}} \overline{\overline{f_T}}'$. By Theorem 18.17, the series $\sum_{T \in \mathcal{L}} (\overline{\overline{f_T}})'$ converges to $l'$ in
$H(D)$. Now, let $\psi = h - l - \overline{f_V}$. Clearly, $\psi$ belongs to $H(D)$ and no hole $T$ (of $D$) included in $d(a, s)$ is a $\psi$-hole. Hence, $\psi$ extends to an element of $H(D \cup d(a, s))$. In $d(a, s)$, $\psi(x)$ is equal to a power series $\phi(x) \in H(d(a, s))$, hence $\phi' \in H(d(a, s))$. Thus, in $D \cap d(a, s)$, $\psi'(x)$ is equal to the series $\phi'(x)$, and then for every hole $T$ of $D$ included in $d(a, s)$, the Mittag-Leffler term of $\psi$ associated with $T$ (with respect to $D$) is zero.

On the other hand, we have $\psi' = h' - l' - (\overline{f_V})' = \sum_{T \in \mathcal{S}} \overline{g_T} - \sum_{T \in \mathcal{L}} (\overline{f_T})' - (\overline{f_V})'$ and then the Mittag-Leffler term of $\psi'$ associated with $V$ (with respect to $D$) is $\overline{g_V} - (\overline{f_V})' \neq 0$. Hence, we have a contradiction with $\psi \in H(d(a, s))$. This finishes proving that (b) is true when $D$ is bounded.

Now, we suppose $D$ unbounded. Since $D \in \text{Alg}$, there exists a disk $d(0, S)$ such that all holes of $D$ are included in this disk. Then for every element $h \in H(D)$, its Mittag-Leffler series in $H(D)$ is the same as in $H(D')$. This is true for both $f, f'$, and therefore (b), which is true in $H(D')$, is obviously true in $H(D)$. That ends the proof of Theorem 18.19. $\square$

When $\mathbb{K}$ has characteristic zero, in most of the cases, we are now able to answer the question “does $f' = 0$ implies $f = ct$?”. When $D$ is not infraconnected, it admits an empty annulus $\Lambda = \Gamma(a, r', r'')$, and hence by Proposition 11.15, we know that there exists $w \in H(D)$ such that $w(x) = 1$ whenever $x \in \mathcal{I}(\Lambda)$, while $w(x) = 0$ whenever $x \in \mathcal{E}(\Lambda)$. Thus, the condition “$D$ is infraconnected” is certainly necessary to be able to answer “yes” to the question above.

The two theorems that follow show this condition to be sufficient too, provided $D$ satisfies a little extra condition like to be closed or to belong to Alg.

By Theorems 15.1, 18.19, and 18.8 and Corollary 18.9, we can derive the following:

**Corollary 18.20.** Let $D$ be an open closed bounded infraconnected subset of $\mathbb{K}$ and let $(f_n)_{n \in \mathbb{N}}$ be a sequence of $H(D)$ converging uniformly to a function $f$. Then, the sequence $(f'_n)_{n \in \mathbb{N}}$ converges uniformly to $f'$.

**Remark.** Let $D$ be an open closed infraconnected bounded subset of $\mathbb{K}$, let $\alpha \in D$, and let $(f_n)_{n \in \mathbb{N}}$ be a sequence of $H(D)$
such that the sequence $f_n(\alpha)$ converges in $\mathbb{K}$ and that the sequence $(f'_n)_{n \in \mathbb{N}}$ converges in $H(D)$ to a function $h \in H(D)$. Comparatively to the Archimedean context, we could expect that the sequence $(f_n)_{n \in \mathbb{N}}$ converges in $H(D)$ to a function $f$ such that $f' = h$. Actually, that's wrong. For example, define the sequence $(f_n)_{n \in \mathbb{N}}$ as $f_n(x) = \sum_{k=0}^{n} \frac{x^{2k}}{p^k}$. Then, $f'_n(x) = 1 + \sum_{k=1}^{n} p^k x^{2k-1}$, and hence, the sequence $(f'_n)_{n \in \mathbb{N}}$ converges in $H(d(0,1))$ to the function $1 + \sum_{k=1}^{+\infty} p^k x^{2k-1}$, whereas $f_n(0) = 1 \forall n \in \mathbb{N}$. However, the function $\sum_{k=0}^{+\infty} \frac{x^{2k}}{p^k}$ is unbounded in $d(0,1)$ and hence does not belong to $H(d(0,1))$. That remark does not contradict Theorem 18.10 which only concerned analytic functions in an “open” disk $d(0, r^-)$.

**Theorem 18.21.** $\mathbb{K}$ is supposed to have characteristic zero. Let $E$ be an open subset of $\mathbb{K}$ such that $\bar{E}$ also is open. Then, $E$ is infraconnected if and only if for every $f \in H(E)$ such that $f'(x) = 0$ whenever $x \in E$, we have $f = ct$.

**Proof.** If $E$ is not infraconnected, it admits at least an empty annulus $\Lambda$, and then, by Proposition 11.15, the characteristic function $u$ of $\mathcal{I}(\Lambda)$ belongs to $H(E)$. Hence, there do exist non-constant elements $f \in H(E)$ whose derivative is identically 0. Now, let $E$ be infraconnected and let $f \in H(E)$ satisfy $f'(x) = 0$ whenever $x \in E$. We just have to prove that $f$ is a constant.

Suppose first that $E$ is bounded. By Theorem 11.5, $f$ is in the form $f^* + \bar{f}$ with $f^* \in R_0(\mathbb{K}\setminus(\bar{E}\setminus E))$ and $\bar{f} \in H(\bar{E})$, so $f^{*'}(x) + \bar{f}'(x) = 0$ whenever $x \in E$, and therefore, $f^{*'} = -\bar{f}'$. Hence, $\bar{f}' \in R_0(\mathbb{K}\setminus(\bar{E}\setminus E)) \cap H(\bar{E})$. Thus, $f^*$ is a rational function that has no pole in $\mathbb{K}$ and then it is a polynomial. Now, as an element of $R_0(\mathbb{K}\setminus(\bar{E}\setminus E))$, it tends to 0 when $|x|$ goes to $+\infty$, hence $f^* = 0$ and therefore, $f^{*'}$ is identically 0. Since $f^*$ belongs to $R_0(\mathbb{K}\setminus(\bar{E}\setminus E))$ clearly, $f^* = 0$, and therefore, $f$ belongs to $H(\bar{E})$.

Let $a \in E$ and let $\sum_{n \in S} h_n$ be the Mittag-Leffler series of $f$ in $\bar{E}$, with $h_0 = \bar{f}_0$, and for each $n \in S$, set $h_n = \overline{f_{T_n}}$, for any $f$-hole $T_n$. Since $\bar{E}$ is open, we can apply Theorem 18.19 to $E$ and then we have $(h_n)' = 0$ for every $n \in \mathbb{N}$. Since $\bar{E}$ is bounded of diameter $r$, then by Theorem 16.8, we know that $h_0$ is a constant. In the same way, by Corollary 18.13, for each $q \in \mathbb{N}$, we know that $h_q = 0$. Hence, $f$ is a constant.
Finally, let $E$ be unbounded. Then, for all $r > 0$, the set $E_r = E \cap d(0, r)$ is such that $\bar{E}_r$ is open, hence $f$ is constant in $E_r$ and therefore in all of $E$. $\square$

**Remark.** In particular, Theorem 18.21 applies to open closed sets.

**Corollary 18.22.** $\mathbb{K}$ is supposed to have characteristic zero. Let $E$ be open and belong to Alg. Then $E$ is infraconnected if and only if for every $f \in H(E)$ such that $f'(x) = 0$ whenever $x \in E$, $f$ is a constant in $E$.

**Proof.** Indeed, since $E$ belongs to Alg, it satisfies Condition (B) in Theorem 17.9: $\bar{D} \setminus D \subset (\bar{D})^o$. But then, as it is also open, we check that $\bar{E}$ is open. $\square$

We now study thoroughly the question of whether all analytic elements in a set $D$ have derivative in $H(D)$.

**Definition.** We call *piercing of* $D$ the number $\delta(\bar{D}, \mathbb{K}\setminus D) > 0$ and $D$ is said to be *well pierced* if $\delta(\bar{D}, \mathbb{K}\setminus D) > 0$.

**Theorem 18.23.** Let $\bar{D}$ be open. Then, $D$ is well pierced if and only if for every $f \in H(D)$, $f'$ also belongs to $H(D)$.

**Proof.** If $D$ is well pierced, by Corollary 18.18, we know that for every $f \in H(D)$, $f'$ belongs to $H(D)$. Now let us suppose $D$ has piercing zero and let $(T_n)_{n \in \mathbb{N}}$ be a sequence of holes $T_n = d(\alpha_n, \rho_n^-)$ such that $\lim_{n \to \infty} \rho_n = 0$. Let $\lambda_n$ be a sequence in $\mathbb{K}$ such that $\lim_{n \to \infty} \frac{|\lambda_n|}{\rho_n^-} = 0$, while $\lim_{n \to \infty} \frac{|\lambda_n|}{\rho_n^+} = +\infty$. It is seen that the series $\sum_{n=1}^{\infty} \frac{\lambda_n}{x-\alpha_n}$ converges in $H(D)$ to an element $f$, while the series $\sum_{n=1}^{\infty} \frac{\lambda_n}{(x-\alpha_n)^2}$ does not. On the other hand, $\sum_{n=1}^{\infty} \frac{\lambda_n}{x-\alpha_n}$ obviously is the Mittag-Leffler series of $f$. If $f'$ belongs to $H(D)$, by Theorem 18.19, its Mittag-Leffler series must be $\sum_{n=1}^{\infty} \frac{\lambda_n}{(x-\alpha_n)^2}$. Since this series does not converge, this is just impossible. $\square$

Before closing this chapter, we note the following result that may be sometimes helpful in differential equations.

**Theorem 18.24.** Let $\bar{D}$ be open. We suppose that both $f$ and $f'$ belong to $H(D)$. For every $\epsilon > 0$, there exists $h \in R(D)$ such that $\|f - h\|_D \le \epsilon$ together with $\|f' - h'\|_D \le \epsilon$.
**Proof.** First, we suppose that $f$ belongs to $H_b(\bar{D})$. We have to introduce a notation. Let $g \in H_b(\bar{D})$. If $D$ is bounded, $\bar{D}$ is a disk $d(a, r)$ and $g$ is of the form $\sum_{m=0}^{\infty} \lambda_m (x-a)^m$. Then, for every $q \in \mathbb{N}$, we put $(g)_q = \sum_{m=0}^{q} \lambda_m (x-a)^m$. If $D$ is unbounded, then $g$ is a constant $\lambda_\circ$, and we put $(g)_q = g$ whenever $q \in \mathbb{N}$. Let $T = d(b, r^-)$ be a hole of $D$ and let $l(x) = \sum_{m=1}^{\infty} \frac{\mu_m}{(x-b)^m}$. For each $q \in \mathbb{N}^*$, we put $(l)_q = \sum_{m=1}^{q} \frac{\mu_m}{(x-b)^m}$. Now, let $\sum_{n=0}^{\infty} f_n$ be the Mittag-Leffler series of $f$, with $f_0 = \overline{f_0}$ and for each $n \in \mathbb{N}^*$, $f_n = \overline{f_{T_n}}$, for any $f$-hole $T_n$. By Theorem 18.19, the Mittag-Leffler series of $f'$ is $\sum_{n=0}^{\infty} f'_n$, and therefore, there exists an integer $N(\epsilon)$ such that

$$ (1) \quad \left\| \sum_{n=0}^{N(\epsilon)} f_n - f \right\|_D \le \epsilon $$

and

$$ (2) \quad \left\| \sum_{n=0}^{N(\epsilon)} f'_n - f' \right\|_D \le \epsilon. $$

Obviously, we have an integer $Q(\epsilon)$ such that $\|f_n - (f_n)_{Q(\epsilon)}\|_D \le \epsilon$ whenever $n = 0, \dots, N(\epsilon)$, and then by (1) and (2), it is easily seen that $\|\sum_{n=0}^{N(\epsilon)} (f_n)_{Q(\epsilon)} - f\|_D \le \epsilon$ and $\|\sum_{n=0}^{\infty} (f'_n)_{Q(\epsilon)} - f'\|_D \le \epsilon$. By putting $h = \sum_{n=0}^{N(\epsilon)} (f_n)_{Q(\epsilon)}$, we obtain the $h \in R(D)$ we want.

Now, we can consider the general case. By Corollary 11.7, $f$ is of the form $l + \psi$, with $l \in H_b(\bar{D})$ and $\psi \in R(D)$. Hence, $f' = l' + \psi'$. Since $f'$ belongs to $H(D)$, so does $\psi'$. Hence, by Theorem 18.21, $\psi'$ belongs to $H_b(\bar{D})$. We have just proven that there exists $t \in H_b(\bar{D})$ such that $\|l - t\|_D \le \epsilon$ and $\|l' - t'\|_D \le \epsilon$. Hence, we just have to consider $h = t + \psi$, and this ends the proof. $\square$

**Theorem 18.25.** Let $\bar{D}$ be open. Suppose $0 \in D$. Let $f \in H(D)$ and let $r \in [\delta(0, D), \text{diam}(D)]$. Then, $|f'(r)| \le \frac{|f|(r)}{r}$.

**Proof.** Suppose first that $|f|(r) \ne 0$. Let $\epsilon > 0$ and let $\eta > 0$ be such that $\frac{|f|+\eta}{r} + \eta < \frac{|f|}{r} + \epsilon$. Now, by Theorem 18.24, we can find $h \in R(D)$ such that $|h'|(r) \le |f'| + \eta$ and $|h|(r) \le |f| + \eta$.
By Corollary 18.3, we have $|h'|(r) \le \frac{|h|(r)}{r}$, hence

$$|f'|(r) \le |h'|(r) + \eta \le \frac{|h|(r)}{r} + \eta \le \frac{|f|(r) + \eta}{r} + \eta \le \frac{|f|(r)}{r} + \epsilon.$$

Now, suppose $|f|(r) = 0$. Then, by Theorem 18.24, we can find $h \in R(D)$ such that $\max(|h'|(r), |h|(r)) \le \epsilon$, hence we have again $|f'| \le \frac{|f|(r)}{r}$. This is true for all $\epsilon > 0$ and hence the claim is proven. □

In the case when $\mathbb{K}$ has a characteristic $p \ne 0$, we have Theorem 18.26.

**Theorem 18.26.** Let $\mathbb{K}$ have characteristic $p \ne 0$, let $D$ be closed, and let $f \in H_b(D)$. Then, $f'(x)$ is identically 0 if and only if there exists $g \in H_b(D)$ such that $f = g^p$.

**Proof.** Indeed, if there exists $g \in H_b(D)$ such that $f = g^p$, of course, we have $f' = 0$. Now, suppose that $f'(x)$ is identically 0. Let $a \in D$ and let $\sum_{n \in S} h_n$ be the Mittag-Leffler series of $f$ in $D$, with $h_0 = \overline{f_0}$ and for each $n \in S$, $h_n = \overline{f_{T_n}}$, for any $f$-hole $T_n$. By Theorem 18.19, we have $(h_n)' = 0$ for every $n \in \mathbb{N}$.

If $D$ is unbounded, $h_0$ is a constant, and then we can find $g_0 \in \mathbb{K}$ such that $(g_0)^p = h_0$. If $D$ is bounded of diameter $r$, then by Corollary 18.13, we can find $g_0 \in H(d(a, r))$ such that $(g_0)^p = h_0$. In the same way, for each $q \in \mathbb{N}$, by Corollary 18.13, we can find $g_q \in H_0(\mathbb{K}\backslash T_q)$ such that $(g_q)^p = h_q$ and then, it is seen that $\lim_{n \to \infty} \|g_n\|_D = 0$ because for each $n \in \mathbb{N}^*$, we have $(\|g_n\|_D)^p = \|h_n\|_D$. So, the series $(\sum_{n=0}^\infty g_n)$ converges in $H_b(D)$ to an element $g$ which clearly satisfies $g^p = f$. This ends the proof of Theorem 18.26. □
# Chapter 19

## Properties of the Function $\Psi$ for Analytic Elements

Throughout this chapter, $D$ is infraconnected.

The function $\Psi(f, \mu)$ was defined for rational functions in Chapter 4. Here, we generalize that function to analytic elements. Its interest is to transform the multiplicative property of the norm $|\cdot|$ into an additive property. Overall, $\Psi$ is piecewise affine. Long ago, such a function was first defined in classical works such as the valuation function of an analytic element [1], [35], [44] denoted by $v(f, \mu)$. However, the function $v(f, \mu)$ has the inconvenient of being contravariant: $\mu = -\log(|x|)$ and $v(f, -\log(|x|)) = -\log(|f|(r))$. Here, we change both senses of variation: $\Psi(f, \mu) = -v(f, -\mu)$ [49], [50].

Among applications, we can show that a set $E$ is infraconnected if and only if for all $f \in H(E)$, $f(E)$ is infraconnected and that an analytic element converges along a monotonous filter $\mathcal{F}$ if and only if $f'$ is vanishing along $\mathcal{F}$.

**Notations.** For every $a \in \tilde{D}$, we put $\lambda(a) = \log(\delta(a, D))$ if $\delta(a, D) > 0$ and $\lambda(a) = -\infty$ if $\delta(a, D) = 0$. We denote by $S$ the diameter of $D$, with $S = +\infty$ if $D$ is not bounded.

Let $a \in \tilde{D}$ and let $\mathcal{F}$ be a circular filter of center $a$ and diameter $r \in [\delta(a, D), S] \cap \mathbb{R}$. By Proposition 3.17, $\mathcal{F}$ is secant with $D$ and then defines an element $\partial_\mathcal{F}\varphi_\mathcal{F}$ of $\text{Mult}(H(D), \mathcal{U}_D)$.

For every $f \in H(D)$ such that $\partial_\mathcal{F}\varphi_\mathcal{F}(f) \neq 0$, we put $\Psi_a(f, \log r) = \log(\partial_\mathcal{F}\varphi_\mathcal{F}(f))$. Next, for an $f \in H(D)$ such that $\partial_\mathcal{F}\varphi_\mathcal{F}(f) = 0$, we put $\Psi_a(f, \log r) = -\infty$.
When $a = 0$, for simplicity, we just put $\Psi(f, \mu) = \Psi_0(f, \mu)$. Then, by definition, we have $\Psi(f, \log r) = \log(|f|(r))$.

In the same way, consider an annulus $\Gamma(0, r, t)$ and $f \in \mathcal{A}(\Gamma(0, r, t))$. Then, for any $s \in [r, t]$, $f$ belongs to $H(C(0, s))$, so we can put consider $\Psi(f, \log(s)) = \log(|f|(s))$. If we consider $f \in \mathcal{A}(a, r^{-})$, so much the more, we can consider $\Psi_a(f, \ell)$ for each $\ell < \log(r)$.

**Remark.** Let $f(x) = \sum_{-\infty}^{+\infty} a_n x^n \in H(C(0, r))$ for some $r > 0$. By Theorem 15.7, we have $\Psi(f, \log r) = \log(C(0,r)\varphi_{0,r}(f)) = \log \|f\|_{C(0,r)} = \sup_{n \in \mathbb{Z}} \Psi(a_n) + n \log r$.

**Proposition 19.1.** Let $a \in \tilde{D}$, let $\mu \in [\lambda(a), \log(S)] \cap \mathbb{R}$, and let $f, g \in H(D)$. Then, $\Psi_a(f + g, \mu) \le \max(\Psi_a(f, \mu), \Psi_a(g, \mu))$, and when $\Psi_a(f, \mu) > \Psi_a(g, \mu)$, then $\Psi_a(f + g, \mu) = \Psi_a(f, \mu)$. Moreover, $\Psi_a(fg, \mu) = \Psi_a(f, \mu) + \Psi_a(g, \mu)$.

Let $r, t \in [0, +\infty[$ be such that $r \le t$. Let $f \in H(D)$ be such that $\Psi_a(f, \mu)$ is bounded in $[\log(r), \log(t)]$. Then, $\Psi_a(f, \mu)$ is continuous and piecewise affine in $[\log(r), \log(t)]$. Further, there exists $h \in R(D)$ such that $\Psi_a(f, \mu) = \Psi_a(h, \mu) \quad \forall \mu \in [\log(r), \log(t)]$.

Inside $D \cap \Gamma(a, r, t)$, the relation $\Psi(f(x)) = \Psi_a(f, \Psi(x-a))$ holds in all classes of all circles $C(a, s)$, except maybe in finitely many classes of finitely many circles $C(a, s)$.

Moreover, if $\Gamma(a, r, t) \subset D$, the function $\Psi_a(f, \mu)$ is convex in $[\log r, \log t]$.

**Proof.** Without loss of generality, we can assume $a = 0$. The first statements concerning operations and inequalities come directly from those of multiplicative semi-norms $D\varphi$. Now, suppose that $\Psi(f, \mu)$ is bounded in $[\log(r), \log(t)]$, hence there exists $\epsilon > 0$ such that $\Psi(f, \mu) > \log \epsilon \quad \forall \mu \in [\log r, \log t]$.

Let $h \in R(D)$ satisfy $\|f-h\|_D < \epsilon$. Particularly, for every circular filter $\mathcal{F}$ secant with $D$, we have $D\varphi_{\mathcal{F}}(f-h) < \epsilon$, and particularly, $D\varphi_{a,\rho}(f-h) < \epsilon \quad \forall \rho \in [r,t]$ i.e. $\Psi(f-h, \mu) < \log(\epsilon) < \Psi(f, \mu) \quad \forall \mu \in [\log r, \log t]$. Consequently, $\Psi(f, \mu) = \Psi(h, \mu) \quad \forall \mu \in [\log r, \log t]$. Now, by Corollary 4.18, the function $\Psi(h, \mu)$ is continuous, piecewise in $[\log r, \log t]$ and so is $\Psi(f, \mu)$. Moreover, if $\Gamma(a, r, t) \subset D$, the function $\Psi(h, \mu)$ is convex in $[\log r, \log t]$, hence so is $\Psi(f, \mu)$.

By Lemma 4.13, the relation $\Psi(h(x)) = \Psi_a(h, \Psi(x-a))$ holds in all classes of all circles $C(a, s)$, except maybe in finitely many classes
of finitely many circles $C(a, s)$. Therefore, the same relation holds for $f$. $\square$

**Proposition 19.2.** Let $a \in D$ and let $f \in H(D)$ satisfy $f(a) \neq 0$. There exists $\mu_0 \in \mathbb{R}$ such that $\Psi_a(f, \mu) = \Psi(f(a))$ whenever $\mu \le \mu_0$. Let $r \in \mathbb{R}_+^*$, let $\Lambda = C(0, r)$, and let $f$ and $g \in H(\Lambda)$ satisfy $\|f - g\|_{\Lambda} < \|f\|_{\Lambda}$. Then, we have $\nu^+(f, \log r) = \nu^+(g, \log r)$, $\nu^-(f, \log r) = \nu^-(g, \log r)$.

**Proof.** Indeed, let us take $r > 0$ such that $|f(x) - f(a)| < |f(a)|$ whenever $x \in d(a, r) \cap D$, hence $|f(x)| = |f(a)|$ whenever $x \in d(a, r) \cap D$, and therefore, $\Psi_a(f, \mu) = \Psi(f(a))$ whenever $\mu \le \log(r)$.

Let $f(x) = \sum_{n=-\infty}^{+\infty} a_n x^n$ and let $g(x) = \sum_{n=-\infty}^{+\infty} b_n x^n$. From the hypothesis, we see that $\|f\|_{\Lambda} = \|g\|_{\Lambda}$. By Corollary 14.9, we have

$$ (1) \sup_{n \in \mathbb{Z}} |a_n|r^n = \|f\|_{\Lambda} = \sup_{n \in \mathbb{Z}} |b_n|r^n $$

$$ \text{and } \|f - g\|_{\Lambda} = \sup_{n \in \mathbb{Z}} |a_n - b_n|r^n. $$

Let $s = \nu^-(f, \log r)$ and let $t = \nu^+(f, \log r)$. We see that $|a_s - b_s|r^s \le \|f - g\|_{\Lambda} < \|f\|_{\Lambda} = |a_s|r^s$, hence

$$ (2) \quad |b_s| = |a_s|. $$

In the same way, we have

$$ (3) \quad |a_t| = |b_t|. $$

Now, for every $n < s$ and for every $n > t$, we have $|a_n|r^n < |a_s|r^s = \|f\|_{\Lambda}$, hence $|b_n|r^n < \|f\|_{\Lambda}$. Finally, by (1), (2), and (3), we see that $\nu^-(g, \log r) = s$, $\nu^+(g, \log r) = t$. $\square$

By Propositions 19.1, 19.2, and 4.19, we can derive Corollary 19.3.

**Corollary 19.3.** Let $f(x) \in H(\Gamma(0, r_1, r_2))$ (respectively $f(x) \in H(\Delta(0, r_1, r_2))$) (with $0 < r_1 < r_2$) and let $\sum_{n=-\infty}^{+\infty} a_n x^n$ be its Laurent series. The function $\mu \to \Psi(f, \mu)$ is bounded in $] \log r_1, \log r_2[$ (respectively in $[\log(r_1), \log(r_2)]$) and equal to $\sup_{n \in \mathbb{Z}}(\Psi(a_n) + n\mu)$. Next, we have $\Psi(f(x)) \le \Psi(f, \Psi(x))$ whenever $x \in \Gamma(0, r_1, r_2)$ (respectively whenever $x \in \Delta(0, r_1, r_2)$) and the equality holds in all of $\Gamma(0, r_1, r_2)$ (respectively un all of $\Delta(0, r_1, r_2)$) except in finitely many classes of finitely many circles $C(0, r)$ ($r_1 < r < r_2$)
(respectively $r_1 \le r \le \le r_2$). 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 $] \log r_1, \log r_2[$ (respectively in $[\log r_1, \log r_2]$). Next, given another $g \in H(\Gamma(0, r_1, r_2))$ (respectively $g \in H(\Delta(0, r_1, r_2))$), the functions $\nu^+$ and $\nu^-$ satisfy $\nu^+(fg, \mu) = \nu^+(f, \mu) + \nu^+(g, \mu)$, $\nu^-(fg, \mu) = \nu^-(f, \mu) + \nu^-(g, \mu)$. Further, 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.

**Proposition 19.4.** Let $a \in \tilde{D}$ and let $f \in H(D)$. If $f(a) \neq 0$, there exists $s > 0$ such that $\Psi(f, \mu) = \Psi(f(a)) \quad \forall \mu \le s$. Let $b \in D$ be such that $|a-b| = r$ and $d(b, r^{-}) \subset D$. Then we have $\Psi_b(f, \mu) = \Psi_a(f, \mu) \quad \forall \mu \le \Psi(b-a)$.

**Proof.** Since $f(a) \neq 0$, the first statement is immediate since $|f(x)|$ is a constant inside a disk of center $a$. Next, by Lemma 4.14, the relation $\Psi_a(f, \mu) = \Psi_b(f, \mu)$ when $\Psi(a-b) \le \mu$ is true for every $f \in R(D)$, hence by (2), is obviously generalized to every $f \in H(D)$. $\square$

**Proposition 19.5.** Let $\mu \in \mathbb{R}$ and let $f(x) = \sum_{n=-\infty}^{+\infty} a_n x^n \in H(C(0, \theta^\mu))$. Then, $\Psi(f, \mu)$ is equal to $\sup_{n \in \mathbb{Z}} \Psi(a_n) + n\mu$ and we have $\Psi(f(x)) \le \Psi(f, \mu)$ for all $x \in C(0, \theta^\mu)$. Moreover, the equality holds in every class except in finitely many classes where $f$ admits zeros. Further, if $\nu^+(f, \mu) = \nu^-(f, \mu)$, then $\Psi(f(x)) = \Psi(f, \mu)$ whenever $x \in C(0, \theta^\mu)$.

If $h \in H(C(0, \theta^\mu))$ satisfies $\Psi(f-h, \mu) < \Psi(f, \mu)$, then $\nu^+(f, \mu) = \nu^+(h, \mu)$ and $\nu^-(f, \mu) = \nu^-(h, \mu)$.

**Proof.** Let $\Lambda = C(0, \theta^\mu)$, let $s = \nu^-(f, \mu)$, and let $t = \nu^+(f, \mu)$. By the remark above, $\Psi(f, \mu)$ is obviously equal to $\sup_{n \in \mathbb{Z}} (\Psi(a_n) + n\mu)$. Let $x \in C(0, \theta^\mu)$. The inequality $\Psi(f(x)) \le \Psi(f, \mu)$ is true because $\Psi(f, \mu) = \log \|f\|_{\Lambda} \ge \Psi(f(x))$. Finally, by Proposition 19.1, the equality holds in all the classes except in finitely many. If $\nu^+(f, \mu) = \nu^-(f, \mu)$, then $\Psi(a_s x^s) = \Psi(a_s) + s\mu > \Psi(a_n x^n)$ whenever $n \neq s$, hence $\Psi(f(x)) = \Psi(f, \mu)$.

Now, let $h \in H(\Lambda)$ satisfy $\Psi(f - h, \mu) < \Psi(f, \mu)$ and let $h(x) = \sum_{n=-\infty}^{+\infty} b_n x^n$. We have $\Psi(a_n - b_n) + n\mu < \Psi(a_s) + s\mu$ whenever