**Notation.** We consider a divisor $T$ on $d(0, R^{-})$ of the form $(a_{i,m}, q_{i,m})_{i \le u_m, m \in \mathbb{N}}$, where the points $a_{i,m}$ lie in the circle $C(0, \rho_m)$ with $0 < \rho_m < \rho_{m+1} \quad \forall m \in \mathbb{N}$ and $\lim_{m \to +\infty} \rho_m = R$. We denote by $(P_m)_{m \in \mathbb{N}}$ the polynomial $P_m = \prod_{i=1}^{u_m} \left(1 - \frac{x}{a_{i,m}}\right)^{q_{i,m}}$ whose zeros by definition belong to the circle $C(0, \rho_m)$, and for each $m \in \mathbb{N}$, we set $d_m = \deg(P_m)$.

We denote by $(Q_m)_{m \in \mathbb{N}}$ a sequence of $K[x]$ satisfying $|Q_m|(\rho_m) \le |T|(\rho_m)$, $\deg(Q_m) < d_m \quad \forall m \in \mathbb{N}$. We note that $|T|(r) = \prod_{m \ge 1}^{s-1} \left(\frac{r}{\rho_m}\right)^{d_m}$ whenever $r \in [\rho_{s-1}, \rho_s]$. Given $q \in \mathbb{N}$, $s \in \mathbb{N}$, $r \in ]0, R[$, we set

$$ \lambda(q, s, r) = |T|(r) \left(\frac{r}{R}\right)^q \min \left(1, \left(\frac{r}{\rho_s}\right)\right), \text{ i.e.} $$

$$ \lambda(q, s, r) = |T|(r) \left(\frac{r}{R}\right)^q \left(\frac{r}{\rho_s}\right) \quad \forall r \in ]0, \rho_s], $$

$$ \lambda(q, s, r) = |T|(r) \left(\frac{r}{R}\right)^q \quad \forall r \in [\rho_s, R[. $$

Particularly, we note that $\lambda(q, s, r) \le |T|(r) \quad \forall r < R$. Now, given $n \in \mathbb{N}$, we set $\nu(q, s, n) = \inf_{r<R} \left(\frac{\lambda(q,s,r)}{r^n}\right)$.

Recall that the Euclidean division in $H(C(0, r))$ is defined in Chapter 27. We denote by $\Lambda(q, s)$ the subset of the $h \in \mathcal{A}(d(0, R^{-}))$ satisfying the following:

(α) $h(0) = 1$.

(β) $|h|(r) \le |T|(r) \quad \forall r < R$.

(γ) $P_m$ divides $h - Q_m$ in $\mathcal{A}(d(0, R^{-})) \quad \forall m = 0, \dots, s$.

(δ) The rest $X_m$ of the Euclidean division of $h$ by $P_m$ in $H(C(0, \rho_m))$ satisfies $|X_m - Q_m|(\rho_m) \le \lambda(q, m, \rho_m) = |T|(\rho_m) \left(\frac{\rho_m}{R}\right)^q \quad \forall m \ge 0$.

**Remark.** By definition, $q$ and $n$ being fixed, the sequence in $s$: $(\nu(q, s, n))_{s \in \mathbb{N}}$ is decreasing.

In the proof of Theorem 28.4, we use Lemmas 28.2 and 28.3. Lemma 28.2 is immediate:

**Lemma 28.2.** $\Lambda(q+1, s) \subset \Lambda(q, s)$ and $\Lambda(q, s)$ is a closed subset of $\mathcal{A}(d(0, R^{-}))$.

**Lemma 28.3.** Let $q \in \mathbb{N}$, $s \in \mathbb{N}^*$, $f \in \Lambda(q, s-1)$. There exists $g \in \Lambda(q, s)$ such that $|g-f|(r) \le \lambda(q, s, r) \quad \forall r < R$.
**Proof.** Let $h = \prod_{m=0}^{s-1} P_m$ and $u = x^{q+1}h$. Then $P_m$ and $x$ have no zero in $C(0, \rho_s)$, hence they are invertible in $H(C(0, \rho_s))$ and hence so is $u$. For each $m \in \mathbb{N}$, we denote by $R_m$ the rest of the Euclidean division of $f$ by $P_m$ in $H(C(0, \rho_m))$. Since $f \in \Lambda(q, s-1)$, by $(\gamma)$, $P_m$ divides $f - Q_m$ for every $m \le s-1$, hence $R_m = Q_m \forall m = 0, \dots, s-1$.

Consider now the Euclidean division of $(Q_s - R_s)u^{-1}$ by $P_s$ in $H(C(0, \rho_s))$: $(Q_s - R_s)u^{-1} = EP_s + S$, with $\deg(S) < d_s$, $E \in H(C(0, \rho_s))$ and $|S(\rho_s)| \le |(Q_s - R_s)u^{-1}|(\rho_s)$, hence

$$ (1) \quad |Su|(\rho_s) \le |(Q_s - R_s)|(\rho_s). $$

We then take $g = f + Su$, hence $g = f + Shx^{q+1}$. We show that $g$ belongs to $\Lambda(q, s)$ and that $|g - f|(r) \le \lambda(q, s, r) \forall r < R$. We note that $g$ belongs to $\mathcal{A}(d(0, R^{-}))$ and that $g(0) = f(0) = 1$.

Next, by hypothesis, $P_s$ divides $f - R_s$ and by construction divides $Su + R_s - Q_s$. But $g - Q_s = f + Su - Q_s = f - R_s + Su + R_s - Q_s \in \mathcal{A}(d(0, R^{-}))$, hence $P_s$ divides $g - Q_s$ in $\mathcal{A}(d(0, R^{-}))$.

Now, by hypothesis, $f$ lies in $\Lambda(q, s-1)$, hence particularly $|f|(r) \le |T|(r)$, and we have

$$ (2) \quad |Q_m - R_m|(\rho_m) \le |T|(\rho_m) \left(\frac{\rho_m}{R}\right)^q \quad \forall m \ge 1. $$

Thus, by (1) and (2), we have

$$ (3) \quad |g - f|(\rho_s) \le |T|(\rho_s) \left(\frac{\rho_s}{R}\right)^q. $$

But since $|S|$ is an increasing function in $r$, we have $|S|(r) \le |S|(\rho_s)$ whenever $r \le \rho_s$. On the other hand, $|h|(r) = |T|(r) \forall r \le \rho_s$. And, as we saw,

$$ \begin{aligned} |g - f|(\rho_s) &= |Su|(\rho_s) = |Shx^{q+1}|(\rho_s) = |S|(\rho_s)|T|(\rho_s)(\rho_s)^{q+1} \\ &\le |T|(\rho_s) \left(\frac{\rho_s}{R}\right)^q. \end{aligned} $$

Consequently, $|S|(r) \le |S|(\rho_s) \le \frac{1}{R^q \rho_s} \forall r \le \rho_s$, and hence

$$ \begin{aligned} |g - f|(r) &= |Shx^{q+1}|(r) = |S|(r)|h|(r)r^{q+1} \\ &\le |T|(r) \left(\frac{\rho_s}{R}\right)^q \left(\frac{r}{\rho_s}\right) \forall r \le \rho_s. \end{aligned} $$
Now, when $r \ge \rho_s$, we have

$$ 
\begin{aligned}
|T|(r) &= \prod_{m \ge 1} |P_m|(r) = |h|(r) \prod_{m \ge s} |P_m|(r) \ge |h|(r) |P_s|(r) \\
&= |h|(r) \left(\frac{r}{\rho_s}\right)^{d_s}.
\end{aligned}
$$

Next, $Sx^{q+1}$ is of the form $\lambda \prod_{j=1}^{m} (x - x_j)$ and is a polynomial of degree $m \le q + d_s$, hence

$$ \frac{|x^{q+1}S|(r)}{|x^{q+1}S|(|\rho_s|)} = \prod_{j=1}^{m} \frac{|x - x_j|(r)}{|x - x_j|(|\rho_s|)} \le \left(\frac{r}{\rho_s}\right)^{q+d_s}. $$

Thus,

$$ 
\begin{aligned}
|g - f|(r) &= |Shx^{q+1}|(r) = |Sx^{q+1}|(r)|h|(r) \\
&\le |Sx^{q+1}|(|\rho_s|) \left(\frac{r}{\rho_s}\right)^{q+d_s} \left(\frac{\rho_s}{r}\right)^{d_s} |T|(r) \\
&= |T|(r) \left(\frac{r}{\rho_s}\right)^q |Sx^{q+1}|(|\rho_s|) = |T|(r) \left(\frac{r}{\rho_s}\right)^q (\rho_s)^{q+1} |S|(\rho_s) \\
&\le |T|(r) r^q (\rho_s) \frac{1}{R^q \rho_s} = |T|(r) \left(\frac{r}{R}\right)^q.
\end{aligned}
$$

And finally, with (4), we obtain $|g - f|(r) \le \lambda(q, s, r) \forall r \in ]0, R[. \quad \square$

**Theorem 28.4.** Suppose $\mathbb{K}$ is spherically complete. Assume that $|Q_m|(\rho_m) \le |T|(\rho_m) \forall m \in \mathbb{N}$. Let $R \in ]0, +\infty[$. There exists $f \in \mathcal{A}(d(0, R^{-}))$ satisfying the following:

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

(ii) $|f|(r) \le |T|(r) \forall r < R$.

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

**Proof.** We mean to construct a sequence of functions $(f_q)_{q \in \mathbb{N}}$ which belong to $\mathcal{A}(d(0, R^{-}))$, converging in $\mathcal{A}(d(0, R^{-}))$ to a function $f$ satisfying the claim.

We first fix $q \in \mathbb{N}$ and take $f_q \in \Lambda(q, 0)$. Let $f_q(x) = \sum_{n=0}^{\infty} a_n x^n$. We construct a sequence $(g_{q,s})_{s \in \mathbb{N}}$ satisfying $g_{q,s} \in \Lambda(q, s)$ and $|g_{q,s} - g_{q,s-1}|(r) \le \lambda(q, s, r)$, with $g_{q,0} = f_q$. Suppose already constructed
the $g_{q,j}$ for $j = 0, \dots, s-1$. By Lemma 28.3, there exists $h \in \Lambda(q, s)$ such that $|h - g_{q,s-1}|(r) \le \lambda(q, s, r)$. So, we can set $g_{q,s} = h$ and the sequence is then defined by induction for all $s \in \mathbb{N}$.

Now, for each $s \in \mathbb{N}$, we set $g_{q,s}(x) = \sum_{n=0}^{\infty} b_{q,s,n}x^n$. Since by construction the sequence $g_{q,s}$ satisfies

$$ (1) \quad |g_{q,s} - g_{q,s-1}|(r) \le \lambda(q, s, r) \forall r < R, $$

then for each fixed $n \in \mathbb{N}$, the sequence $(b_{q,s,n})_{s \in \mathbb{N}}$ satisfies $|b_{q,s,n} - b_{q,s-1,n}| \le \nu(q, s, n)$. Thus, for each fixed $n \in \mathbb{N}$, we consider the sequence of disks $(D_{q,s,n})_{s \in \mathbb{N}}$ defined as $D_{q,s,n} = d(b_{q,s,n}, \nu(q, s, n))$. Since the sequence $(\nu(q, s, n))_{s \in \mathbb{N}}$ is decreasing and since $|b_{q,s,n} - b_{q,s-1,n}| \le \nu(q, s, n)$, the sequence of disks $(D_{q,s,n})_{s \in \mathbb{N}}$ is decreasing with respect to the inclusion. Consequently, since $\mathbb{K}$ is spherically complete, for each $n \in \mathbb{N}$, there exists $a_{q+1,n} \in \bigcap_{s=0}^{\infty} D_{q,s,n}$. Particularly, since $\nu(q, s, n) = 0 \forall n \le q$, we note that $b_{q,s,0} = b_{q,s-1,0} = 1$ because $g_{q,s-1} \in \Lambda(q, s-1)$. Consequently, $a_{q+1,0} = 1$.

Now, we show that $f_{q+1}$ belongs to $\Lambda(q+1, 0)$. Let $f_{q+1}(x) = \sum_{n=0}^{\infty} a_{q+1,n}x^n$. Since $a_{q+1,0} = 1$, $f_{q+1}$ satisfies Relation (α). Next, by construction, we have $|a_{q+1,n} - b_{q,s-1,n}| \le \nu(q, s, n) \le \nu(q, 1, n) \forall n \in \mathbb{N}$, hence obviously, $|a_{q+1,n} - a_{q,n}| \le \nu(q, 1, n) \forall n \in \mathbb{N}$. Consequently, $|a_{q+1,n} - a_{q,n}|r^n \le \lambda(q, 1, r) \forall r < R$, hence

$$ (2) \quad |f_{q+1} - f_q|(r) \le \lambda(q, 1, r) \le |T|(r) \left(\frac{r}{R}\right)^q \quad \forall r < R. $$

Now since, by hypothesis, $|f_q|(r) \le |T|(r)$, by (2), we can see that $|f_{q+1}|(r) \le |T|(r)$, and therefore, $f_{q+1}$ satisfies Relation (β). Since $(\gamma)$ is trivial when $s = 0$, it only remains to show that $f_{q+1}$ satisfies (δ).

For each $m \in \mathbb{N}$, let $S_{m,q+1}$ be the rest of the Euclidean division of $f_{q+1}$ by $P_m$ in $H(C(0, \rho_m))$. For each $s \ge m$, since $P_m$ divides $g_{q,s} - Q_m$, the rest of the Euclidean division of $g_{q,s} - f_{q+1}$ by $P_m$ in $H(C(0, \rho_m))$ is equal to $Q_m - S_{m,q+1}$. Consequently, by (1), we have

$$ (3) \quad |Q_m - S_{m,q+1}|(\rho_m) \le |g_{q,s} - f_{q+1}|(\rho_m) \le \lambda(q, s, \rho_m), $$

and hence

$$ (4) \quad |Q_m - S_{m,q+1}|(\rho_m) \le |T|(\rho_m) \left(\frac{\rho_m}{R}\right)^q \left(\min\left(1, \frac{\rho_m}{\rho_s}\right)\right) \quad \forall s \ge m. $$
Now, since $\lim_{s \to \infty} \rho_s = R$, by (4), we have $|Q_m - S_{m,q+1}|(\rho_m) \le |T|(\rho_m)(\frac{\rho_m}{R})^{q+1}$. This finishes showing that $(\delta)$ is satisfied by $g_{q+1,s}$, and therefore, $g_{q+1,s}$ belongs to $\Lambda(q+1,0)$. This true for all $s$, hence by Lemma 28.2, $f_{q+1}$ also belongs to $\Lambda(q+1,0)$.

Thus, we have constructed a sequence $(f_q)_{q \in \mathbb{N}}$ of $\mathcal{A}(d(0, R^{-}))$ satisfying $f_q \in \Lambda(q, 0) \forall q \in \mathbb{N}$. By (2), we can see that the sequence $(f_q)_{q \in \mathbb{N}}$ converges in all $H(d(0, \rho))$, for every $\rho < R$, to a limit $f$ which thereby belongs to $H(d(0, \rho))$ for all $\rho < R$. Consequently, that function $f$ belongs to $\mathcal{A}(d(0, R^{-}))$. Moreover, since $\Lambda(q, 0)$ is closed, $f$ belongs to $\Lambda(q, 0)$ for every $q \in \mathbb{N}$. Consequently, by Relation $(\delta)$ true for every $q$, the rest $X_m$ of the Euclidean division of $f_q$ by $P_m$ in $H(C(0, \rho_m))$ satisfies $|X_m - Q_m|(\rho_m) \le |T|(\rho_m)(\frac{\rho_m}{R})^q$ for every $q \in \mathbb{N}$, hence $X_m = Q_m$. So, $P_m$ divides $f - Q_m$ for every $m \in \mathbb{N}$. And by construction, $f$ satisfies $(\alpha)$ and $(\beta)$, which completes the proof. $\square$

**Corollary 28.5.** Suppose $\mathbb{K}$ is spherically complete. Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of $d(0, R^{-})$ such that $|a_n| \le |a_{n+1}| \forall n \in \mathbb{N}$ and $\lim_{n \to +\infty} |a_n| = R$ and let $(b_n)_{n \in \mathbb{N}}$ be a sequence of $\mathbb{K}$. There exists $f \in \mathcal{A}(d(a, R^{-}))$ such that $f(a_n) = b_n \forall n \in \mathbb{N}$.

**Proof.** Indeed, we can define a sequence of integers $(s_n)_{n \in \mathbb{N}}$ such that the divisor $T = (a_n, s_n)_{n \in \mathbb{N}}$ satisfies $|T|(|a_n|) \ge |b_n| \forall n \in \mathbb{N}$. $\square$

**Theorem 28.6.** Suppose $\mathbb{K}$ is spherically complete. Let $T$ be a divisor on $d(a, R^{-})$. There exists $f \in \mathcal{A}(d(a, R^{-}))$ such that $\mathcal{D}(f) = T$.

**Proof.** Without loss of generality, we may obviously assume $a = 0$. Take $Q_m = 0 \forall m \in \mathbb{N}$. By Theorem 28.4, there exists $f \in \mathcal{A}(d(0, R^{-}))$ such that

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

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

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

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

$$|T|(\rho_s) = \prod_{j=1}^{s} \prod_{i=1}^{u_m} \left|\left(1 - \frac{x}{a_{i,m}}\right)^{q_{i,m}}\right|(\rho_s) = \prod_{j=1}^{s} \prod_{i=1}^{u_m} \left(\frac{\rho_s}{\rho_m}\right)^{q_{i,m}}.$$
Now, suppose that $T \neq \mathcal{D}(f)$. Then there exists $\alpha \in d(0, R^{-})$ such that $\omega_{\alpha}(f) > T(\alpha)$. Let $s$ be such that $\rho_{s} > |\alpha|$. Since $f(0) = 1$, we have

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

a contradiction to (iii). □

Similar to $\mathcal{A}(\mathbb{K})$, the algebra $\mathcal{A}(d(a, R^{-}))$ is provided with the natural topology of uniform convergence on each disk $d(0, r)$ whenever $0 < r < R$. Such a topology makes $\mathcal{A}(d(a, R^{-}))$ a topological $\mathbb{K}$-algebra.

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

**Theorem 28.7.** Suppose $\mathbb{K}$ is spherically complete. All closed ideals of $\mathcal{A}(d(a, R^{-}))$ are principal.

**Proof.** Let $I$ be a closed ideal of $\mathcal{A}(d(a, R^{-}))$ and let $E = \mathcal{D}(I)$. By Theorem 27.10, we have $I = \mathcal{T}_{a,R}(E)$. Now, by Theorem 28.6, there exists $g \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(g) = E$ and of course $g$ belongs to $\mathcal{T}_{a,R}(E)$ hence to $I$. Consequently, $g\mathcal{A}(d(a, R^{-})) \subset I$. Conversely, by Corollary 27.2, we have $I = g\mathcal{A}(d(a, R^{-}))$. □
# Chapter 29

## Motzkin Factorization and Roots of Analytic Functions

In this chapter, $D$ is a closed infraconnected set and $f$ belongs to $H(D)$.

The idea of factorizing semi-invertible analytic elements into a product of singular factors is a remarkable idea due to Motzkin [75]. This factorization has tight links with the Mittag-Leffler series, as shown in [36].

**Lemma 29.1.** Let $T = d(a, r^{-})$, with $a \in \mathbb{K}$ and $r > 0$, let $E = \mathbb{K}\setminus T$ and take $b \in T$. Let $g \in H(E)$ be invertible in $H(E)$. Then there exist $\lambda \in \mathbb{K}$, $q \in \mathbb{Z}$ and $h \in H(E)$ invertible in $H(E)$, satisfying $\|h - 1\|_{E} < 1$, $\lim_{|x| \to +\infty} h(x) = 1$ and $g(x) = \lambda(x - b)^{q}h(x)$. Moreover, $\lambda$, $q$, and $h$ are respectively unique, satisfying those relations. Further, both $\lambda$ and $q$ do not depend on $b$ in $T$ and $\frac{q'}{g}$ belongs to $H_{0}(E)$.

**Proof.** Without loss of generality, we may obviously assume $a = 0$. As $g$ is invertible, if $g$ belongs to $H_0(E)$, then $\frac{1}{g}$ does not. So, we may clearly assume that $g$ does not belong to $H_0(E)$. By Theorem 11.5, $g$ is of the form $\tilde{g} + \hat{g}$, with $\tilde{g} \in \mathbb{K}[x]$, $\tilde{g} \neq 0$, and $\hat{g} \in H_0(E)$. Let $q = \deg(\tilde{g})$ and let $\lambda$ be its coefficient of degree $q$. Now, we put $h(x) = \frac{g(x)}{\lambda(x-b)^q}$. By definition both $\lambda$, $q$ do not depend on $b$ in $T$. Hence, we may also assume $b = 0$. Clearly, $h$ satisfies $\lim_{|x| \to +\infty} h(x) = 1$. Since $H(E)$ is a $\mathbb{K}$-algebra and since $g$ is invertible, $h$ is invertible in $H(E)$. In particular, we note that $h$ is bounded and admits no zero in $E$. Now, we check that $\|h - 1\|_{E} < 1$. Let $s = \frac{1}{r}$, let $A = d(0, s)$
and let $\phi(u) = h(\frac{1}{u})$ whenever $u \in d(0, s)$, $u \neq 0$. Then $\phi$ belongs to $H(d(0, s)\setminus\{0\})$. But since $h$ is bounded in $E$, $\phi$ is bounded in $d(0, s)\setminus\{0\}$. Moreover, the condition $\lim_{|x|\to+\infty} h(x) = 1$ shows that $\lim_{x\to 0} \phi(x) = 1$, hence $\phi$ belongs to $H(d(0, s))$.

Thus, $\phi(u)$ is of the form $\sum_{n=0}^{\infty} a_n u^n$ with $a_0 = 1$ and hence, by Theorem 22.7, we have $|a_n|s^n < 1 \quad \forall n > 0$. Let $\epsilon = \sup\{|a_n|s^n|n > 0\}$. Then we have $\|\phi - 1\|_{d(0,s)} = \|h - 1\|_E = \epsilon$. Now, $h$, $q$, and $\lambda$ are easily seen to be unique. Indeed, let $g(x) = \alpha x^l l(x)$ with $l$ invertible in $H(E)$, satisfying $\lim_{|x|\to+\infty} l(x) = 1$. Then we have $1 = x^{q-t} \frac{\lambda h(x)}{\alpha l(x)}$. Consequently, considering the limit when $|x|$ tends to $+\infty$, we have $q = t$, $\lambda = \alpha$, and therefore, $h = l$. Finally, we check that $\frac{g'}{g}$ belongs to $H_0(E)$. Indeed, $\frac{g'}{g} = \frac{q}{x-b} + \frac{h'}{h}$. Obviously, $\frac{q}{x-b}$ belongs to $H_0(E)$. Since $\lim_{|x|\to\infty} |h(x)| = 1$, it is seen that $h(x)$ is of the form $1 + \sum_{n=1}^{\infty} \frac{a_n}{x^n}$ with $\lim_{n\to\infty} |\frac{a_n}{s^n}| = 0$, and therefore, $h'$ is an element of $H(E)$ such that $\lim_{|x|\to\infty} |h'(x)| = 0$. As a consequence, $\frac{h'}{h}$ belongs to $H_0(E)$. Hence, so does $\frac{g'}{g}$ and this ends the proof of Lemma 29.1. $\square$

**Definition.** Let $E = \mathbb{K}\backslash d(a, r^{-})$ with $a \in \mathbb{K}$ and $r > 0$. Let $f \in H(E)$ be invertible in $H(E)$ and let $\lambda(x-a)^q h(x)$ be the factorization given in Lemma 29.1. The integer $q$ will be named the index of $f$ associated to $d(a, r^{-})$ and will be denoted by $mo(f, d(a, r^{-}))$. If $\lambda = 1$, the element $f$ will be called a pure factor associated to $d(a, r^{-})$. Let $\mathcal{G}^T$ be the group of invertible elements of $H(\mathbb{K}\backslash T)$.

The following Corollary is then immediate.

**Corollary 29.2.** Let $T = d(a, r^{-})$. The set of pure factors associated to $T$ is a sub-multiplicative group of the group $\mathcal{G}^T$. Further, every element of $\mathcal{G}^T$ is of the form $\lambda h$ with $h$ a pure factor associated to $T$ and $\lambda \in \mathbb{K}^*$.

**Lemma 29.3.** Let $T = d(a, r^{-})$, let $E = \mathbb{K}\backslash T$ with $a \in \mathbb{K}$ and let $f$ be a pure factor associated to $T$ such that $\|f - 1\|_E < 1$. Then $mo(f, T) = 0$.

**Proof.** Without loss of generality, we can assume $a = 0$. Let $q = mo(f, T)$. By Lemma 29.1, there exists a unique element $h$ invertible in $H(E)$ such that $f = x^q h$ and $\lim_{|x|\to+\infty} |h(x)| = 1$. Therefore, by Theorem 14.6, $h(x)$ is of the form $1 + \sum_{n=1}^{+\infty} \frac{a_n}{x^n}$, hence
$\|h\|_E < 1$. So, if $q > 0$, then $f$ is unbounded, a contradiction. Next, by Corollary 29.2, $\frac{1}{j}$ is a pure factor satisfying again the hypothesis of Corollary 29.2, hence the hypothesis $q < 0$ gets to a contradiction again. □

**Definition.** Let $f$ belong to $H(D)$. Let $T$ be a hole of $D$ and let $h$ be a pure factor associated to $T$. If $\frac{f}{h}$ belongs to $H(D \cup T)$ and has no zero inside $T$, $h$ is called the *Motzkin factor of $f$ in the hole $T$*.

**Theorem 29.4.** Let $T$ be a hole of $D$ and let $f$ have a Motzkin factor $h$ in $T$. Then $h$ is unique. Further, if $T$ is not an $f$-hole, $h$ is the polynomial of the zeros of $f$ inside $T$. Moreover, if $E$ is another infraconnected set included in $D$ admitting $T$ as a hole and if $g$ denotes the restriction of $f$ to $E$, then $g$ admits a Motzkin factor in the hole $T$ as an element of $H(E)$ and this Motzkin factor is equal to $h$.

**Proof.** Let $f$ have another Motzkin factor $l$ in $T$, let $F = \frac{f}{h}$ and let $G = \frac{f}{h}$. Since $G$ has no zeros inside $T$, by Theorem 24.9, there exists a closed bounded infraconnected set $D'$ satisfying $T \subset D' \subset (D \cup T)$, $T \neq D'$, such that $G$ is invertible in $H(D')$. Hence, in $H(D')$, we have $\frac{F}{G} = \frac{l}{h}$, and hence, $\frac{l}{h}$ belongs to $H(D')$. Since $T \neq D'$, it is seen that $D' \cap (\mathbb{K}\setminus T)$ is an infraconnected closed bounded set included in $D$ that admits $T$ as a hole. Moreover, we have $D' \cup (\mathbb{K}\setminus T) = \mathbb{K}$. Therefore, by Theorem 15.10, we see that $\frac{l}{h}$ belongs to $H(\mathbb{K})$ and is hence a polynomial $P$. Since $\frac{F}{G}$ belongs to $H(D')$ and has no zeros in $T$, it is seen that $mo(h, T) = mo(l, T)$, so we have $\lim_{|x|\to\infty} \frac{h(x)}{l(x)} = 1$. Hence, $P = 1$ and this proves that $h$ is unique.

Now, we assume that $T$ is not an $f$-hole, hence $f$ belongs to $H(D \cup T)$. Let $Q$ be the polynomial of the zeros of $f$ inside $T$. Then by Corollary 16.6, $\frac{f}{Q}$ belongs to $H(D \cup T)$ and has no zeros inside $T$. Since its Motzkin factor $h$ is unique, we have $h = Q$. The last statement about $g$ is obvious because $\frac{g}{h}$ clearly belongs to $H(E \cup T)$ and has no zero inside $T$. This ends the proof of Theorem 29.4. □

**Definition and notation:** We will call the *f-supergquence of $D$* the sequence of the holes $(T_n)_{n \in I}$ such that either $T_n$ is an $f$-hole or $f$ belongs to $H(D \cup T)$ and has at least one zero inside $T_n$. If $f$ admits a Motzkin factor $h$ in a hole $T$, it will be denoted by $f^T$ and $mo(h, T)$ will be called the *Motzkin index of $f$ in $T$*. For every hole which does not belong to the $f$-supergquence, we put $f^T = 1$.
Lemma 29.5 is immediate.

**Lemma 29.5.** Let $D \in \text{Alg}$, let $T$ be a hole of $D$ and let $f, g \in H(D)$ admitting Motzkin factors in $T$. Then $(fg)^T = f^T g^T$ and $\text{mo}(fg, T) = \text{mo}(f, T) + \text{mo}(g, T)$. Moreover, if $f$ is invertible in $H(D)$, then $(f^{-1})^T = (f^T)^{-1}$ and $\text{mo}(f^{-1}, T) = -\text{mo}(f, T)$.

**Lemma 29.6.** Let $f \in H(D)$ and let $(T_n)_{n \in \mathbb{N}}$ be the $f$-supersequence. Suppose that for each $n \in \mathbb{N}$, $f$ admits Motzkin factors in $T_n$. Then there exists $N \in \mathbb{N}$ such that $\text{mo}(f^{T_n}, T_n) = 0$ whenever $n > N$. Moreover, if $D \in \text{Alg}$, the product $(\prod_{n=1}^t f^{T_n}) (\prod_{n=t+1}^\infty f^{T_n})$ does not depend on $t$ whenever $t \ge N$.

**Proof.** Indeed, there exists $N \in \mathbb{N}$ such that we have $\|f^{T_n} - 1\|_D < 1$ whenever $n \ge N$ and therefore, by Corollary 29.2, $\text{mo}(f^{T_n}, T_n) = 0$. Now, in $H_b(D)$, we have $(\prod_{n=N+1}^\infty f^{T_n}) = (\prod_{n=N+1}^t f^{T_n}) (\prod_{n=t+1}^\infty f^{T_n})$.

But then, if $D$ belongs to Alg, we have $(\prod_{n=1}^t f^{T_n}) (\prod_{n=t+1}^\infty f^{T_n}) = (\prod_{n=1}^N f^{T_n}) (\prod_{n=N+1}^t f^{T_n}) (\prod_{n=t+1}^\infty f^{T_n}) = (\prod_{n=1}^N f^{T_n}) (\prod_{n=N+1}^\infty f^{T_n})$. $\square$

**Definition.** Let $(T_n)_{n \in I}$ be the $f$-supersequence of $D$ with $I$ a subset of $\mathbb{N}$ which is either finite or equal to $\mathbb{N}$.

If $I$ is finite, $f$ is said to have a *finite Motzkin factorization* if it factorizes in $H(D)$ in the form $(f^0 \prod_{n \in I} f^{T_n})$ with $f^0$ an element of $H(\tilde{D})$ whose zeros belong to $D$ and for each $n \in \mathbb{N}$, $f^{T_n}$ is a Motzkin factor in $T_n$.

If $I$ is infinite and equal to $\mathbb{N}$, $f$ is said to have an *infinite Motzkin factorization* if it admits a sequence of Motzkin factors $f^{T_n}$ satisfying $\lim_{n \to \infty} f^{T_n} - 1 = 0$ such that $f$ factorizes in $H(D)$ in the form $(f^0 \prod_{n=1}^t f^{T_n}) (\prod_{n=t+1}^\infty f^{T_n})$, with $f^0$ an element of $H(\tilde{D})$ whose zeros belong to $D$. In both cases, $f^0$ will be called the *principal factor of f*.

**Corollary 29.7.** Let $D$ be bounded and let $f$ have an infinite Motzkin factorization with a $f$-supersequence $(T_n)_{n \in \mathbb{N}}$. Then we have $f = f^0 (\prod_{n=1}^\infty f^{T_n})$.

**Corollary 29.8.** Let $f$ have an infinite Motzkin factorization with a $f$-supersequence $(T_n)_{n \in \mathbb{N}}$ such that $\text{mo}(f^{T_n}, T_n) = 0$ for all $n > 0$. Then we have $f = f^0 (\prod_{n=1}^\infty f^{T_n})$.
**Remark 1.** Let $f \in H(D)$ be unbounded and have Motzkin factorization of the form $(f^0 \prod_{n=1}^N f^{T_n}) (\prod_{n=N+1}^\infty f^{T_n})$. One cannot claim that the product $(f^0 \prod_{n=1}^\infty f^{T_n})$ converges in $H(D)$ even if $D$ is closed and belongs to Alg. Indeed, let $r \in ]0, 1[$, let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $d(0, 1)$ such that $|a_n - a_m| = 1$ whenever $n \neq m$ and $a_1 = 0$. For every $n \in \mathbb{N}^*$, we put $T_n = d(a_n, r^-)$ and $E = \mathbb{K} \setminus (\bigcup_{n=1}^\infty T_n)$. The holes of $E$ are $T_n$. Let $(\lambda_n)_{n \ge 2}$ be a sequence in $d(0, r^-)$ such that $\lim_{n \to \infty} \lambda_n = 0$. For every $n \ge 2$, we put $g_n = 1 + \frac{\lambda_n}{x-a_n}$. The sequence $(g_n)_{n \ge 2}$ is seen to satisfy $\|g_n - 1\|_E \le \frac{|\lambda_n|}{\rho} < 1$, and therefore, we have $\lim_{n \to \infty} \|g_n - 1\|_E = 0$. Hence, the product $h = \prod_{n=2}^\infty g_n$ obviously converges in $H(E)$.

Since $E$ clearly belongs to Alg, we see that $x^2h$ belongs to $H(E)$ and is invertible in $H(E)$. Now, $f$ clearly has Motzkin factorization with $f^{T_n} = g_n$ for every $n \ge 2$, $f^{T_1} = x^2$ and $f^0 = 1$. However, we will check that the sequence $(f_n)_{n \in \mathbb{N}^*}$ defined by $f_n = x^2 \prod_{j=2}^n g_j$ does not converge in $H(E)$. Indeed, we have $f_{n+1}(x) - f_n(x) = x^2 (\prod_{j=2}^n g_j(x))(g_{n+1}(x) - 1)$. For every $x \in \mathbb{K} \setminus d(0, 1)$, we have $|x^2 \prod_{j=2}^n g_j(x)| = |x^2|$ and $|g_{n+1}(x) - 1| = \frac{|\lambda_{n+1}|}{x}$, hence $|f_{n+1}(x) - f_n(x)| = |x| |\lambda_{n+1}|$. Thus, $f_{n+1} - f_n$ is not bounded in $H(E)$, and therefore, the sequence $(f_n)_{n \in \mathbb{N}^*}$ does not converge in $H(E)$. According to Theorem 4 in [75], the product $\prod_{n=1}^\infty f_n$ should converge to $x^2h$ in $H(E)$. Here, we see that this is not true in the general case. Actually, the proof given in [75] only shows the simple convergence of the sequence $(f_n)$ and the uniform convergence on bounded subsets of $D$.

By Lemma 29.5, Lemma 29.9 is immediate.

**Lemma 29.9.** Let $D \in \text{Alg}$, let $f$, $g \in H(D)$ have Motzkin factorization. Then so does $fg$. Moreover, we have $(fg)^0 = f^0 g^0$. Further, if $f$ is invertible, $f^{-1}$ also has Motzkin factorization and it satisfies $(f^{-1})^0 = (f^0)^{-1}$.

**Corollary 29.10 (Boussaf) [13].** Let $D \in \text{Alg}$, let $f$ have an infinite Motzkin factorization of the form $(f^0 \prod_{n=1}^t f^{T_n}) (\prod_{n=t+1}^\infty f^{T_n})$. Let $N \in \mathbb{N}$ be such that $\text{mo}(f^{T_n}, T_n) = 0$ for all $n > N$. Then we have

$$f = f^0 \left( \prod_{n=1}^N f^{T_n} \right) \left( \prod_{n=N+1}^\infty f^{T_n} \right).$$
**Proposition 29.11.** Let $f \in H(D)$ satisfy $\|f - 1\|_D < 1$ and have Motzkin factorization of the form $f^0(\prod_{n=1}^{\infty} f^{T_n})$ with $(T_n)_{n \in \mathbb{N}^*}$ the f-supergroup of $D$. Then for each $n \ge 1$, we have $\mathrm{mo}(f^{T_n}, T_n) = 0$.

**Proof.** For every $n \in \mathbb{N}^*$, we put $q_n = \mathrm{mo}(f^{T_n}, T_n)$. By Lemma 29.6, we may assume the $(T_n)_{n \in \mathbb{N}^*}$ ranged in such a way that $q_n \ne 0$ for $n \le N$ while $q_n = 0$ whenever $n > N$. When $n \le N$, $f^{T_n}$ is of the form $(x - \alpha_n)^{q_n}(1 + \omega_n)$ with $\omega_n \in H_0(\mathbb{K}\backslash T_n)$, $\|\omega_n\|_{\mathbb{K}\backslash T_n} < 1$ and $\alpha_n \in T_n$. When $n > N$, $f^{T_n}$ is just in the form $(1 + \omega_n)$ with $\omega_n \in H_0(\mathbb{K}\backslash T_n)$ and $\|\omega_n\|_{\mathbb{K}\backslash T_n} < 1$. On the other hand, since $f$ has no zero in $D$, obviously $f^0$ has no zero in $D$, and therefore, it has no zero in $\tilde{D}$. Hence, by Theorem 22.7, $f^0$ is of the form $A(1 + \omega_0(x))$ with $\omega_0 \in H(\tilde{D})$, $\|\omega_0\|_D < 1$. Let $h(x) = A \prod_{n=1}^N (x - \alpha_n)^{q_n}$. We see that $f$ factorizes in the form $h(\prod_{n=0}^{\infty} (1 + \omega_n))$. Since $\|\omega_n\|_D < 1$ for every $n \in \mathbb{N}$ and since $\lim_{n \to \infty} \omega_n = 0$, it is seen that $h$ satisfies (1) $\|h-1\|_D < 1$ as does $f$. Let us suppose $q_1 \ne 0$. We may obviously assume $\alpha_1 = 0$. Let $T_1 = d(0, r^-)$. Thus, in $T_1$, $h$ admits 0 as a zero of order $q_1$ if $q_1 > 0$ (respectively a pole of order $-q_1$ if $q_1 < 0$) and has neither any zero nor any pole different from 0. Anyway, when $x \in T_1$, we have

$$ (2) \quad |h(x)| = B|x^{q_1}| \quad \text{with} \quad B = |A|\prod_{n=2}^{N} |\alpha_n|^{q_n}. $$

We will show that (2) contradicts (1) except if $q_1 = 0$.

Suppose $q_1 > 0$. In $T_1$, $h(x)$ is a series of the form $\sum_{n=q_1}^{+\infty} c_n x^n$ and $h-1 = (\sum_{n=q_1}^{+\infty} c_n x^n) - 1$, hence $1 \le \|h-1\|_{T_1} \le \|h-1\|_D$, which contradicts (1).

Now, suppose $q_1 < 0$. By definition, $h$ is obviously invertible in $R(D)$. Hence, we put $F = \frac{1}{h}$ and we see that $F$ satisfies $\|F-1\|_D < 1$ and admits 0 as a unique zero in $T_1$ while it has no pole in $T_1$. Hence, the same process lets us get to the same contradiction and finishes showing that $q_1 = 0$, and similarly, $q_n = 0$ for every $n \ge 1$. $\square$

**Proposition 29.12.** Let $f \in H(D)$ be invertible in $H(D)$ and have Motzkin factorization and let $a \in D$. Then $f$ satisfies $\left\| \frac{f}{f(a)} - 1 \right\|_D < 1$ if and only if for every hole $T$ of the f-supergroup of $D$, we have $\mathrm{mo}(f, T) = 0$.
**Proof.** Without loss of generality, we may obviously assume $f(a)=1$. By Proposition 29.11, we already know that if $f$ satisfies $\|f-1\|_D < 1$, then for every hole of the $f$-supersequence, we have $mo(f, T) = 0$. Now, we suppose that for every hole $T$ of the $f$-supersequence, we have $mo(f, T) = 0$ and we will prove that $\|f-1\|_D < 1$. Indeed, by Lemma 29.3, for each hole of the $f$-supersequence, we have $\|f^T-1\|_D < 1$. Moreover, since $f$ is invertible, $f^0$ must also be invertible, hence by Theorem 23.1, it is of the form $(1+\psi(x))$, with $\|\psi\|_D < 1$. Then $\|f-1\|_D < 1$. $\square$

We will show that all semi-invertible elements have Motzkin factorization step after step, and first, we consider rational functions.

**Proposition 29.13.** Let $f \in R(D)$. Then $f$ admits the Motzkin factorization.

**Proof.** The $f$-supersequence is obviously finite. Let $T_1, \dots, T_s$ be this $f$-supersequence. We can obviously factorize $f$ in a unique way in the form $\prod_{j=1}^s \left(\frac{h_j}{t_j}\right)$, whereas for each $j=1, \dots, s$, both $h_j$ and $l_j$ are monic polynomials whose zeros lie in $T_j$. Thus, we can check that $\frac{h_j}{l_j}$ is the Motzkin factor $f^{T_j}$ of $f$ in the hole $T_j$. Therefore, putting $f^0 = \frac{f}{\prod_{j=1}^s \left(\frac{h_j}{t_j}\right)}$, we have the Motzkin factorization: $f = f^0 \prod_{j=1}^s f^{T_j}$. $\square$

**Proposition 29.14.** Let $\phi \in H(D)$ satisfy $\|\phi - 1\|_D < 1$. Then $\phi$ admits the Motzkin factorization $\phi^0(\prod_{n=1}^\infty \phi^{T_n})$ with $(T_n)_{n \in \mathbb{N}^*}$ the $f$-supersequence. For every $\phi$-hole $T$, we have $\|\phi^T - 1\|_D = \|\overline{\phi_T}\|_D$. Moreover, $\phi^0$ satisfies $\|\phi^0 - 1\|_D = \|\overline{\phi_0} - 1\|_D$.

**Proof.** First, we suppose $\phi \in R(D)$. Then by Proposition 29.13, $\phi$ admits the Motzkin factorization. Now, by Proposition 29.12, for each $n > 0$, we have $mo(\phi, T_n) = 0$, and therefore, $\phi^{T_n}$ is of the form $1 + \omega_n$ with $\|\omega_n\|_D < 1$ whenever $n > 0$, while $\phi^0 = 1 + \omega_0$ with $\|\omega_0\|_D < 1$. Hence, we see that $\overline{\phi_{T_n}} = (\omega_n \prod_{j \neq n} (1 + \omega_j))_{T_n}$. Clearly, $\prod_{j \neq n} (1 + \omega_j)$ is of the form $1 + \phi_n$ with $\|\phi_n\|_D < 1$, hence $\|(\omega_n \phi_n)\|_D < \|\omega_n\|_D$, and we obtain (1) $\|(\overline{\omega_n \phi_n})_{T_n}\|_D \le \|\omega_n \phi_n\|_D < \|\omega_n\|_D$.
But $\omega_n$ is clearly equal to $\overline{(\phi^{T_n})}_{T_n}$ and then we have (2) $\|\overline{(\omega_n)_{T_n}}\|_D = \|\omega_n\|_D > \|\omega_n\phi_n\|_D \ge \|\overline{(\omega_n\phi_n)_{T_n}}\|_D$.

Moreover, $\overline{(\omega_n + \omega_n\phi_n)_{T_n}} = \overline{(\omega_n)_{T_n}} + \overline{(\omega_n\phi_n)_{T_n}}$, hence by (1) and (2) we have $\|\overline{(\omega_n(1 + \phi_n))}_{T_n}\|_D = \|\omega_n\|_D$, and finally,

(3) $\|\overline{\phi}_{T_n}\|_D = \|\overline{(\omega_n(1 + \phi_n))}_{T_n}\|_D = \|\omega_n\|_D = \|\phi^{T_n} - 1\|_D$. In the same way, we put $\prod_{n=1}^{\infty}(1 + \omega_n) = 1 + \psi$ with

(4) $\|\psi\|_D < 1$.

It is seen that $\psi$ belongs to $H_0(\mathbb{K}\setminus(\bigcup_{n=1}^{\infty}T_n))$. Hence, Theorem 15.1, when applied to $\psi$, shows that

(5) $\overline{\psi}_0 = 0$.

Next, we have $\phi = (1 + \omega_0)(1 + \psi) = 1 + \omega_0 + \psi + \omega_0\psi$, hence $\overline{\phi_0} = 1 + \overline{(\omega_0)_0} + \overline{\phi_0} + \overline{(\omega_0\psi)_0}$. By definition, $\omega_0 \in H(\tilde{D})$, hence $\omega_0 = \overline{(\omega_0)_0}$, and then by (5), we have $\overline{\phi_0} = 1 + \omega_0 + \overline{(\omega_0\psi)_0}$. But by (4), it is seen that $\|\overline{(\omega_0\psi)_0}\|_D < \|\omega_0\|_D$, and hence finally, we obtain $\|\overline{\phi_0} - 1\|_D = \|\omega_0\|_D = \|\phi^0 - 1\|_D$. Thus, we have proven the inequalities satisfied by the $\phi^{T_n}$ and by $\phi^0$ when $\phi$ belongs to $R(D)$.

Now, we consider the general case when $\phi \in H(D)$. Let $(f_m)_{m \in \mathbb{N}}$ be a sequence in $R(D)$ such that $\lim_{m \to \infty} \|\phi - f_m\|_D = 0$. Let $\varepsilon \in ]0, 1[$ and let $N \in \mathbb{N}$ be such that $\|f_m - \phi\|_D \le \varepsilon$ whenever $m \ge N$. Let $T$ be a hole of the $\phi$-supersequence. We will show that the sequence $((f_m)^T)_{m \in \mathbb{N}}$ converges in $H(D)$ and that this convergence is uniform with respect to the $\phi$-supersequence. We fix $m \ge N$. It is seen that $\|f_m - 1\|_D < 1$ and then by Lemma 29.1 and Proposition 29.12, we have $\|(f_m)^T - 1\|_D < 1$, and in particular, $\|(f_m)^T\|_D = 1$. Moreover, we remember that in $H(\mathbb{K}\setminus T)$, the norm $\|\cdot\|_D$ is multiplicative and actually equal to $\phi_T$. Now, let $s \ge N$. We have $\|(f_m)^T - (f_s)^T\|_D = \|\frac{(f_m)^T}{(f_s)^T} - 1\|_D$. But by Lemma 29.5, we have $\frac{(f_m)^T}{(f_s)^T} = (\frac{f_m}{f_s})^T$ and then by (3), in $R(D)$, we have $\|\left(\frac{f_m}{f_s}\right)^T - 1\|_D = \|\overline{\left(\frac{f_m}{f_s}\right)}_T\|_D$. Finally, by Theorem 15.1, we obtain

(6) $\|(f_m)^T - (f_s)^T\|_D \le \varepsilon$. Relation (6) does not depend on the hole $T$ and it shows that, for each fixed $n \in \mathbb{N}^*$, the sequence $((f_m)^{T_n})_{m \in \mathbb{N}}$ is
a Cauchy sequence which converges in $H(\mathbb{K}\setminus T_n)$ to an element whose index is equal to 0 and this convergence is uniform with respect to $n$. For each $n \in \mathbb{N}^*$, we put $\phi_n = \lim_{m\to\infty} (f_m)^{T_n}$. Then it is seen that $\prod_{n=1}^\infty \phi_n = \lim_{m\to\infty} \prod_{n=1}^\infty (f_m)^{T_n}$. As a consequence, the sequence $(f_m)^0$ is also convergent in $H(D)$ and actually in $H_b(\tilde{D})$. Let $\phi_0$ be its limit. Then we have this factorization: $\phi = \prod_{n=0}^\infty \phi_n$. We recognize the Motzkin factorization for $\phi$. Obviously, for each fixed $n > 0$, the equality satisfied by the $(f_m)^{T_n}$ holds for $\phi^{T_n}$ and shows that $\|(\overline{\phi})_{T_n}\|_D = \|(\phi)^{T_n} - 1\|_D$. In the same way, the equality satisfied by the $(f_m)^0$ shows that $\|\overline{(\phi)_0} - 1\|_D = \|(\phi)^0 - 1\|_D$. This ends the proof of Proposition 29.14. $\square$

Theorem 29.15 is given in [36, 42, 50].

**Theorem 29.15.** Let $a \in D$. Let $\phi \in H_b(D)$ be such that $|\phi(a)| \neq 0$. The following statements (i), (ii), and (iii) are equivalent:

(i) $\|\phi - \phi(a)\|_D < |\phi(a)|$.

(ii) For every hole T, we have $\|\overline{\phi_T}\|_D < |\phi(a)|$ and $\|\overline{\phi_0} - \overline{\phi_0}(a)\|_D < |\phi(a)|$.

(iii) $\phi$ is invertible, admits a Motzkin factorization and for every hole T, $\phi^T$ satisfies $\|\phi^T - 1\|_D < 1$ and $\phi^0$ satisfies $\|\phi^0 - \phi^0(a)\|_D < |\phi(a)|$.

Further, if statements (i)-(iii) are satisfied, then we have the following:

(u) $m(\phi, T) = 0$ for every hole T.

(v) $\|\overline{\phi_T}\|_D = \|\phi^T - 1\|_D |\phi(a)|$ for every hole T.

(w) $\|\phi^0 - \phi^0(a)\|_D = \|\overline{\phi_0} - \overline{\phi_0}(a)\|_D$.

**Proof.** Without loss of generality, we may obviously assume $|\phi(a)| = 1$ and

(1) $|\phi(a) - 1| < 1$. Let $(T_m)_{m \in I}$ be the $\phi$-supersequence of $D$. We note that when (i) is satisfied, $\phi$ is obviously invertible.

First, we suppose (i) is satisfied and will show that so is (ii). By Theorem 15.1, we have

$$ (2) \quad \|(\overline{\phi - \phi(a)})_{T_m}\|_D \le \|\phi - \phi(a)\|_D. $$

But it is seen that $\overline{(\phi - \phi(a))}_{T_m} = \overline{\phi_{T_m}}$. Hence, by (2), we have
$$ (3) \quad \|\overline{\overline{\phi}_{T_m}}\|_D \le \|\phi - \phi(a)\|_D < 1, \text{ whenever } m \in I. $$

In the same way, $(\overline{\overline{\phi - \phi(a)}})_0 = \overline{\overline{\phi}_0} - \phi(a)$, and then by Theorem 15.1, we have

$$ (4) \quad \|\overline{\overline{\phi}_0} - \phi(a)\|_D < 1. $$

Besides by (3), we see that $\|\sum_{m=1}^{\infty} \overline{\overline{\phi}_{T_m}}\|_D < 1$, hence $\|\phi - \overline{\overline{\phi}_0}\|_D = \|\sum_{m=1}^{\infty} \overline{\overline{\phi}_{T_m}}\|_D < 1$, and therefore, $|\phi(a) - \overline{\overline{\phi}_0}(a)| < 1$, hence by (4), we see that

$$ (5) \quad \|\overline{\overline{\phi}_0} - \overline{\overline{\phi}_0}(a)\|_D < 1. $$

Finally by (3) and (5), statement (ii) is clearly proven.

Now, we will show that each of the statements (ii) and (iii) separately imply (i). We suppose that (ii) is satisfied. Hence, we have

$$ (6) \quad \|\sum_{m \in I} \overline{\overline{\phi}_{T_m}}\|_D < 1. $$

If $D$ is bounded, by statement (ii), and by (6), we obtain (i). Now, let $D$ be not bounded. Then $\overline{\overline{\phi}_0}$ is a constant $\lambda$. Hence, $\phi$ is in the form $\lambda + \sum_{m=1}^{\infty} \overline{\overline{\phi}_{T_m}}$ with $\|\overline{\overline{\phi}_{T_m}}\|_D < 1$ whenever $m \ge 1$, hence

$$ (7) \quad \|\sum_{m=1}^{\infty} \overline{\overline{\phi}_{T_m}}\|_D < 1. $$

Now, we have $\phi - \phi(a) = \sum_{m=0}^{\infty} \overline{\overline{\phi}_{T_m}} - \overline{\overline{\phi}_{T_m}}(a) = \sum_{m=1}^{\infty} (\overline{\overline{\phi}_{T_m}} - \overline{\overline{\phi}_{T_m}}(a))$. By (7), we see that $\|\sum_{m=1}^{\infty} (\overline{\overline{\phi}_{T_m}} - \overline{\overline{\phi}_{T_m}}(a))\|_D < 1$, hence finally (i) $\|\phi - \phi(a)\|_D < 1$.

We now suppose (iii) is satisfied. Hence, we have (8) $\|\phi^{T_m} - 1\|_D < 1$ for all $m \in I$.

If $D$ is bounded, we have $\|\phi^0 - \phi^0(a)\|_D < 1$, hence by (8), we directly have (i). If $D$ is not bounded, then $\phi^0$ is a constant $B$ such that $\phi(a) = B \prod_{m \in I} \phi^{T_m}(a)$, hence by (1) and (8), we see that $|B - 1| < 1$, and hence by (8), we obtain (i) again.

Thus, (i) is also implied by (ii) as by (iii). Obviously, by (1), (i) implies $\|\phi - 1\|_D < 1$, and therefore, we may apply Proposition 29.14. Next, we suppose that either (ii) or (iii) is satisfied. Hence, so is (i) and so are (u) and (v) by Proposition 29.14.

Finally, we will show (w), and at the same time, we will finish proving the equivalence between (ii) and (iii). Let $\psi = (\overline{\overline{\phi_0}(a)})^{-1}\phi$. We may apply Proposition 29.14 to $\psi$ and we have

$$ (9) \quad \|\overline{\overline{\psi_0}} - 1\|_D = \|\psi^0 - 1\|_D. $$
But we have

$$ (10) \quad \|\psi^0 - 1\|_D \ge \|\psi^0 - \psi^0(a)\|_D = \|\phi^0 - \phi^0(a)\|_D. $$

$$ (11) \quad \|\bar{\phi}_0 - \bar{\phi}_0(a)\|_D = \|\bar{\psi}_0 - \bar{\psi}_0(a)\|_D = \|\bar{\psi}_0 - 1\|_D. $$

Hence by (9), (10), (11) we obtain

$$ (12) \quad \|\phi^0 - \phi^0(a)\|_D \le \|\bar{\phi}_0 - \bar{\phi}_0(a)\|_D. $$

Now, let $\gamma = \phi^0(a)$ and let $\chi = \gamma^{-1}\phi$. By (1) and (7), we see that $|\gamma - 1| < 1$, hence we may apply Proposition 29.14 to $\chi$ and we have

$$ (13) \quad \|\chi^0 - 1\|_D = \|\bar{\phi}_0 - \bar{\phi}_0(a)\|_D $$

while $\|\bar{\phi}_0 - \bar{\phi}_0(a)\|_D = \|\bar{\chi}_0 - \bar{\chi}_0(a)\|_D \le \|\bar{\chi}_0 - 1\|_D$ and $\|\chi^0 - 1\|_D = \|\chi^0 - \chi^0(a)\|_D = \|\phi^0 - \phi^0(a)\|_D$. Hence, by (13), we see that $\|\bar{\phi}_0 - \bar{\phi}_0(a)\|_D \le \|\phi^0 - \phi^0(a)\|_D$, and therefore, by (12), we obtain (w). This finishes proving the equivalence between (ii) and (iii), and ends the proof of Theorem 29.15. $\square$

**Remark 2.** If $D$ is not bounded, as $\phi$ is bounded, both $\phi^0$ and $\bar{\phi}_0$ are constant, and therefore, the statements $\|\bar{\phi}_0 - \bar{\phi}_0(a)\|_D < |\phi(a)|$ and $\|\phi^0 - \phi^0(a)\|_D < |\phi(a)|$ are automatically satisfied. Statement (ii) is then equivalent to

(ii') For every hole T, we have $\|\bar{\phi}_T\|_D < |\phi(a)|$ and statement (iii) is equivalent to

(iii') $\phi$ is invertible and for every hole T, $\phi^T$ satisfies $\|\phi^T - 1\|_D < 1$.

**Theorem 29.16.** Let $D \in \text{Alg.}$ Then $f$ has the Motzkin factorization if and only if it is semi-invertible.

**Proof.** Without loss of generality, we may assume the $f$-supersequence to be infinite. We denote it by $(T_n)_{n \in \mathbb{N}^*}$. Let $f$ have the Motzkin factorization

$$ f^0 \left( \prod_{n=1}^{t} f^{T_n} \right) \left( \prod_{n=t+1}^{\infty} f^{T_n} \right), $$

where the product $(\prod_{n=t+1}^{\infty} f^{T_n})$ converges in $H(D)$. By definition, $f^0$ is semi-invertible in $H(\tilde{D})$ and hence in $H(D)$. Moreover,
($\prod_{n=1}^{t} f^{T_n}$) ($\prod_{n=t+1}^{\infty} f^{T_n}$) is clearly invertible in $H(D)$. So, $f$ is semi-invertible.

Now, we suppose $f$ to be semi-invertible and will show it to have the Motzkin factorization. By Lemma 29.9, we may clearly suppose that $f$ is invertible without loss of generality.

First, we suppose that there exists $M$ in $\mathbb{R}_+^*$ satisfying

$$ (1) \quad M \le |f(x)| \quad \text{whenever } x \in D. $$

Let $h \in R(D)$ satisfy

$$ (2) \quad \|f - h\|_D < \frac{M}{2}, $$

and let $h^0 \prod_{n=1}^N h^{T_n}$ be the Motzkin factorization of $h$. For every $n = 1, \dots, N$, let $q_n = \text{mo}(h, T_n)$, let $a_n \in T_n$, let $h_n = (x - a_n)^{-q_n} h^{T_n}$ and let $h_0 = h^0$. Let $u(x) = \prod_{n=1}^N (x - a_n)^{q_n}$ and let $l(x) = h^0 \prod_{n=1}^N h_n$. By (1) and (2), it is seen that $h$ has no zero in $D$. Let $a \in D$. Then, by Theorem 23.1, $h^0$ satisfies $\|h^0 - h^0(a)\|_D < |h^0(a)|$, and of course for every $n > 0$, $h_n$ satisfy $\|h_n - 1\|_D < 1$. Hence, we have

$$ (3) \quad \|l - l(a)\|_D < |l(a)|. \quad \text{Let } b = |l(a)|. $$

In particular, we have $|l(x)| = b$ whenever $x \in D$. Moreover, we note that we have

$$ (4) \quad \frac{M}{|l(a)|} \le |u(x)|. $$

Let $F = \frac{f}{u}$. Then $F$ does belong to $H_b(D)$. By (3) and (4), we check that $|F(x) - l(x)| < \frac{b}{2}$, and therefore, by (3) again, we have $|F(x)| = b$ and $\|F - F(a)\|_D \le \frac{|F(a)|}{2}$. Now, we can apply Theorem 29.15 to $F$ and then $F$ has Motzkin factorization $F^0 \prod_{n=1}^{\infty} F^{T_n}$, with $\text{mo}(F, T_n) = 0$ whenever $n > 0$. As a consequence $f$ also has the Motzkin factorization, $(f^0 \prod_{n=1}^N f^{T_n}) (\prod_{n=N+1}^{\infty} f^{T_n})$ with $f^0 = F^0$ and for each $n = 1, \dots, N$, $f^{T_n} = (x - a_n)^{q_n} F^{T_n}$ and finally for each $n > N$, $f^{T_n} = F^{T_n}$.

Now, we suppose that $\inf\{|f(x)| \mid x \in D\} = 0$. Since $D$ is closed and since $f$ is invertible, we see that $D$ is unbounded and that the element $G = \frac{1}{f}$ is not bounded in $D$. Hence, by Corollary 11.7, there exists $q \in \mathbb{N}^*$ such that $\frac{x-q}{f}$ has a non-zero limit when $|x|$ tends to $+\infty$, $(x \in D)$. Then it is easily seen that there exists $c > 0$ such that $|G(x)| \ge c$ for all $x \in D$. Indeed, on one hand, there exists $r$ such that $|G(x)| \ge 1$ for all $x \in D \setminus d(0, r)$, and on the other hand, $f$ is bounded in $D \cap d(0, r)$, hence we can find $c \in ]0, 1[$ such that
$|G(x)| \ge c$ whenever $x \in D \setminus d(0, r)$. Thus, $G$ admits the Motzkin factorization, and then by Lemma 29.9, so does $f$. This ends the proof of the theorem. □

**Remark 3.** If a closed set $B$ does not belong to Alg, there are counter-examples of invertible elements $F$ which admit certain Motzkin factor $F^T$ such that $\frac{F}{F^T}$ does not belong to $H(B)$ (and obviously does not belong to $H(B \cup T)$). Indeed, don't let $B$ belong to Alg. Since hypothesis $B$ is closed, we know that $\tilde{B} \setminus \overline{B}$ is not bounded. Hence, by Lemma 29.6, there exists a quasi-minorated element $f \in H_b(B)$ satisfying

$$ (1) \quad \lim_{|x| \to \infty} f(x) = 0, $$

such that $xf$ does not belong to $H(B)$. Since $f \in H_b(B)$, we can take it such that $\|f\|_B < 1$. Without loss of generality, we may assume that 0 belongs to a hole of $B$. Let $T = d(a, r^-)$ be another hole of $B$ and let $F = \frac{x(1+f)}{(x-a)}$. Then it is seen that $F$ belongs to $H_b(B)$ and is invertible in $H_b(B)$ because both $\frac{x}{(x-a)}$, $1+f$ are invertible in $H_b(B)$. Hence, $F$ admits the Motzkin factorization. In particular, we see that $F^T = \frac{1}{(x-a)}$. However, we check that $(x-a)F$ does not belong to $H(B)$ because $(x-a)F = x(1+f)$ and by hypothesis, $xf$ does not belong to $H(B)$.

In the same way, let $G = \frac{1}{F}$. Since $F$ is invertible in $H_b(B)$, so is $G$. But then we see that $\frac{1}{x-a}G$ does belong to $H_b(B)$ and has no zero in $B$, but obviously, its inverse does not belong to $H(B)$. Therefore, $\frac{1}{x-a}G$ is not semi-invertible in $H(B)$. Thus, there exist invertible elements $h$ and $g$ in $H(B)$ such that $hg$ is not semi-invertible, although it belongs to $H(B)$. This contradicts Theorem 1 in [75], which states that $\frac{f}{f^T}$ extends to an element of $H(D \cup T)$.

**Theorem 29.17 (Boussaf) [13].** Let $D$ belong to Alg and let $T = d(a, r^-)$ be a hole of $D$. Then $f$ admits a Motzkin factor in the hole $T$ if and only if $\phi_{D, \varphi_{a,r}}(f) \neq 0$.

**Proof.** On the one hand, we suppose that $f$ admits a Motzkin factor in the hole $T$. Let $f = gf^T$. Since $g$ belongs to $H(D \cup T)$ and has no zero in $T$, of course, by Theorem 22.16, we have $\phi_{D, \varphi_{a,r}}(g) \neq 0$. Next, as an invertible element of $H(D)$, it is seen that $\phi_{D, \varphi_{a,r}}(f^T) \neq 0$. Hence, $\phi_{D, \varphi_{a,r}}(f) \neq 0$.
On the other hand, we suppose $\mathcal{D}\varphi_{a,r}(f) \neq 0$. Let $\mathcal{F}$ be the circular filter of center $a$ of diameter $r$ and let $M = \mathcal{D}\varphi_{a,r}(f)$. There do exist $a_1, \dots, a_q \in d(a,r)$ and $s, t$ satisfying $s < r < t$, such that $|f(x)| \ge M$ whenever $x \in \mathcal{D} \cap (\bigcap_{j=1}^q \Gamma(a_j, s, t))$. Let $F = \mathcal{D} \cap (\bigcap_{j=1}^q \Gamma(a_j, s, t))$. Then $T$ is clearly a hole of $F$. Next, the restriction $g$ of $f$ to $F$ is invertible in $H(F)$, and therefore, by Theorem 29.16, it admits a Motzkin factor $g^T$ in the hole $T$. But then, $\frac{f}{g^T}$ belongs to $H(D)$ and $H(F \cup T)$. Let $E = F \cup T$. Clearly, a hole of $D \cap E$ is either a hole of $D$ included in $E$ or a hole of $E$. Hence, $D$ and $E$ are infraconnected sets that satisfy the hypothesis of Theorem 15.10 and then $\frac{f}{g^T}$ belongs to $H(D \cup F) = H(D \cup T)$. Finally, as $g^T$ is the Motzkin factor of $g$ in $T$, $\frac{g}{g^T}$ has no zero inside $T$. This ends the proof. $\square$

**Theorem 29.18.** Let $D \in \text{Alg}$ and let $\mathcal{G}$ be the multiplicative group of the invertible elements in $H(D)$. Let $\mathcal{T}$ be the set of the holes of $D$. Let $\mathcal{G}^0$ be the subgroup of the elements invertible in $H(\tilde{D})$. Let $\mathcal{H} = \mathcal{G}^0 \prod_{T \in \mathcal{T}} \mathcal{G}^T$. The product $\mathcal{H}$ is a direct product and is dense in $\mathcal{G}$.

**Proof.** The product is direct because for each element, the Motzkin factorization is unique. Thus, $\mathcal{H}$ is the set of the invertible elements whose Motzkin factorization is finite. Since every element of $\mathcal{G}$ has the Motzkin factorization, it obviously belongs to the closure of $\mathcal{H}$. $\square$

Thanks to the Motzkin factorization, the question on whether the $n$th root of an analytic element is an analytic element appears to be linked to the number of zeros of each Motzkin factor.

**Theorem 29.19.** Let $D$ be a closed bounded infraconnected set and let $f \in H(D)$ be semi-invertible. Let $T$ be a hole of $D$. We assume $f^s$ to have continuation to an element of $H(D \cup T)$ for some $s \in \mathbb{N}^*$. Then the number of the zeros of $f^s$ inside $T$ is a multiple of $s$ (taking multiplicities into account). Moreover, if $f$ does not belong to $H(D \cup T)$, then the number of the zeros of $f^s$ inside $T$ is different from 0.

**Proof.** By Lemma 29.9, we have $(f^s)^T = (f^T)^s$. But as $f^s$ belongs to $H(D \cup T)$, by Theorem 29.4, $(f^s)^T$ is the polynomial of the zeros of $f^s$ inside $T$. Let $Q = (f^s)^T$. Then we have $\text{deg}(Q) = \text{mo}((f^s), T) = \text{smo}(f, T)$. So, $s$ divides $\text{deg}(Q)$. Now,
assume deg($Q$) = 0. We have $Q = 1$, $mo(f, T) = 0$, and therefore, $(f^T)^s = 1$ and $\lim_{|x|\to\infty} f^T(x) = 1$. Thus, $f^T$ is just the constant 1, and therefore, $f$ belongs to $H(D \cup T)$, which ends the proof of Theorem 29.19. $\square$

In particular, Theorem 29.19 applies to open disks.

**Theorem 29.20.** Let $r \in |\mathbb{K}|$ and let $f \in H(d(0, r^-))\H(d(0, r))$ satisfy $f^s \in H(d(0, r))$. Then $f$ has continuation to an analytic element in a set $D$ of the form $d(0, r)\setminus\left(\bigcup_{i=1}^t d(a_i, r^-)\right)$ with $|a_i| = r = |a_i - a_j|$ whenever $i \neq j$, such that for each $i = 1, \dots, t$, the number of zeros of $f^s$ in $d(a_i, r^-)$ is a multiple of $s$ different from 0.

**Proof.** The Mittag-Leffler series of $f$ in $d(0, r^-)$ is of the form $\sum_{n=0}^{\infty} f_n$ with $f_0 = \bar{f}_0 \in H(d(0, r))$, and for every $n > 0$, $f_n = \bar{f}_{T_n}$, with $T_n = d(a_n, r^-)$ and $|a_n - a_j| = |a_n| = r$ whenever $n \neq j$. Since $f \notin H(d(0, r))$, at least one of the $f_n$ is different from 0. Let $l$ be an integer such that $f_l \neq 0$. Now, since $f^s$ belongs to $H(d(0, r))$, by Theorem 29.19, $f^s$ has a number of zeros inside $T_l$ which is different from 0 and a multiple of $s$. Since any element of $H(d(0, r))$ has finitely many zeros in $d(0, r)$, we see that there are finitely many integers $l$ such that $f_l \neq 0$. Let $I$ be the finite set of the $l \in \mathbb{N}^*$ such that $f_l \neq 0$ and let $D = d(0, r)\setminus\left(\bigcup_{l \in I} d(a_n, r^-)\right)$. Then by definition, $f$ belongs to $H(D)$ and for every $l \in I$, the number of zeros of $f^s$ in $T_n$ is different from 0 and a multiple of $s$. This ends the proof of Theorem 29.20. $\square$

**Corollary 29.21.** Let $f$ be a power series whose radius of convergence is $r$ though $f$ does not belong to $H(d(0, r))$. If for some $s \in \mathbb{N}^*$, $f^s$ has a radius of convergence $r'$ strictly superior to $r$ and if $f^s$ has strictly less than $s$ zeros inside $C(0, r)$ (taking multiplicities into account), then $f$ does not belong to $H(d(0, r^-))$.

**Proof.** Since $r' > r$, obviously we have $s > 1$. We assume that $f$ belongs to $H(d(0, r^-))$, and therefore, $r$ must belong to $|\mathbb{K}|$. Since $r' > r$, $f^s$ belongs to $H(d(0, r))$, and then by Theorem 29.20, its number of zeros inside $C(0, r)$ is different from 0 and is a multiple of $s$, which contradicts the hypothesis. Hence, finally, $f$ does not belong to $H(d(0, r^-))$. $\square$

We have now got to recall the definition of the function $\sqrt[q]{u}$ when $u \in d(1, 1^-)$.
**Notation.** Henceforth, we suppose that $\mathbb{K}$ has characteristic zero and residue characteristic $p$. Let $q \in \mathbb{N}^*$. Let $(1+x)^q = 1+qx+\sum_{j=2}^q b_j x^j$. So, $|q| = 1$ and $|b_j| \le 1$ whenever $j = 2, \dots, q$. Recall that $r_k$ was defined in Chapter 7.

**Theorem 29.22.** Let $q \in \mathbb{N}^*$. If $q$ is prime to $p$, the mapping $g_q(x) = (1+x)^q$ is injective in $d(1, 1^{-})$ and maps $d(1, 1^{-})$ onto $d(1, 1^{-})$. If $p \ne 0$ and if $q = p$, the mapping $g_p(x) = (1+x)^p$ is injective in $d(1, r_1^{-})$ and maps $d(1, (r_1)^{-})$ onto $d(1, (r_2)^{-})$.

**Proof.** Suppose first $q$ prime to $p$. Since $|q| = 1 \ge |b_j| \forall j \ge 2$, by Corollary 22.10 and Corollary 23.14, the mapping $g_q$ defines a bijection from $d(0, 1^{-})$ onto $d(1, 1^{-})$.

Suppose now $p \ne 0$ and take $q = p$. By Theorems 25.3 and 25.5, inside $d(0, (r_1)^{-})$, we can write $g_p(x) = \exp(p\text{Log}(1+x))$. This way, we note that when $x \in d(0, (r_1)^{-})$, $g_p$ is injective and that we have $|\text{Log}(1+x)| = |x|$ and $|\exp(x) - 1| = |x|$, hence

$$|(1+x)^p - 1| = |\exp(p\text{Log}(1+x)) - 1| = \frac{x}{p}.$$

Consequently, the image of $d(0, r_1^{-})$ by $g_p$ is the disk $d(1, (r_1/p)^{-}) = d(1, (r_2)^{-})$. $\square$

**Notation.** Suppose $q \in \mathbb{N}^*$ prime to $p$. The mapping $\eta_q$ defined in $d(1, 1^{-})$ by $\eta_q(u) = u^q$ is a bijection from $d(1, 1^{-})$ onto $d(1, 1^{-})$. We denote by $\sqrt[q]{u}$ the inverse mapping from $d(1, 1^{-})$ onto $d(1, 1^{-})$ and we put $\phi_q(x) = \sqrt[q]{1+x}$ whenever $x \in d(0, 1^{-})$.

Suppose now $p \ne 0$ and take $q = p$. The mapping $\eta_p$ defined in $d(1, (r_1)^{-})$ by $\eta_p(u) = u^p$ is a bijection from $d(1, (r_1)^{-})$ onto $d(1, (r_2)^{-})$. So, we can denote by $\sqrt[p]{u}$ that inverse mapping from $d(1, (r_2)^{-})$ onto $d(1, (r_1)^{-})$ and we put $\phi_p(x) = \sqrt[p]{1+x}$ whenever $x \in d(0, (r_2)^{-})$.

**Theorem 29.23.** Let $q \in \mathbb{N}^*$. If $q$ is prime to $p$, $\phi_q$ belongs to $A_b(d(0, 1^{-}))$ but does not belong to $H(d(0, 1^{-}))$. Next, suppose now $p > 0$. Then $\phi_p$ belongs to $A_b(d(0, r_1^{-}))$ but does not belong to $H(d(0, r_1^{-}))$.

**Proof.** Suppose first that $q$ is prime to $p$. By construction and by Corollary 23.15, $\phi_q$ belongs to $A_b(d(0, 1^{-}))$. Suppose that $\phi_q$ belongs to $H(d(0, 1))$. Then it must satisfy $(\phi_q(-1))^q = 0$, hence $\phi_q(-1) = 0$,
and therefore, $\phi_q(x)^q$ admits a zero of order $q$ at $-1$. But this contradicts the identity $(\phi_q(x))^q = 1+x$, and therefore, $\phi_q(x)$ does not belong to $H(d(0,1))$. Finally, since $(\phi'_q x)^q$ has a unique zero in $C(0,1)$, by Corollary 29.21, we see that $\phi_q$ does not belong to $H(d(0,1^-))$.

Suppose now that $p > 0$. By Theorem 29.22, the function $f(x) = (1+x)^p$ is strictly injective inside $d(0, r_1^-)$ and maps $d(0, r_1^-)$ onto itself. So, by Corollary 23.15, it admits an inverse mapping $\phi_p$ defined inside $d(0, r_1^-)$ that belongs to $A_b(d(0, r_1^-))$.

Let us show that $\phi_p$ does not belong to $H(d(0, r_1^-))$. Indeed, suppose $\phi_p$ belongs to $H(d(0, r_1^-))$. Consider the Mittag-Leffler decomposition of $\phi_p$ on the infraconnected set $d(0, r_1^-)$. It is of the form $\sum_{n=0}^{\infty} g_n$ with $g_0 \in H(d(0, r_1)$ and $g_n \in H_0(K\setminus d(a_n, r_n^-))$ with $a_n \in C(0, r_1)$. Now, by Theorem 8.4, we can consider a complete algebraically closed extension $\tilde{K}$ of $K$ whose residue class field is not countable.

Let $\chi$ be the residue class field of $K$. Thus, we can find a class $G$ of $\hat{C}(0, r_1)$ that has an empty intersection with $\bigcup_{n=1}^{\infty} d(a_n, r_1^-)$ and then $\phi_p$ has continuation to an element of $H(d(0, r_1^-) \cup G$. Let $c \in G$. Since $|c| = r_1$, the function $h(x) = (1+x)^p - c$ satisfies $\nu^-(h, \log(r_1) = 0, \nu^+(h, \log(r_1) > 1$, hence $h$ admits a zero $a \in \hat{C}(0, r_1)$. Then of course $a$ does not belong to $K$. Now, let $\zeta$ be a $p$th root of 1 different from 1 and let $t = \zeta(1+a)$. By Corollary 7.7, we have $|\zeta - 1| = r_1$, hence $t$ is of the form $1+b$ with $b \in \hat{C}(0, r_1)$. We then have $(1+a)^p = (1+b)^p$. Set $E = \hat{d}(a, r_1^-)$, $F = \hat{d}(b, r_1^-)$, $D' = d(0, r_1^-) \cup E \cup F$ and $D'' = d(0, r_1^-) \cup G$. Since the image of $d(0, r_1^-)$ by the function $f$ is $d(0, r_1^-)$, we can check that for each $u \in \hat{C}(0, r_1)$, the image of $d(u, r_1^-)$ by $f$ is $d(1, r_1^-)$. Consequently, both images of $E$ and $F$ by $f$ are equal to $G$. Now, since $f$ belongs to $\hat{H}(D')$, the image of $D'$ by $f$ is $D''$ and $\phi_p$ belongs to $\hat{H}(D'')$. Consequently, by Corollary 12.3, $\phi_p \circ f$ belongs to $\hat{H}(D')$ and we have $\phi_p \circ f(x) = 1+x \forall x \in d(0, r_1^-)$. But since $D'$ has no pierced filter, by Corollary 21.4, it is an analytic set. Consequently, the equality $\phi_p \circ f(x) = 1+x \forall x \in d(0, r_1^-)$ holds in all $D'$, a contradiction since $f$ is not injective in $D'$. This finally shows that $\phi_p$ does not belong to $H(d(0, r_1^-))$. $\square$

**Remark 4.** Let $q$ be prime to $p$. Since $\phi_q$ belongs to $A_b(d(0, 1^-))$, obviously $\phi_q$ belongs to $H(d(0, r))$, whenever $r \in ]0, 1[$. Now, let $E$
be a closed bounded set in $\mathbb{K}$ and let $h \in H(E)$ satisfy $\|h\|_E < 1$. Then by Corollary 12.3, $\phi_q \circ h$ belongs to $H(E)$. In other words, if $g \in H(E)$ and if $\|g - 1\|_E < 1$, then $\sqrt[q]{g}$ also belongs to $H(E)$.

**Remark 5.** Theorems 29.19 and 29.20 couldn't be significantly improved as this example shows. Let $q$ be an integer prime to $p$, let $a, b \in d(1, 1^{-})$, with $a \neq b$, and let $P(x) = (a-x)^{q-1}(b-x)$. It is easily seen that $|P(x) - 1| < 1$ whenever $x \in d(0, 1^{-})$ and then we can consider $f(x) = \sqrt[q]{P(x)}$. We will show that $f \in H(d(0, 1^{-})) \setminus H(d(0, 1))$. Indeed, we have

$$ 
\begin{aligned}
f(x) &= \sqrt[q]{(a-x)^q \left(\frac{b-x}{a-x}\right)} = (a-x) \sqrt[q]{1 + \left(\frac{b-a}{a-x}\right)} \\
&= (a-x) \sum_{n=0}^{\infty} \left(\frac{1}{n}\right) \left(\frac{b-a}{a-x}\right)^n = a - x + \frac{1}{q}(b-a) \\
&\quad + \sum_{j=1}^{\infty} \left(\frac{1}{j+1}\right) \frac{(b-a)^{j+1}}{(a-x)^j}.
\end{aligned}
$$

This is just a Mittag-Leffler series of the form $f_0 + f_1 \in H(d(0, 1^{-}))$, with

$$ f_0 = -x + a + \frac{1}{q}(b-a) \in H(d(0, 1)), $$

$$ f_1 = - \sum_{j=1}^{\infty} \left( \frac{1}{j+1} \right) \frac{(a-b)^{j+1}}{(x-a)^j} \in H_0(\mathbb{K} \setminus d(1, 1^{-})). $$

Thus, we see that $f$ belongs to $H(d(0, 1^{-}))$, and more precisely, $f \in H(\mathbb{K}\setminus d(1, 1^{-}))$, but $f \notin H(d(0, 1))$. Actually, $f^q$ has exactly $q$ zeros in $d(1, 1^{-})$.
Chapter 30

Order of Growth for
Entire Functions

Here we mean to introduce and study the notion of order of growth
of an entire function on $\mathbb{K}$ in relation with the distribution of zeros
in disks and in relation with the question whether an entire function
can be divided by its derivative inside the algebra of entire functions.
First, results were published in [16], [18], and [50]. The notion of order
of growth was defined by G. Valiron [90].

**Definition and notation.** Let $f \in \mathcal{A}(\mathbb{K})$. Similar to the definition known on complex entire functions, $\limsup_{r \to +\infty} \frac{\log(\log(|f(r)|))}{\log(r)}$ is called the order of growth of $f$ or the growth order of $f$ in brief and is denoted by $\rho(f)$. We say that $f$ has finite order if $\rho(f) < +\infty$. In this chapter and in Chapters 30, 31, 32, and 33 we denote by Log the Neperian logarithm function and by $e$ the number such that $\text{Log}(e) = 1$. Next, the function $\Psi$ is now defined by $\Psi(x) = \text{Log}(|x|)$.

Theorem 30.1 is easily proven:

**Theorem 30.1.** Let $f, g \in \mathcal{A}(\mathbb{K})$. Then

(i) if $c(|f(r)|)^\alpha \ge |g|(r)$ with $\alpha$ and $c > 0$, when $r$ is big enough, then $\rho(f) \ge \rho(g)$,

(ii) $\rho(f+g) \le \max(\rho(f), \rho(g))$ and if $\rho(g) < \rho(f)$, then $\rho(f+g) = \rho(f)$,

(iii) $\rho(fg) = \max(\rho(f), \rho(g))$.
**Proof.** Suppose $c(|f|(r))^{\alpha} \ge |g|(r)$ with $\alpha$ and $c > 0$ when $r$ is big enough. Then we check that $(\limsup_{r \to +\infty} \frac{\log(\log(c(|f|(r))^{\alpha}))}{\log(r)}) \ge \limsup_{r \to +\infty} \frac{\log(\log(|g|(r)))}{\log(r)}$, hence $\rho(f) \ge \rho(g)$.

Next, we have $|f + g|(r) \le \max(|f|(r), |g|(r))$, hence

$$ \frac{\log(\log(|f + g|(r)))}{\log(r)} \le \max\left(\frac{\log(\log(|f|(r)))}{\log(r)}, \frac{\log(\log(|g|(r)))}{\log(r)}\right), $$

and hence $\rho(f + g) \le \max(\rho(f), \rho(g))$.

Now, suppose $\rho(f) > \rho(g)$. There exists a sequence $(r_n)_{n \in \mathbb{N}}$ of limit $+\infty$ such that $\lim_{n \to +\infty} \frac{\log((\log(|f|(r_n)) / \log(r_n))}{\log(r_n)} = \rho(f)$ and $|f|(r_n) > |g|(r_n)$, hence $|f + g|(r_n) = |f|(r_n)$ and hence $\rho(f + g) \ge \rho(f)$. But we have proved that $\rho(f + g) \le \max(\rho(f), \rho(g)) = \rho(f)$.

Let us now show that $\rho(fg) = \max(\rho(f), \rho(g))$. It is clear that $\rho(fg) \ge \max(\rho(f), \rho(g))$, since $|fg|(r) = |f|(r)|g|(r)$. Now, let $t = \max(\rho(f), \rho(g))$. Then there exists a function $\omega$ defined in $\mathbb{R}_+$, of limit 0 at $\infty$, such that $\frac{\log(\log(|f|(r)))}{\log(r)} \le t + \omega(r)$ and $\frac{\log(\log(|g|(r)))}{\log(r)} \le t + \omega(r)$. Hence, we have

$$ \log(|f|(r)) \le r^{t+\omega(r)}, \quad \log(|g|(r)) \le r^{t+\omega(r)}, $$

hence

$$ \log(|f|(r)) + \log(|g|(r)) \le 2r^{t+\omega(r)}, $$

therefore

$$ \log(\log(|f|(r)) + \log(|g|(r))) \le \log(2) + (t + \omega(r))\log(r), $$

hence,

$$ \log(\log(|f|(r)) + \log(|g|(r))) \le \log(2) + (t + \omega(r))\log(r), $$

hence

$$ \frac{\log(\log(|f|(r)).|g|(r)))}{\log(r)} \le \frac{\log(2)}{\log(r)} + t + \omega(r), $$
and hence

$$ \limsup_{r \to +\infty} \frac{\text{Log}(\text{Log}(|f(r)| \cdot |g(r))}}{\text{Log}(r)} \le t. $$

Consequently, $\rho(fg) \le \max(\rho(f), \rho(g))$, which ends the proof. □

**Corollary 30.2.** Let $f, g \in \mathcal{A}(\mathbb{K})$. Then, $\rho(f^n) = \rho(f) \forall n \in \mathbb{N}^*$.

**Notation.** Given a number $t \ge 0$, we denote by $\mathcal{A}(\mathbb{K}, t)$ the $\mathbb{K}$-algebra of entire functions of order inferior or equal to $t$ and we put $\mathcal{A}^*(\mathbb{K}) = \bigcup_{t>0} \mathcal{A}(\mathbb{K}, t)$.

**Corollary 30.3.** For any $t \ge 0$, $\mathcal{A}(\mathbb{K}, t)$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}(\mathbb{K})$.

**Corollary 30.4.** Consider the differential equation

$$ (\mathcal{E}) \quad f^{(n)} + a_{n-1}(x)f^{(n-1)}(x) + \dots + a_0(x)f(x) = 0 $$

with $a_j \in \mathcal{A}^*(D)$, $j = 0, \dots, n-1$ and $\rho(a_j) < \rho(a_0) \forall j = 1, \dots, n-1$. Then every non-trivial solution $f$ of $(\mathcal{E})$ satisfies $\rho(f) \ge \rho(a_0)$.

**Theorem 30.5.** Let $f \in \mathcal{A}(\mathbb{K})$ and let $P \in \mathbb{K}[x]$. Then, $\rho(P \circ f) = \rho(f)$ and $\rho(f \circ P) = \deg(P)\rho(f)$.

**Proof.** Let $n = \deg(P)$. For $r$ big enough, we have

$$ \begin{aligned} \text{Log}(\text{Log}(|f(r)|)) &\le \text{Log}(\text{Log}(|P \circ f(r)|)) \le \text{Log}((n+1)\text{Log}(|f(r)|)) \\ &= \text{Log}(n+1) + \text{Log}(\text{Log}(|f(r)|)). \end{aligned} $$

Consequently,

$$ \begin{aligned} \limsup_{r \to +\infty} \left( \frac{\text{Log}(\text{Log}(|f(r)|))}{\text{Log}(r)} \right) &\le \limsup_{r \to +\infty} \left( \frac{\text{Log}(\text{Log}(|P \circ f(r)|))}{\text{Log}(r)} \right) \\ &\le \limsup_{r \to +\infty} \left( \frac{\text{Log}(n+1) + \text{Log}(\text{Log}(|f(r)|))}{\text{Log}(r)} \right), \end{aligned} $$

and therefore, $\rho(P \circ f) = \rho(f)$.
Next, for $r$ big enough, we have

$$ \frac{\mathrm{Log}(\mathrm{Log}(|f|(r)))}{\mathrm{Log}(r)} \le \frac{\mathrm{Log}(\mathrm{Log}(|f \circ P|(r)))}{\mathrm{Log}(r)} \\ = \left( \frac{\mathrm{Log}(\mathrm{Log}(|f \circ P|(r)))}{\mathrm{Log}(|P|(r))} \right) \left( \frac{\mathrm{Log}(|P|(r))}{\mathrm{Log}(r)} \right). $$

Now,

$$ \limsup_{r \to +\infty} \left( \frac{\mathrm{Log}(\mathrm{Log}(|f \circ P|(r)))}{\mathrm{Log}(|P|(r))} \right) = \limsup_{r \to +\infty} \left( \frac{\mathrm{Log}(\mathrm{Log}(|f|(r)))}{\mathrm{Log}(r)} \right) $$

because the function $h$ defined in $[0, +\infty[$ as $h(r) = |P|(r)$ is obviously an increasing continuous bijection from $[0, +\infty[$ onto $[|P(0)|, +\infty[$. On the other hand, it is obviously seen that $\limsup_{r \to +\infty} \left( \frac{\mathrm{Log}(|P|(r))}{\mathrm{Log}(r)} \right) = n$. Consequently,

$$ \limsup_{r \to +\infty} \left( \frac{\mathrm{Log}(\mathrm{Log}(|f \circ P|(r)))}{\mathrm{Log}(|P|(r))} \right) = n \limsup_{r \to +\infty} \left( \frac{\mathrm{Log}(\mathrm{Log}(|f|(r)))}{\mathrm{Log}(r)} \right), $$

and hence $\rho(f \circ P) = n\rho(f)$. $\square$

**Theorem 30.6.** Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental. Then, $\rho(f \circ g) \ge \max(\rho(f), \rho(g))$. If $\rho(f) \ne 0$, then $\rho(f \circ g) = +\infty$.

**Proof.** Let $f(x) = \sum_{j=0}^{\infty} a_n x^n$ and $g(x) = \sum_{j=0}^{\infty} b_n x^n$. Since $g$ is transcendental, for every $n \in \mathbb{N}$, there exists $r_n$ such that $s(r_n, g) \ge n$. Then, $|g|(r) \ge |b_n|r^n \forall r \ge r_n$, and hence, by Theorem 30.5, we have

$$ (1) \qquad \rho(f \circ g) \ge n\rho(f). $$

Therefore, $\rho(f \circ g) \ge \rho(f)$.

Now, let $k \in \mathbb{N}$ be such that $a_k \ne 0$ and let $s_0$ be such that $s(s_0, f) \ge k$. Then, $|f|(r) \ge |a_k|r^k \forall r \ge s_0$, hence $|f \circ g|(r) \ge |a_k|(|g|(r))^{k} \forall r \ge s_0$, hence by Theorems 30.1 and 30.5, we have $\rho(f \circ g) \ge \rho(g)$. Next, Relation (1) is true for every $n \in \mathbb{N}$. Suppose now that $\rho(f) \ne 0$. Then, by (1), we have $\rho(f \circ g) = +\infty$. $\square$

**Notation.** Let $f \in \mathcal{A}(d(0, R^{-}))$. For each $r \in ]0, R[$, we denote by $s(r, f)$ the number of zeros of $f$ in $d(0, r)$, taking multiplicity into account.
**Theorem 30.7.** Let $f \in \mathcal{A}(\mathbb{K})$ be not identically zero and such that for some $t \ge 0$, $\limsup_{r \to +\infty} \frac{s(r,f)}{r^t}$ is finite. Then, $\rho(f) \le t$.

**Proof.** Set $\limsup_{r \to +\infty} \frac{s(r,f)}{r^t} = b \in [0, +\infty[$. Let us fix $\epsilon > 0$. We can find $R > 0$ such that $\frac{s(r,f)}{r^t} \le b + \epsilon \quad \forall r \ge R$, and hence, by Corollary 22.30, we have $\frac{|f|(r)}{|f|(R)} \le (\frac{r}{R})^{s(r,f)} \le (\frac{r}{R})^{r^t(b+\epsilon)}$. Therefore, putting $M = |f|(R)$, we have

$$ \log(|f|(r)) \le \log(M) + r^t(b + \epsilon)(\log(r) - \log(R)). $$

Now, when $u > 2$, $v > 2$, we know that $\log(u + v) \le \log(u) + \log(v)$. Applying that inequality with $u = M$ and $v = r^t(b + \epsilon)(\log(r) - \log(R))$ when $r^t(b + \epsilon)(\log(r) - \log(R)) > 2$, since $\log R \ge 0$, yields

$$ \begin{aligned} & \log(\log(|f|(r))) \\ & \le \log(\log(M)) + t\log(r) + \log(b + \epsilon) + \log(\log(r) - \log R) \\ & \le \log(\log(M)) + t\log(r) + \log(b + \epsilon) + \log(\log(r)). \end{aligned} $$

Consequently,

$$ \frac{\log(\log(|f|(r)))}{\log(r)} \le \frac{\log(\log(M)) + t\log(r) + \log(b + \epsilon) + \log(\log(r))}{\log(r)}, $$

and hence we can check that

$$ \limsup_{r \to +\infty} \frac{\log(\log(|f|(r)))}{\log(r)} \le t. \quad \square $$

**Theorem 30.8.** Let $f \in \mathcal{A}(\mathbb{K})$ be not identically zero. If there exists $q \ge 0$ such that

$$ \limsup_{r \to +\infty} \left( \frac{s(r, f)}{r^q} \right) < +\infty, $$

then $\rho(f)$ is the lowest bound of the set of $q \in [0, +\infty[$ such that

$$ \limsup_{r \to +\infty} \left( \frac{s(r, f)}{r^q} \right) = 0. $$
Moreover, if $\limsup_{r \to +\infty} \left(\frac{s(r,f)}{r^t}\right)$ is a number $b \in ]0, +\infty[$, then $\rho(f) = t$.

If there exists no $q$ such that $\limsup_{r \to +\infty} \left(\frac{s(r,f)}{r^q}\right) < +\infty$, then $\rho(f) = +\infty$.

**Proof.** The proof holds in two statements. First, we prove that given $f \in \mathcal{A}(\mathbb{K})$ non-constant and such that for some $t \ge 0$, $\limsup_{r \to +\infty} \frac{s(r,f)}{r^t}$ is finite, then $\rho(f) \le t$.

Set $\limsup_{r \to +\infty} \left(\frac{s(r,f)}{r^t}\right) = b \in [0, +\infty[$. Let us fix $\epsilon > 0$. We can find $R > 1$ such that $|f|(R) > e^2$ and $\frac{s(r,f)}{r^t} \le b + \epsilon \quad \forall r \ge R$, and hence, by Corollary 22.30, we have $\frac{|f|(r)}{|f|(R)} \le \left(\frac{r}{R}\right)^{s(r,f)} \le \left(\frac{r}{R}\right)^{r^t(b+\epsilon)}$. Therefore, since $R > 1$, we have

$$ \log(|f|(r)) \le \log(|f|(R)) + r^t(b + \epsilon)(\log(r)). $$

Now, when $u > 2$, $v > 2$, we check that $\log(u + v) \le \log(u) + \log(v)$. Applying that inequality with $u = \log(|f|(R))$ and $v = r^t(b + \epsilon)(\log(r))$ when $r^t(b + \epsilon)(\log(r)) > 2$ yields

$$ \begin{aligned} \log(\log(|f|(r))) &\le \log(\log(|f|(R))) + t\log(r) + \log(b + \epsilon) + \log(\log(r)) \\ &\quad + \log(b + \epsilon) + \log(\log(r)). \end{aligned} $$

Consequently,

$$ \frac{\log(\log(|f|(r)))}{\log(r)} \le \frac{\log(\log(|f|(R))) + t\log(r) + \log(b + \epsilon) + \log(\log(r))}{\log(r)}, $$

and hence we can check that

$$ \limsup_{r \to +\infty} \frac{\log(\log(|f|(r)))}{\log(r)} \le t $$

which proves the first claim.

Second, we prove that given $f \in \mathcal{A}(\mathbb{K})$ not identically zero and such that for some $t \ge 0$, we have $\limsup_{r \to +\infty} \frac{s(r,f)}{r^t} > 0$, then $\rho(f) \ge t$.
By hypotheses, there exists a sequence $(r_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} r_n = +\infty$ and such that $\lim_{n \to +\infty} \frac{s(r_n, f)}{r_n^2} > 0$. Thus, there exists $b > 0$ such that $\lim_{n \to +\infty} \frac{s(r_n, f)}{r_n^2} \ge b$. We can assume that $|f|(r_0) \ge 1$, hence by Corollary 22.27, $|f|(r_n) \ge 1 \forall n$. Let $\lambda \in ]1, +\infty[$. By Corollary 22.30, we have

$$ \frac{|f|(\lambda r_n)}{|f|(r_n)} \ge (\lambda)^{s(r_n, f)} \ge (\lambda)^{[b(r_n)^t]}, $$

hence

$$ \mathrm{Log}(|f|(\lambda r_n) \ge \mathrm{Log}(|f|(r_n)) + b(r_n)^t \mathrm{Log}(\lambda). $$

Since $|f|(r_n) \ge 1$, we have $\mathrm{Log}(\mathrm{Log}(|f|(\lambda r_n))) \ge \mathrm{Log}(b\mathrm{Log}(\lambda)) + t\mathrm{Log}(r_n)$, therefore

$$ \frac{\mathrm{Log}(\mathrm{Log}(|f|(\lambda r_n)))}{\mathrm{Log}(r_n)} \ge t + \frac{\mathrm{Log}(b\mathrm{Log}(\lambda))}{\mathrm{Log}(r_n)} \quad \forall n \in \mathbb{N}, $$

and hence

$$ \limsup_{r \to +\infty} \frac{\mathrm{Log}(\mathrm{Log}(|f|(r)))}{\mathrm{Log}(r)} \ge t $$

which ends the proof of the second claim. □

**Definition and notation.** Let $t \in [0, +\infty[$ and let $f \in \mathcal{A}(\mathbb{K})$ be of order $t$. We set $\psi(f) = \limsup_{r \to +\infty} \frac{s(r, f)}{r^t}$ and call $\psi(f)$ the *cotype* of growth of $f$, or just the *cotype* of $f$ in brief and we put $\tilde{\psi}(f) = \liminf_{r \to +\infty} \frac{s(r, f)}{r^t}$.

**Lemma 30.9.** Let $f, g \in \mathcal{A}(\mathbb{K})$. Then, $\max(s(r, f), s(r, g)) \le s(r, fg) = s(r, f) + s(r, g)$.

**Proof.** Let $r' > r$ be such that both $s(u, f)$ and $s(u, g)$ are constant in $[r, r']$. Then, when $\mu = \mathrm{Log}(u)$, by Corollary 19.3, we have

$$ s(u, f) = \frac{d\Psi(f, \mu)}{d\mu}, \quad s(u, g) = \frac{d\Psi(g, \mu)}{d\mu} \quad \forall \mu \in ]\mathrm{Log}(r), \mathrm{Log}(r')]. $$
But since $\Psi(fg, \mu) = \Psi(f, \mu) + \Psi(g, \mu)$, the inequalities

$$ \max(s(r, f), s(r, g)) \leq s(r, fg) = s(r, f) + s(r, g) $$

are clear.

<figure><img src="image_9.png" /></figure>

**Theorem 30.10.** Let $f, g \in \mathcal{A}^*(\mathbb{K})$. Then $\psi(fg) \leq \psi(f) + \psi(g)$. Moreover, if $\rho(f) \geq \rho(g)$, then $\psi(f) \leq \psi(fg)$. If $\rho(f) = \rho(g)$, then $\max(\psi(f), \psi(g)) \leq \psi(fg)$.

**Proof.** Set $q = \rho(f)$, $t = \rho(g)$ and suppose $q \geq t$. By Theorem 30.1, we have $\rho(f.g) = \rho(f) = q$. By Lemma 30.9, for each $r > 0$, we have $\max(s(r, f), s(r, g)) \leq s(r, f.g) = s(r, f) + s(r, g)$. Consequently,

$$ \begin{aligned} \limsup_{r \to +\infty} \frac{s(r, f.g)}{r^q} &\leq \limsup_{r \to +\infty} \frac{s(r, f)}{r^q} + \limsup_{r \to +\infty} \frac{s(r, g)}{r^q} \\ &\leq \limsup_{r \to +\infty} \frac{s(r, f)}{r^q} + \limsup_{r \to +\infty} \frac{s(r, g)}{r^t} \\ &= \psi(f) + \psi(g). \end{aligned} $$

Moreover, assuming again $q \geq t$, then

$$ \psi(f) = \limsup_{r \to +\infty} \frac{s(r, f)}{r^q} \leq \limsup_{r \to +\infty} \frac{s(r, fg)}{r^q} = \limsup_{r \to +\infty} \frac{s(r, fg)}{r^{\rho(fg)}} = \psi(fg). $$

Consequently, if $\rho(f) = \rho(g)$, then $\max(\psi(f), \psi(g)) \leq \psi(fg)$. $\square$

<figure><img src="image_10.png" /></figure>

**Theorem 30.11.** Let $f(x) = \sum_{n=0}^{+\infty} a_n x^n \in \mathcal{A}(\mathbb{K})$. Then,

$$ \rho(f) = \limsup_{n \to +\infty} \left( \frac{n\mathrm{Log}(n)}{-\mathrm{Log}|a_n|} \right). $$

**Proof.** We follow a similar way as this of [85] when $\rho(f) < +\infty$.

Let $t = \rho(f)$ and suppose first that $t < +\infty$. Let $\alpha = \limsup_{n \to +\infty} \frac{n\mathrm{Log}(n)}{-\mathrm{Log}|a_n|}$. Take $q > t$. For all $n \in \mathbb{N}$, we have $|a_n|r^n \leq |f|(r)$, and therefore, $|a_n|r^n \leq e^{(r^q)}$ and hence $|a_n| \leq r^{-n}e^{(r^q)}$ i.e.

$$ (1) \qquad \mathrm{Log}|a_n| \leq r^q - n\mathrm{Log}(r), $$

when $r$ is big enough.
Now, choose $r = \left(\frac{n}{s}\right)^{\frac{1}{q}}$. So, we have $\text{Log}|a_n| \le \frac{n}{q} - \frac{n}{q}\text{Log}(\frac{n}{q})$, i.e.

$$-\text{Log}(|a_n|) \geq -\frac{n}{q} + \frac{n}{q}\text{Log}\left(\frac{n}{q}\right).$$

Consequently, when $n$ is big enough, we have

$$\frac{n\text{Log}(n)}{(-\text{Log}|a_n|)} \leq \frac{n\text{Log}(n)}{\frac{n}{q}\text{Log}(\frac{n}{q}) - \frac{n}{q}} \leq q + O(1).$$

Therefore, we have $\alpha \le q$ and since this is true for each $q > t$, that shows that $\alpha \le t$.

Now, take $\beta > \alpha$ so that $\frac{n\text{Log}(n)}{(-\text{Log}|a_n|)} < \beta$ for $n$ big enough. Then, when $n$ is big enough, we have $n\text{Log}(n) \le \beta(-\text{Log}|a_n|)$, hence $n^{\frac{n}{\beta}} \le \frac{1}{|a_n|}$, and hence $|a_n| \le \frac{1}{n^{\frac{n}{\beta}}}$. Consequently, $|a_n|r^n \le \frac{r^n}{n^{\frac{n}{\beta}}}$. Now, for $r$ big enough, $|f|(r) = \sup_{n \in \mathbb{N}} |a_n|r^n \le \sup_{n \in \mathbb{N}} \frac{r^n}{n^{\frac{n}{\beta}}}$. Putting $\varphi(n) = \frac{n}{\beta}$ and $R = \frac{r}{\beta}$, we have

$$|f|(r^{\frac{1}{\beta}}) \leq \sup_{n \in \mathbb{N}} \frac{r^{\varphi(n)}}{n^{\varphi(n)}} \leq \sup_{x>0} \frac{R^x}{x^x}.$$

Now, we check that the maximum on $[0, +\infty[$ of the function $g(x) = \frac{R^x}{x^x}$ is reached when $x = \frac{R}{e}$ and hence is $e^{\frac{R}{e}} = e^{\frac{r}{\beta e}}$. Therefore, we have $|f|(r^{\frac{1}{\beta}}) \le e^{\frac{r}{\beta e}}$. Putting now $u = r^{\frac{1}{\beta}}$, we can derive $|f|(u) \le e^{\frac{u^\beta}{\beta e}}$, hence

$$\text{Log}(\text{Log}(|f|(u))) \leq \beta\text{Log}(u) - \text{Log}(e\beta).$$

Consequently,

$$\limsup_{r \to +\infty} \frac{\text{Log}(\text{Log}(|f|(r)))}{\text{Log}(r)} \leq \beta.$$

So, we have $t \le \beta$ and since this is true for all $\beta > \alpha$, we have proven that $t \le \alpha$, which ends the proof when $t < +\infty$.
Suppose now that $t = +\infty$ and suppose that $\limsup_{n \to +\infty} \frac{n\text{Log}(n)}{(-\text{Log}|a_n|)} < +\infty$. Let us take $q \in \mathbb{N}$ such that

$$ (2) \qquad \frac{n\text{Log}(n)}{(-\text{Log}|a_n|)} < q \quad \forall n \in \mathbb{N}. $$

By Theorem 30.8, we have $\limsup_{r \to +\infty} \frac{s(r, f)}{r^q} = +\infty$. So, we can take a sequence $(r_m)_{m \in \mathbb{N}}$ such that

$$ (3) \qquad \lim_{m \to +\infty} \frac{s(r_m, f)}{(r_m)^q} = +\infty. $$

For simplicity, set $u_m = s(r_m, f)$, $m \in \mathbb{N}$. By (2), for $m$ big enough, we have

$$ u_m \text{Log}(u_m) < q(-\text{Log}(|a_{u_m}|)) = q \text{Log} \left( \frac{1}{|a_{u_m}|} \right), $$

hence

$$ \frac{1}{(u_m)^{u_m}} > |a_{u_m}|^q, $$

therefore

$$ |a_{u_m}|^q (r_m)^{q u_m} < \frac{(r_m)^{q u_m}}{(u_m)^{u_m}} $$

i.e.

$$ (|f|(r_m))^s < \left( \frac{(r_m)^s}{u_m} \right)^{u_m}. $$

But by Corollary 22.27, we have $\lim_{r \to +\infty} |f|(r_m) = +\infty$, hence $(r_m)^q > u_m$ when $m$ is big enough, and therefore, $\limsup_{m \to +\infty} \frac{s(r_m, f)}{(r_m)^q} \le 1$, a contradiction to (3). Consequently, (2) is impossible, and therefore,

$$ \limsup_{n \to +\infty} \left( \frac{n \text{Log}(n)}{-\text{Log}|a_n|} \right) = +\infty = \rho(f). \quad \square $$
**Remark.** Of course, polynomials have a growth order equal to 0. On $\mathbb{K}$ as on $\mathbb{C}$, we can easily construct transcendental entire functions of order 0 or of order $\infty$.

**Example 1.** Suppose that for each $r > 0$ we have $s(r, f) \in [r^t \text{Log}r, r^t \text{Log}r + 1]$. Then of course, for every $s > t$, we have $\limsup_{r \to +\infty} \frac{s(r, f)}{r^s} = 0$ and $\limsup_{r \to +\infty} \frac{s(r, f)}{r^t} = +\infty$, so there exists no $t > 0$ such that $\frac{s(r, f)}{r^t}$ have non-zero superior limit $b < +\infty$. Consequently, $\rho(f) = +\infty$.

**Example 2.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $-\text{Log}|a_n| \in [n(\text{Log}(n))^2, n(\text{Log}(n))^2 + 1]$. Then clearly, $\lim_{n \to +\infty} \frac{\text{Log}|a_n|}{n} = -\infty$, hence the function $\sum_{n=0}^{\infty} a_n x^n$ has radius of convergence equal to $+\infty$. On the other hand, $\lim_{n \to +\infty} \frac{n\text{Log}(n)}{-\text{Log}|a_n|} = 0$, hence $\rho(f) = 0$.

**Example 3.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $-\text{Log}|a_n| \in [n\sqrt{\text{Log}(n)}, n\sqrt{\text{Log}(n)} + 1]$. Then, $\lim_{n \to +\infty} \frac{\text{Log}|a_n|}{n} = -\infty$ again, and hence the function $\sum_{n=0}^{\infty} a_n x^n$ has radius of convergence equal to $+\infty$. On the other hand,

$$ \lim_{n \to +\infty} \left( \frac{n\text{Log}(n)}{-\text{Log}|a_n|} \right) = +\infty \text{ hence } \rho(f) = +\infty. $$

**Theorem 30.12.** Let $f \in \mathcal{A}^*(\mathbb{K})$. Then,

$$ \rho(f) = \inf \left\{ q \in ]0, +\infty[ \mid \lim_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r^q} = 0 \right\}. $$

**Proof.** Indeed, let $M = \inf\{q \in ]0, +\infty[ \mid \lim_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r^q} = 0\}$. First, we prove that $\rho(f) \le M$. Let $q$ be such that $\lim_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r^q} = 0$. Let us fix $\epsilon > 0$. For $r$ big enough, we have $\frac{\text{Log}(|f|(r))}{r^q} \le \epsilon$, hence $\text{Log}(|f|(r)) \le \epsilon r^q$, therefore $\text{Log}(\text{Log}(|f|(r))) \le \text{Log}\epsilon + q\text{Log}(r)$, hence $\frac{\text{Log}(\text{Log}(|f|(r)))}{\text{Log}(r)} \le q + \frac{\epsilon}{\text{Log}(r)}$. This is true for every $\epsilon > 0$, therefore $\limsup_{r \to +\infty} \frac{\text{Log}(\text{Log}(|f|(r)))}{\text{Log}(r)} \le q$ i.e. $\rho(f) \le q$, and hence, $\rho(f) \le M$.

On the other hand, we note that

$$ M = \sup \left\{ q \in ]0, +\infty[ \mid \limsup_{r \to +\infty} \frac{\text{Log}(|f|(r))}{r^q} > 0 \right\}. $$
Now, suppose that for some $q > 0$, we have $\limsup_{r \to +\infty} \frac{\log(|f|(r))}{r^q} = b > 0$. Let us fix $\epsilon \in ]0, b[$. There exists a sequence $(r_n)_{n \in \mathbb{N}}$ such that, when $n$ is big enough, we have $b - \epsilon \le \frac{\log(|f|(r_n))}{(r_n)^q} \le b + \epsilon$, hence $q\log(r_n) + \log(b - \epsilon) < \log(\log(|f|(r_n))) < q\log(r_n) + \log(b - \epsilon)$, therefore

$$q + \frac{\log(b - \epsilon)}{\log(r_n)} < \frac{\log(\log(|f|(r_n))}{\log(r_n)} < q + \frac{\log(b + \epsilon)}{\log(r_n)}.$$

Consequently, $\lim_{n \to +\infty} \frac{\log(\log(|f|(r_n))}{\log(r_n)} = q$, and therefore, $\rho(f) \ge q$, hence $\rho(f) \ge M$. Finally, $\rho(f) = M$. □
Chapter 31

Type of Growth for Entire Functions

**Definition and notation.** In complex analysis, the type of growth is defined for an entire function having a finite order of growth $t$ as $\sigma(f) = \limsup_{r \to +\infty} \frac{\log(M_f(r))}{r^t}$, with $t < +\infty$ [90]. Of course, the same notion may be defined for $f \in \mathcal{A}(\mathbb{K})$. Here, as in Chapters 30, we denote by Log the Neperian logarithm and by $e$ the number such that $\log(e) = 1$. Then, given $f \in \mathcal{A}^*(\mathbb{K})$ of order $t$, we set $\sigma(f) = \limsup_{r \to +\infty} \frac{\log(|f(r)|)}{r^t}$. Moreover, we put $\tilde{\sigma}(f) = \liminf_{r \to +\infty} \frac{\log(|f(r)|)}{r^t}$.

A function $f \in \mathcal{A}(\mathbb{K})$ is said to be *clean* if $\rho(f) < +\infty$ and $\sigma(f) = \tilde{\sigma}(f)$ and $f$ is said to be *regular* if $\rho(f) = \tilde{\rho}(f)$.

**Theorem 31.1.** Let $f, g \in \mathcal{A}^*(\mathbb{K})$. Then, $\sigma(fg) \le \sigma(f) + \sigma(g)$. If $\rho(f) \ge \rho(g)$, then $\sigma(f) \le \sigma(fg)$, and if $\rho(f) > \rho(g)$, then $\sigma(fg) = \sigma(f)$. If $f$ is clean and such that $\rho(f) > \rho(g)$, then $fg$ is clean. If $f$ and $g$ are clean and if $\rho(f) = \rho(g)$, then $fg$ is clean and $\sigma(fg) = \sigma(f) + \sigma(g)$ and $\psi(fg) = \psi(f) + \psi(g)$. If $\rho(f) = \rho(g)$, then $\max(\sigma(f), \sigma(g)) \le \sigma(fg)$. If $\rho(f) = \rho(g)$ and $\sigma(f) > \sigma(g)$, then $\rho(f+g) = \rho(f)$ and $\sigma(f+g) = \sigma(f)$. If $\rho(f+g) = \rho(f) \ge \rho(g)$, then $\sigma(f+g) \le \max(\sigma(f), \sigma(g))$.

**Proof.** Let $s = \rho(f) \ge t = \rho(g)$. Then, $\rho(fg) = s$, hence

$$ \sigma(fg) = \limsup_{r \to +\infty} \frac{\log(|fg|(r))}{r^s} = \limsup_{r \to +\infty} \frac{\log(|f|(r) \cdot |g|(r))}{r^s}. $$
Now, if $s \ge t$, then

$$ \sigma(fg) = \limsup_{r \to +\infty} \left( \frac{\log(|f|(r))}{r^s} + \frac{\log(|g|(r))}{r^s} \right) \\ \le \limsup_{r \to +\infty} \frac{\log(|f|(r))}{r^s} + \limsup_{r \to +\infty} \frac{\log(|g|(r))}{r^t} = \sigma(f) + \sigma(g). $$

Then we note that when $t < s$, we have

$$ \limsup_{r \to +\infty} \frac{\log(|g|(r))}{r^s} = 0, $$

hence

$$ \limsup_{r \to +\infty} \frac{\log(|fg|(r))}{r^s} = \limsup_{r \to +\infty} \frac{\log(|f|(r))}{r^s} = \sigma(f). $$

Particularly, if $f$ is clean, we have limits instead of limitsup as long as $f$ is concerned. Consequently, if $t < s$, then $fg$ is clean.

Now, suppose that $f$ and $g$ are clean and that $s = t$. Then,

$$ \limsup_{r \to +\infty} \frac{\log(|f|(r)) + \log(|g|(r))}{r^t} = \lim_{r \to +\infty} \frac{\log(|f|(r)) + \log(|g|(r))}{r^t} \\ = \sigma(f) + \sigma(g). $$

Thus, $fg$ is clean. Moreover, since $f$ and $g$ are clean, we can see that $\sigma(fg) = \sigma(f) + \sigma(g)$ and $\psi(fg) = \psi(f) + \psi(g)$.

Now, suppose again $s > t$. Then, by Theorem 30.1, $\rho(f + g) = \rho(f) = s$. Consequently,

$$ \sigma(f + g) = \limsup_{r \to +\infty} \left( \frac{\log|f + g|(r)}{r^s} \right) \\ \le \limsup_{r \to +\infty} \left( \frac{\max(\log|f|(r), \log|g|(r))}{r^s} \right) \\ = \max \left( \limsup_{r \to +\infty} \left( \frac{\log|f|(r)}{r^s} \right), \limsup_{r \to +\infty} \left( \frac{\log|g|(r)}{r^s} \right) \right) \\ \le \max \left( \limsup_{r \to +\infty} \left( \frac{\log|f|(r)}{r^s} \right), \limsup_{r \to +\infty} \left( \frac{\log|g|(r)}{r^t} \right) \right) \\ = \max(\sigma(f), \sigma(g)). $$
Now, suppose $\rho(f) = \rho(g) = s$. Then,

$$ \max \left( \limsup_{r \to +\infty} \left( \frac{\log(|f(r)|)}{r^s} \right), \limsup_{r \to +\infty} \left( \frac{\log(|g(r)|)}{r^s} \right) \right) \\ \le \limsup_{r \to +\infty} \left( \frac{\log(|f \cdot g(r)|)}{r^s} \right) $$

because the two both $|f|(r)$ and $|g|(r)$ tend to $+\infty$ with $r$. Consequently, $\sigma(fg) \ge \max(\sigma(f), \sigma(g))$.

Now, suppose again that $\rho(f) = \rho(g)$ and suppose $\sigma(f) > \sigma(g)$. Let $s = \rho(f)$, $b = \sigma(f)$. Then, $b > 0$. Let $(r_n)_{n \in \mathbb{N}}$ be a sequence such that $\lim_{n \to +\infty} r_n = +\infty$ and $\lim_{n \to +\infty} \frac{\log(|f(r_n)|)}{(r_n)^s} = b$. Since $\sigma(g) < \sigma(f)$, we note that when $n$ is big enough we have $|g(r_n)| < |f(r_n)|$. Consequently, when $n$ is big enough, we have $|f + g(r_n)| = |f(r_n)|$, and hence

$$ (1) \qquad \lim_{n \to +\infty} \frac{\log(|f + g|(r_n))}{(r_n)^s} = b. $$

Now, by definition of $\sigma$, we have $\sigma(f + g) \ge \lim_{n \to +\infty} \frac{\log(|f + g|(r_n))}{(r_n)^\rho}$.
By Theorem 30.1, we have $\rho(f + g) \le s$, hence

$$ \sigma(f+g) \ge \lim_{n \to +\infty} \frac{\log(|f+g|(r_n))}{(r_n)^{\rho(f+g)}} \ge \lim_{n \to +\infty} \frac{\log(|f+g|(r_n))}{(r_n)^s} \\ = \lim_{n \to +\infty} \frac{\log(|f|(r_n))}{(r_n)^s} = \sigma(f), $$

therefore by (1), $\sigma(f + g) \ge \sigma(f)$.

Suppose that $\sigma(f + g) > \sigma(f)$. Putting $h = f + g$, we have $f = h - g$ with $\sigma(g) < \sigma(h)$, hence $\sigma(h - g) \ge \sigma(h)$ i.e. $\sigma(f) > \sigma(f + g)$, a contradiction. Consequently, $\sigma(f + g) = \sigma(f)$. Thus, $\limsup_{r \to +\infty} \frac{\log(|f+g|(r))}{r^s} = b > 0$. But then, $\limsup_{r \to +\infty} \frac{\log(|f+g|(r))}{r^m} = 0 \forall m > s$. Therefore, by Theorem 30.12, we have $\rho(f + g) = \rho(f)$.
Finally, suppose now that $\rho(f + g) = \rho(f) \ge \rho(g)$. Then,

$$ \begin{aligned} \sigma(f + g) &= \limsup_{r \to +\infty} \frac{\log(|f + g|(r))}{r^s} \\ &\le \max \left( \limsup_{r \to +\infty} \frac{\log(|f|(r))}{r^s}, \limsup_{r \to +\infty} \frac{\log(|g|(r))}{r^s} \right) \\ &\le \max \left( \limsup_{r \to +\infty} \frac{\log(|f|(r))}{r^s}, \limsup_{r \to +\infty} \frac{\log(|g|(r))}{r^t} \right) \\ &= \max(\sigma(f), \sigma(g)). \end{aligned} $$

The last statement derives from the previous ones and from Theorem 30.1(iii). □

**Corollary 31.2.** Let $f, g \in \mathcal{A}(\mathbb{K})$ be such that $\rho(f) \ne \rho(g)$. Then, $\sigma(f + g) \le \max(\sigma(f), \sigma(g))$. The set of clean functions $f \in \mathcal{A}(\mathbb{K})$ is a multiplicative semi-group. The set $C(t), \cdot$ of clean functions $f \in \mathcal{A}(\mathbb{K})$ of order $t$ is a submultiplicative semi-group and $\sigma$ and $\psi$ are semi-group morphisms from $(C(t), \cdot)$ into $(\mathbb{R}^+, +)$.

**Proof.** Indeed, assuming that $\rho(f) > \rho(g)$, we have $\rho(f + g) = \rho(f)$ and hence the conclusion comes from the last statement of Theorem 31.1. □

Now we show that $\sigma(f)$ may be computed by the same formula as on $\mathcal{C}$.

**Theorem 31.3 (16).** Let $f(x) = \sum_{n=0}^{\infty} b_n x^n \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$. Then, $\sigma(f)\rho(f) = \limsup_{n \to +\infty} (n^{\frac{1}{n}}|b_n|^{\rho(f)})$.

**Proof.** Let $t = \rho(f)$. First, let us show that $\limsup_{n \to +\infty} n|b_n|^{\frac{t}{n}}$. We follow a similar way as in [86]. Let $u = \sigma(f)$ and let us take $w > \sigma(f)$. For $r$ big enough, we have $\log(|f|(r)) \le wr^t$, hence for all $n \in \mathbb{N}$, we can derive

$$ (1) \qquad |b_n| \le \frac{|f|(r)}{r^n} \le \frac{e^{wr^t}}{r^n}. $$

Now, let us take $r$ such that the derivative of the logarithm of the function $\frac{e^{wr^t}}{r^n}$ vanishes: we have $wtr^{t-1} - \frac{n}{r} = 0$. So, we can choose
$r_n = \left(\frac{n}{wt}\right)^{\frac{1}{t}}$ and we can check that

$$|b_n| \le \frac{e^{\frac{n}{t}}}{\left(\frac{n}{wt}\right)^{\frac{n}{t}}} = \left(\frac{ewt}{n}\right)^{\frac{n}{t}}.$$

Consequently, we have $n|b_n|^{\frac{t}{n}} \le ewt$, therefore $\limsup_{n \to +\infty} n|b_n|^{\frac{t}{n}} \le etw$. This is true for all $w > \sigma(f)$, and hence $\limsup_{n \to +\infty} n|b_n|^{\frac{t}{n}} \le et\sigma(f)$.

Now, let us show the reverse inequality. Take $c > \frac{1}{\frac{ect}{n}}$ and $\limsup_{n \to +\infty} n|b_n|^{\frac{t}{n}}$. When $n$ is big enough, we have $|b_n| \le \left(\frac{ect}{n}\right)^{\frac{t}{n}}$, hence $|b_n|r^n \le \left(\frac{ect}{n}\right)^{\frac{n}{t}}r^n$, and consequently, $|f|(r) \le \sup_{n \ge 1} \left(\frac{ect}{n}\right)^{\frac{n}{t}}r^n$. Therefore, $|f|(r) \le \sup_{x>1} \frac{(ect)^{\frac{x}{t}}r^x}{x^{\frac{x}{t}}}$. Now, set $y = \frac{x}{t}$ and $R = ecr$. Then,

$$|f|(r^{\frac{1}{t}}) \le \sup_{y>0} \frac{(ect)^y r^y}{(ty)^y} = \sup_{y>0} \left(\frac{ecr}{y}\right)^y = \sup_{y>0} \frac{R^y}{y^y} = e^{\frac{R}{e}} = e^{cr}.$$

Thus, we have $|f|(r) \le e^{cr^t}$, and hence $\limsup_{r \to +\infty} \frac{\log(|f|(r))}{r^t} \le c$. Therefore, $\sigma(f) \le c$, which ends the proof. □

In the proof of Theorem 31.5, we use the following trivial lemma:

**Lemma 31.4.** Let $g, h$ be the real functions defined in $]0, +\infty[$ as $g(x) = \frac{e^{tx}-1}{x}$ and $h(x) = \frac{1-e^{-tx}}{x}$ with $t > 0$. Then,

(i) $\inf\{|g(x)| |x > 0\} = t,$

(ii) $\sup\{|h(x)| |x > 0\} = t.$

**Notations.** Let $f \in \mathcal{A}(\mathbb{K})$ and let $(a_n)_{n \in \mathbb{N}}$ be the sequence of zeros of $f$ with $|a_n| \le |a_{n+1}|$, $n \in \mathbb{N}$ and for each $n \in \mathbb{N}$, let $w_n$ be the multiplicity order of $a_n$. For every $r > 0$, let $k(r)$ be the integer such that $|a_n| \le r \ \forall n \le k(r)$ and $|a_n| > r \ \forall n > k(r)$. Then, by Corollary 22.27, $\log(|f|(r))$ is of the form $\sum_{n=0}^{k(r)} w_n(\log(r) - \log(|a_n|))$. We put $\sigma(f, r) = \frac{\sum_{n=0}^{k(r)} w_n(\log(r) - \log(|a_n|))}{r^{\rho(f)}}$. In the same way, for any $r > 0$ and $n \in \mathbb{N}$, $\psi(f, r) = \frac{s(r, f)}{r^{\rho(f)}}$.

**Theorem 31.5.** Let $f \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$. Then, $\sigma(f) < +\infty$ if and only if $\psi(f) < +\infty$. Suppose that these hypotheses
are satisfied. Then,

$$ \rho(f)\sigma(f) \le \psi(f) \le \rho(f)(e\sigma(f) - \tilde{\sigma}(f)) $$

and

$$ \rho(f)(\tilde{\sigma}(f) - \frac{\sigma(f)}{e}) \le \tilde{\psi}(f) \le \rho(f)\tilde{\sigma}(f). $$

Further, the hypotheses $\sigma(f) = \tilde{\sigma}(f)$ and $\psi(f) = \tilde{\psi}(f)$ are equivalent, and if they are satisfied, then $\psi(f) = \rho(f)\sigma(f)$.

**Proof.** Without loss of generality, we can assume that $f(0) = 1$. Let $u = \rho(f)$. Let $(a_n)_{n \in \mathbb{N}}$ be the sequence of zeros of $f$ with $|a_n| \le |a_{n+1}|$, $n \in \mathbb{N}$, and for each $n \in \mathbb{N}$, let $w_n$ be the multiplicity order of $a_n$. For every $r > 0$, let $k(r)$ be the integer such that $|a_n| \le r \forall n \le k(r)$ and $|a_n| > r \forall n > k(r)$. We put $c_n = |a_n|$, $n \in \mathbb{N}$.

We first show the inequality $\rho(f)\sigma(f) \le \psi(f)$. By the definition of $\sigma(f, r)$, we can derive

$$ \sigma(f, r) = \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\text{Log}(r) - \text{Log}(re^{-\alpha}))}{r^u} + \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\text{Log}(re^{-\alpha}) - \text{Log}(c_n))}{r^u} + \sum_{k(re^{-\alpha}) < n \le k(r)} \frac{w_n(\text{Log}(re^{-\alpha}) - \text{Log}(c_n))}{r^u}, $$

hence

$$ \sigma(f, r) \le \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\text{Log}(r) - \text{Log}(re^{-\alpha}))}{r^u} + \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\text{Log}(re^{-\alpha}) - \text{Log}(c_n))}{r^u} + \alpha \sum_{k(re^{-\alpha}) < n \le k(r)} \frac{w_n}{r^u} $$
because $\mathrm{Log}(r) - \mathrm{Log}(c_n) \le \alpha \ \forall n \in [k(re^{-\alpha}), k(r)] \cap \mathbb{N}$. Consequently,

$$ \sigma(f, r) \le \alpha \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n}{r^u} + \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\mathrm{Log}(re^{-\alpha}) - \mathrm{Log}(c_n))}{r^u} \\ + \alpha \sum_{k(re^{-\alpha})<n \le k(r)} \frac{w_n}{r^u}, $$

therefore

$$ \sigma(f, r) \le \alpha \sum_{n=0}^{k(r)} \frac{w_n}{r^u} + \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\mathrm{Log}(re^{-\alpha}) - \mathrm{Log}(c_n))}{r^u}, $$

hence

$$ \sigma(f, r) \le e^{-u\alpha} \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\mathrm{Log}(re^{-\alpha}) - \mathrm{Log}(c_n))}{(re^{-\alpha})^u} + \alpha \sum_{0 \le n \le k(r)} \frac{w_n}{r^u}. $$

Thus, we have

$$ (1) \qquad \sigma(f, r) \le e^{-u\alpha}\sigma(f, re^{-\alpha}) + \alpha\psi(f, r). $$

Suppose first that $\sigma(f) < +\infty$. We check that we can pass to superior limits on both sides, so we obtain $\sigma(f) \le e^{-u\alpha}\sigma(f) + \alpha\psi(f)$, therefore $\sigma(f) \frac{(1-e^{-u\alpha})}{\alpha} \le \psi(f)$. That holds for every $\alpha > 0$, hence, by Lemma 31.4 (ii), we can derive

$$ (2) \qquad \psi(f) \ge \rho(f)\sigma(f). $$

Now, by (1), we have

$$ \sigma(f, r) - e^{-u\alpha}\sigma(f, re^{-\alpha}) \le \alpha\psi((r, f)), $$

hence passing to inferior limits on both sides, we deduce

$$ \frac{\tilde{\sigma}(f) - e^{-u\alpha}\sigma(f)}{\alpha} \le \tilde{\psi}(f), $$

hence

$$ \frac{u(\tilde{\sigma}(f) - e^{-u\alpha}\sigma(f))}{u\alpha} \le \tilde{\psi}(f), $$
therefore when $\alpha u = 1$, we obtain

$$ (3) \qquad \rho(f) \left( \tilde{\sigma}(f) - \frac{\sigma(f)}{e} \right) \le \tilde{\psi}(f). $$

We now show the inequality

$$ \psi(f) \le \rho(f)(e\sigma(f) - \tilde{\sigma}(f)). $$

Let us fix $\alpha > 0$. We can write

$$ \begin{aligned} \sigma(f, r) = & \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n(\log(r) - \log(re^{-\alpha}))}{r^u} \\ & + \sum_{j=0}^{k(re^{-\alpha})} \frac{w_j(\log(re^{-\alpha}) - \log(c_n))}{r^u} \\ & + \sum_{k(re^{-\alpha}) < j \le k(r)} \frac{w_j(\log(r) - \log(c_j))}{r^u}, \end{aligned} $$

hence

$$ \sigma(f, r) \ge \alpha \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n}{r^u} + \sum_{j=0}^{k(re^{-\alpha})} \frac{w_j(\log(re^{-\alpha}) - \log(c_n))}{r^u}, $$

hence

(4)

$$ \sigma(f, r) \ge \alpha e^{-u\alpha} \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n}{(re^{-\alpha})^u} + e^{-u\alpha} \sum_{j=0}^{k(re^{-\alpha})} \frac{w_n(\log(re^{-\alpha}) - \log(c_n))}{(re^{-\alpha})^u} $$

and hence

$$ (5) \qquad \sigma(f, r) \ge \alpha e^{-u\alpha} \psi(f, re^{-\alpha}) + e^{-u\alpha} \sigma(f, re^{-\alpha}). $$

Therefore, we can deduce

$$ \alpha e^{-u\alpha} \psi(f) \le \limsup_{r \to +\infty} \left( \sigma(f, r) - e^{-u\alpha} \sigma(f, re^{-\alpha}) \right), $$
and therefore

$$ (6) \qquad \alpha e^{-u\alpha} \psi(f) \le \sigma(f) - e^{-u\alpha} \tilde{\sigma}(f). $$

That holds for every $\alpha > 0$ and hence, when $u\alpha = 1$, by (6), we obtain

$$ (7) \qquad \psi(f) \le \rho(f)(e\sigma(f) - \tilde{\sigma}(f)) $$

which is the left-hand inequality of the general conclusion.

Moreover, we see that if $\psi(f) = +\infty$, then $\sigma(f) = +\infty$.

Now, on (4), we can also take the inferior limit on both sides and we deduce

$$ \tilde{\sigma}(f) \ge \alpha e^{-u\alpha} \tilde{\psi}(f) + e^{-u\alpha} \tilde{\sigma}(f), $$

therefore

$$ \alpha e^{-u\alpha} \tilde{\psi}(f) \le \tilde{\sigma}(f)(1 - e^{-u\alpha}). $$

Then when $u\alpha$ tends to 0, we have

$$ (8) \qquad \tilde{\psi}(f) \le \rho(f) \tilde{\sigma}(f). $$

Now, suppose that $\sigma(f) = +\infty$. We can find an increasing sequence $r_n$ of limit $+\infty$ such that $\sigma(f, r_n) = \sup\{\sigma(f, r) \mid r \le r_n\}, n \in \mathbb{N}$. Consider (1) when $\min(\alpha, u\alpha) > 1$. Then, $\sigma(f, r_n e^{-\alpha}) < \sigma(f, r)_n$, hence of course,

$$ (1 - e^{-u\alpha})\sigma(f, r_n e^{-\alpha}) \le (1 - e^{-u\alpha})\sigma(f, r_n), $$

and hence $\sigma(f, r_n) - e^{-u\alpha}\sigma(f, r_n e^{-\alpha}) \ge (1 - e^{-u\alpha})\sigma(f, r_n)$, therefore $(1 - e^{-u\alpha})\sigma(f, r_n) \le \psi(f, r_n)$, which proves that $\psi(f) = +\infty$.

Thus, $\sigma(f) < +\infty$ is equivalent to $\psi(f) < +\infty$. Consequently, Relations (2), (3), (5), (7), and (8) still apply and hence hold as soon as $\sigma(f) < +\infty$ or $\psi(f) < +\infty$.

Now, suppose that $\tilde{\psi}(f) = \psi(f)$. Then we have

$$ \rho(f)\sigma(f) \le \psi(f) \le \rho(f)\tilde{\sigma}(f), $$

therefore $\sigma(f) = \tilde{\sigma}(f)$ since $\rho(f) > 0$.
Conversely, suppose that $\sigma(f) = \tilde{\sigma}(f)$. Then, by (6), we have
$\psi(f) \le \sigma(f) \left(\frac{e^{u\alpha}-1}{\alpha}\right)$. That holds for every $\alpha > 0$, and then, $\psi(f) \le u\sigma(f)$, i.e. $\psi(f) \le \rho(f)\sigma(f)$, hence by (2), we have $\psi(f) = \rho(f)\sigma(f)$.

But now, by (1), we see that $\alpha\psi(f, r) \ge \sigma(f, r) - e^{-u\alpha}\sigma(f, re^{-\alpha})$, hence, passing to the inferior limit, $\alpha\tilde{\psi}(f) \ge \sigma(f)(1 - e^{-u\alpha}) \quad \forall \alpha > 0$, therefore $\tilde{\psi}(f) \ge u\sigma(f)$. But we just showed that $\psi(f) = u\sigma(f)$, hence $\tilde{\psi}(f) = \psi(f)$.

Thus, we have shown that the hypotheses $\tilde{\psi}(f) = \psi(f)$ and $\tilde{\sigma}(f) = \sigma(f)$ are equivalent.

Now, suppose again that $\psi(f) = \tilde{\psi}(f)$. Then, by the inequalities already proved, on the one hand, we have $\psi(f) \ge \rho(f)\sigma(f)$, and on the other hand, we have $\psi(f) \le \rho(f)\sigma(f)$. Consequently, $\psi(f) = \rho(f)\sigma(f)$. $\square$

By Theorem 31.3, we obtain Corollary 31.6:

**Corollary 31.6.** Let $f(x) = \sum_{n=0}^{\infty} b_n x^n \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$. Then,

$$ \rho(f)\sigma(f) \le \psi(f) \le \limsup_{n \to +\infty} \left( n^{\frac{n}{2}} \sqrt{|b_n|^{\rho(f)}} \right) - \rho(f)\tilde{\sigma}(f) $$

$$ \text{and } e\sigma(f) = \frac{\limsup_{n \to +\infty} \left( n^{\frac{n}{2}} \sqrt{|b_n|^{\rho(f)}} \right)}{\rho(f)}. $$

**Theorem 31.7.** Let $f(x) \in \mathcal{A}(\mathbb{K})$ be such that $\tilde{\sigma}(f) > 0$. Then, $f$ is regular.

**Proof.** Let $a = \tilde{\sigma}(f)$, hence $a = \liminf_{r \to +\infty} \frac{\log(|f|(r))}{r^{\rho(f)}}$, then there exists $R > 0$ such that $\frac{\log(|f|(r))}{r^{\rho(f)}} \ge \frac{a}{2} \quad \forall r > R$, hence $\log(|f|(r)) > \frac{a}{2}r^{\rho(f)} \quad \forall r > R$ and hence

$$ \frac{\log(\log(|f|(r)))}{\log(r)} \ge \frac{\log(a) - \log(2)}{\log(r)} + \rho(f) \quad \forall r \ge R. $$

Thus, $\tilde{\rho}(f) \ge \rho(f)$, which ends the proof. $\square$

**Theorem 31.8.** Let $f(x) = \sum_{n=0}^{\infty} b_n x^n \in \mathcal{A}(\mathbb{K})$ be clean. Then,

$$ e\psi(f) = \limsup_{n \to +\infty} \left( n^{\frac{n}{2}} \sqrt{|b_n|^{\rho(f)}} \right) $$

and $f$ is regular.
**Proof.** The first statement is an immediate consequence of Corollary 31.6. Now, suppose that $f$ is clean. Let $u = \rho(f)$ and $s = \sigma(f)$. Then, we have

$$ \lim_{r \to +\infty} \frac{\log(|f|(r))}{r^u} = s, $$

hence $\lim_{r \to +\infty}(\log(\log(|f|(r))) - u\log(r)) = \log(s)$, therefore

$$ \lim_{r \to +\infty} \frac{\log(\log(|f|(r)))}{\log(r)} - u = 0 $$

i.e.

$$ \rho(f) = \lim_{r \to +\infty} \frac{\log(\log(|f|(r)))}{\log(r)}, $$

hence $f$ is regular. □

**Remark.** In [32], the authors claimed that if $f$ is a regular complex entire function, then $\sigma(f) = \tilde{\sigma}(f)$. We see in the following that such a statement does not hold for functions in $\mathcal{A}(\mathbb{K})$ and that the proof in [4] is put in doubt.
Chapter 32

Growth of the Derivative
of an Entire Function

Similar to the situation in complex entire functions, here we see
that the order and the type of the derivative of an entire function *f*
are respectively equal to those of *f*. As in Chapters 30 and 31, we
denote by Log the Neperian logarithm and by *e* the number such
that Log(*e*) = 1.

Throughout this chapter, $\mathbb{K}$ is supposed to have characteristic 0.

**Theorem 32.1.** Let $f \in \mathcal{A}(\mathbb{K})$ be not identically zero. Then, $\rho(f) = \rho(f')$.

**Proof.** Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$. By Theorem 30.11, we have $\rho(f') = \lim \sup_{n \to +\infty} \left( \frac{n \log(n)}{-\log\left(\frac{n+1}{a_{n+1}}\right)} \right)$. But since $\frac{1}{n} \le |n| \le 1$, we have

$$
\mathrm{Log}(|a_{n+1}|) - \mathrm{Log}(n+1) \leq \mathrm{Log}(|(n+1)a_{n+1}|) \leq \mathrm{Log}(|a_{n+1}|),
$$

hence

$$
- \frac{\mathrm{Log}(|a_{n+1}|)}{n \mathrm{Log}(n)} \leq - \frac{\mathrm{Log}(|(n+1)a_{n+1}|)}{n \mathrm{Log}(n)} \leq - \frac{\mathrm{Log}(|a_{n+1}|) - \mathrm{Log}(n+1)}{n \mathrm{Log}(n)},
$$
hence

$$ 
\begin{aligned}
& \liminf_{n \to +\infty} \left( - \frac{\mathrm{Log}(|a_{n+1}|)}{n \mathrm{Log}(n)} \right) \\
& \le \liminf_{n \to +\infty} \left( - \frac{\mathrm{Log}(|(n+1)a_{n+1}|)}{n \mathrm{Log}(n)} \right) \\
& \le \liminf_{n \to +\infty} \left( - \frac{\mathrm{Log}(|a_{n+1}|) - \mathrm{Log}(n+1)}{n \mathrm{Log}(n)} \right).
\end{aligned}
$$

But since

$$ \lim_{n \to +\infty} \frac{\mathrm{Log}(n+1)}{n \mathrm{Log}(n)} = 0, $$

we have

$$ \liminf_{n \to +\infty} \left( - \frac{\mathrm{Log}(|a_{n+1}|)}{n \mathrm{Log}(n)} \right) = \liminf_{n \to +\infty} \left( - \frac{\mathrm{Log}(|a_{n+1}|) - \mathrm{Log}(n+1)}{n \mathrm{Log}(n)} \right), $$

therefore

$$ \liminf_{n \to +\infty} \left( \frac{-\mathrm{Log}(|a_{n+1}|)}{n \mathrm{Log}(n)} \right) = \liminf_{n \to +\infty} \left( - \frac{\mathrm{Log}((n+1)|a_{n+1}|)}{n \mathrm{Log}(n)} \right). $$

But since all quantities are positive, we can derive

$$ \limsup_{n \to +\infty} \frac{n \mathrm{Log}(n)}{-\mathrm{Log}(|a_{n+1}|)} = \limsup_{n \to +\infty} \left( \frac{n \mathrm{Log}(n)}{-\mathrm{Log}(|(n+1)a_{n+1}|)} \right), $$

therefore

$$ \limsup_{n \to +\infty} \frac{n \mathrm{Log}(n)}{-\mathrm{Log}(|a_{n+1}|)} = \limsup_{n \to +\infty} \left( \frac{(n+1)\mathrm{Log}(n+1)}{-\mathrm{Log}(|(n+1)a_{n+1}|)} \right) = \rho(f), $$

and hence $\rho(f') = \rho(f)$. $\square$

<figure><img src="image_11.png" /></figure>

**Corollary 32.2.** Consider the differential equation

$$ (\mathcal{E}) \quad f^{(n)} + a_{n-1}(x)f^{(n-1)}(x) + \cdots + a_0(x)f(x) = 0 $$

with $a_j \in \mathcal{A}^*(\mathbb{K})$, $j = 0, \dots, n-1$, and $\rho(a_j) < \rho(a_0)$ $\forall j = 1, \dots, n-1$. Then, every non-trivial solution $f$ of $(\mathcal{E})$ satisfies $\rho(f) \ge \rho(a_0)$.
**Corollary 32.3.** The derivation on $\mathcal{A}(\mathbb{K})$ restricted to the algebra $\mathcal{A}(\mathbb{K}, t)$ (respectively to $\mathcal{A}^*(\mathbb{K})$) provides that algebra with a derivation.

In complex analysis, it is known that if an entire function $f$ has order $t < +\infty$, then $f$ and $f'$ have the same type. We check that it is the same here.

**Theorem 32.4.** Let $f \in \mathcal{A}(\mathbb{K})$ of order $t \in ]0, +\infty[$. Then, $\sigma(f) = \sigma(f')$.

**Proof.** By Theorem 31.3, we have

$$ 
\begin{aligned}
e\rho(f')\sigma(f') &= \limsup_{n \to +\infty} \left( n \left( \frac{|n+1||a_{n+1}|}{n} \right)^{\frac{t}{n}} \right) \\
&= \limsup_{n \to +\infty} \left( \left( (n+1) \left( \frac{|n+1||a_{n+1}|}{n} \right)^{\frac{t}{n}} \right)^{\frac{n}{n+1}} \left( \frac{n}{n+1} \right) \right) \\
&= \limsup_{n \to +\infty} \left( (n+1) \left( \frac{|n+1||a_{n+1}|}{n} \right)^{\frac{t}{n+1}} \right) = e\rho(f)\sigma(f).
\end{aligned}
$$

But since $\rho(f) = \rho(f')$ and since $\rho(f) \neq 0$, we can see that $\sigma(f') = \sigma(f)$. $\square$

Theorem 32.4 shows a way to compare the growth of an entire function $f$ to this of its derivative. Of course, we know that the inequality $|f'|(r) \le |f|(r)$ holds always. But we don't have an inequality in the other side. However, thanks to Theorem 32.4, we can get this corollary:

**Corollary 32.5.** Let $f \in \mathcal{A}^*(\mathbb{K})$ be not identically zero of order $t \in ]0, +\infty[$. Given $\epsilon > 0$, there exists a sequence of intervals $[r'_n, r''_n]$, with $\lim_{n \to +\infty} r'_n = +\infty$, such that $|f'|(r) \ge |f|(r)e^{-(er^t)}$ $\forall r \in \bigcup_{n \in \mathbb{N}} [r'_n, r''_n]$.

By Theorems 31.5 and 32.4, we can now derive Corollary 32.6:

**Corollary 32.6.** Let $f \in \mathcal{A}^*(\mathbb{K})$ be not identically zero of order $t \in ]0, +\infty[$. If $f$ is clean, then $\psi(f) \le \psi(f')$, and if $f'$ is clean, then $\psi(f) \ge \psi(f')$.

**Corollary 32.7.** Let $f \in \mathcal{A}^*(\mathbb{K})$ be not identically zero of order $t \in ]0, +\infty[$. If $f$ and $f'$ are clean, then $\psi(f) = \psi(f')$.
**Remark.** More generally, if $\psi(f) = \rho(f)\sigma(f)$ and $\psi(f') = \rho(f')\sigma(f')$, then $\psi(f') = \psi(f)$.

By Theorem 31.5 and Corollary 32.7, the following conjecture is natural and actually would just be an immediate consequence of the conjecture following Corollary 31.6.

**Conjecture.** For every $f \in \mathcal{A}(\mathbb{K})$ such that $0 < \rho(f) < +\infty$, we have $\psi(f') = \psi(f)$.

Applying the above remark, in residue characteristic 0, we have Theorem 32.8:

**Theorem 32.8.** Suppose that $\mathbb{K}$ has residue characteristic 0. Then for every $f \in \mathcal{A}(\mathbb{K})$ such that $0 < \rho(f) < +\infty$, we have $\psi(f') = \psi(f)$.

**Proof.** Indeed, we can check that for every $r > 0$, $s(r, f') = s(r, f) - 1$, hence $\psi(f') = \psi(f)$. $\square$

Of course, polynomials have a growth order 0. On $\mathbb{K}$ as on $\mathbb{C}$, we can easily construct transcendental entire functions of order 0 or of order $\infty$.

**Example 1.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $-\text{Log}|a_n| \in [n(\text{Log}(n))^2, n(\text{Log}(n))^2 + 1]$. Then clearly, $\lim_{n \to +\infty} \frac{\text{Log}|a_n|}{n} = -\infty$, hence the function $\sum_{n=0}^{\infty} a_n x^n$ has a radius of convergence equal to $+\infty$. On the other hand, $\lim_{n \to +\infty} \frac{n\text{Log}(n)}{-\text{Log}|a_n|} = 0$ hence $\rho(f) = 0$.

**Example 2.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $-\text{Log}|a_n| \in [n\sqrt{\text{Log}(n)}, n\sqrt{\text{Log}(n)}+1]$. Then, $\lim_{n \to +\infty} \frac{\text{Log}|a_n|}{n} = -\infty$ again, and hence the function $\sum_{n=0}^{\infty} a_n x^n$ has a radius of convergence equal to $+\infty$. On the other hand, $\lim_{n \to +\infty} \left( \frac{n\text{Log}(n)}{-\text{Log}|a_n|} \right) = +\infty$ hence $\rho(f) = +\infty$.

Similarly, comparing the number of zeros of $f'$ to this of $f$ inside a disk is very uneasy. Now, we can give some precisions thanks to Theorems 30.9 and 32.1.

**Theorem 32.9.** Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental and of order $t \in ]0, +\infty[$. Then for every $\epsilon > 0$,

$$ \limsup_{r \to +\infty} \left( \frac{r^\epsilon s(r, g)}{s(r, f)} \right) = +\infty. $$
**Proof.** Suppose first $t = 0$. The proof then is almost trivial. Indeed, for all $\epsilon > 0$, we have $\lim_{r \to +\infty} \frac{s(r,f)}{r^\epsilon} = 0$, hence $\lim_{r \to +\infty} \frac{r^\epsilon}{s(r,f)} = +\infty$, therefore $\lim_{r \to +\infty} \frac{r^\epsilon s(r,g)}{s(r,f)} = +\infty$.

Now, suppose $t > 0$. By Theorem 30.8, there exists $\lambda > 0$ such that

$$ (1) \qquad s(r, f) \le \lambda r^t \quad \forall r > 1. $$

Let us fix $s \in ]0, t[$. By hypothesis, $\rho(g) = \rho(f)$, and hence by Theorem 30.8, we have $\limsup_{r \to +\infty} \frac{s(r,g)}{r^s} = +\infty$, so there exists an increasing sequence $(r_n)_{n \in \mathbb{N}}$ of $\mathbb{R}_+$ such that $\lim_{n \to +\infty} r_n = +\infty$ and $\frac{s(r_n,g)}{(r_n)^s} \ge n$. Therefore, by (1), we have

$$ \frac{\lambda(r_n)^t s(r_n, g)}{(r_n)^s s(r_n, f)} > \frac{s(r_n, g)}{(r_n)^s} > n, $$

and hence

$$ \lambda \lim_{n \to +\infty} \left( \frac{(r_n)^{t-s} s(r_n, g)}{s(r_n, f)} \right) = +\infty. $$

Consequently,

$$ (2) \qquad \limsup_{r \to +\infty} \left( \frac{(r)^{t-s} s(r, g)}{s(r, f)} \right) = +\infty. $$

Now, since that holds for all $s \in ]0, t[$, the statement comes from (2).

□

Finally, by Theorem 32.1, we can derive Corollary 32.10:

**Corollary 32.10.** Let $f \in \mathcal{A}(\mathbb{K})$ be transcendental and of order $t \in ]0, +\infty[$. Then for every $\epsilon > 0$, we have $\limsup_{r \to +\infty} \left( \frac{r^\epsilon s(r, f')}{s(r, f)} \right) = +\infty$ and $\limsup_{r \to +\infty} \left( \frac{r^\epsilon s(r, f)}{s(r, f')} \right) = +\infty$.

The conjecture $\psi(f') = \psi(f)$ in the general context then seems natural, as suggested in [16] and [18], but a proof does not seem easy. However, by Theorems 31.5, 32.1, and 32.4, we can write Corollary 32.11:

**Corollary 32.11.** Suppose that $\mathbb{K}$ is of characteristic 0. Let $f \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$. Then,

$$ |\psi(f) - \psi(f')|_{\infty} \le \rho(f)[(e-1)\sigma(f) - \tilde{\sigma}(f)]. $$
# Chapter 33

## Growth of an Analytic Function in an Open Disk

In Chapters 30, 31, and 32, we defined the order of growth and the type of growth for entire functions in $\mathbb{K}$ in a similar way as it is done for complex entire functions and we also defined a cotype of growth strongly linked to the order and the type: in most of the cases, the cotype is the product of the order of growth by the type of growth.

Here we consider analytic functions in an “open” disk $d(a, R^{-})$ that we denote by $E$ throughout this chapter.

**Notations and definitions.** Let $f = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(E)$. In order to define a growth order similarly as it was done in the algebra of entire functions in $\mathbb{K}$, we can define in $\mathcal{A}(E)$ a growth order in the following way: given $r \in ]0, R[$, as it was done in complex analysis, given an unbounded function $f \in \mathcal{A}(E)$, when $r$ is close enough to $R$, we put $\rho(f, r) = \frac{\text{Log}(\text{Log}(|f(r)|))}{-\text{Log}(R-r)}$ and $\rho(f) = \lim \sup_{r \to R^{-}} \rho(f, r)$, hence $\rho(f) = \lim \sup_{r \to R^{-}} \frac{\text{Log}(\text{Log}(|f(r)|))}{-\text{Log}(R-r)}$. Then $\rho(f)$ is called the order of growth of $f$ [20].

On the other hand, for every $r \in ]0, R[$, if the set of the $u > 0$ such that $\lim_{r \to R^{-}} s(r, f)(R-r)^u = 0$ is empty, we put $\theta(f) = +\infty$. Else, we then denote by $\theta(f)$ the lowest bound of the $u > 0$ such that $\lim_{r \to R^{-}} s(r, f)(R-r)^u = 0$. Similarly, if the set of $u > 0$ such that $\lim_{r \to R^{-}} \text{Log}(|f(r)|)(R-r)^u = 0$ is empty, we put $\lambda(f) = +\infty$. Else, we denote by $\lambda(f)$ the lowest bound of $u > 0$ such that $\lim_{r \to R^{-}} \text{Log}(|f(r)|)(R-r)^u = 0$. And if $0 < \rho(f) < +\infty$, we put $\sigma(f, r) = \text{Log}(|f(r)|)(R-r)^\rho(f)$, $\sigma(f) = \lim \sup_{r \to R^{-}} \sigma(f, r)$,
$\psi(f, r) = s(r, f)(R - r)^{\rho(f)}$, and $\psi(f) = \limsup_{r \to R^{-}} \psi(f, r)$. We call $\sigma(f)$ the type of growth of f and $\psi(f)$ the cotype of growth of f.

Let us recall that, as far as ultrametric entire functions are concerned, the order of growth is equal to the lowest bound of $s > 0$ such that $\lim_{r \to +\infty} \frac{\log(|f(r)|)}{r^s} = 0$ and to the lowest bound of $s > 0$ such that $\lim_{r \to +\infty} \frac{s(r, f)}{r^s} = 0$. Here we try to prove similar results. This chapter is aimed at showing relations between these expressions $\rho(f)$, $\sigma(f)$, $\psi(f)$.

**Notation.** We denote by $\mathcal{A}^*(E)$ the set of unbounded functions $f \in \mathcal{A}(E)$ such that $0 < \rho(f) < +\infty$.

Theorems 33.1 and 33.3 are easy and don't need any proof:

**Theorem 33.1.** Let $f, g \in \mathcal{A}^*(E)$. Then, $\rho(f+g) \le \max(\rho(f), \rho(g))$ and $\rho(fg) = \max(\rho(f), \rho(g))$.

**Corollary 33.2.** Let $f, g \in \mathcal{A}^*(E)$. Then, $\rho(f^n) = \rho(f) \quad \forall n \in \mathbb{N}^0$. If $\rho(f) > \rho(g)$, then $\rho(f + g) = \rho(f)$.

**Theorem 33.3.** Let $f \in \mathcal{A}^*(E)$ and let $P \in \mathbb{K}[x]$ be non-constant. Then, $\rho(P \circ f) = \rho(f)$.

**Theorem 33.4.** Let $f, g \in \mathcal{A}^*(E)$. Then, $\psi(fg) \le \psi(f) + \psi(g)$. Moreover, if $\rho(f) = \rho(g)$, then $\max(\psi(f), \psi(g)) \le \psi(fg)$.

**Proof.** Set $\rho(f) = u$, $\rho(g) = t$. Without loss of generality, we can assume $u \ge t$. By Theorem 30.1, we have $\rho(f.g) = \rho(f) = u$. Now, for each $r > 0$, we have $s(f.g, r) = s(r, f) + s(g, r)$, hence

$$ \begin{aligned} \psi(fg) &= \limsup_{r \to R^{-}} (s(r, f) + s(g, r))(R - r)^s \\ &\le \limsup_{r \to R^{-}} s(r, f)(R - r)^s + \limsup_{r \to R^{-}} s(g, r)(R - r)^t, \end{aligned} $$

hence $\psi(fg) \le \psi(f) + \psi(g)$. Now, suppose $u = t$. Then,

$$ \begin{aligned} \psi(fg) &= \limsup_{r \to R^{-}} (s(r, f) + q(g, r))(R - r)^s \\ &\ge \limsup_{r \to R^{-}} \max(s(r, f), s(g, r))(R - r)^s \\ &= \max(\psi(f), \psi(g)). \end{aligned} $$

□
**Remark 1.** Let $f \in A^*(E)$. If $u > \theta(f)$, then by definition, $\lim_{r \to R^{-}} s(r, f)(R-r)^u = 0$. But if $u < \theta(f)$, then $\limsup_{r \to R^{-}} s(r, f)(R-r)^u = +\infty$ because if $\limsup_{r \to R^{-}} s(r, f)(R-r)^u < +\infty$, we can find $u' \in ]s, \theta(f)]$ and then we can check that $\lim_{r \to R^{-}} s(r, f)(R-r)^{u'} = 0$, a contradiction.

Thanks to the classical inequality $|f'|(r) \le \frac{|f|(r)}{r}$, Theorem 33.5 is then immediate:

**Theorem 33.5.** Suppose $\mathbb{K}$ has characteristic 0. Let $f \in A^*(E)$. Then, $\rho(f') \le \rho(f)$.

**Remark 2.** In a field of characteristic $p \ne 0$, certain analytic functions have a null derivative. This is why we must suppose that $\mathbb{K}$ has characteristic 0 in all statement involving derivatives.

In complex analysis, many estimates were given concerning the growth order of solutions of linear differential equations. Here, by Corollary 33.2 and Theorem 33.5, we can immediately obtain Corollary 33.6:

**Corollary 33.6.** Suppose $\mathbb{K}$ has characteristic 0. Consider the differential equation

$$ (\mathcal{E}) \quad f^{(n)} + a_{n-1}(x)f^{(n-1)}(x) + \dots + a_0(x)f(x) = 0 $$

with $a_j \in A^*(E)$, $j = 0, \dots, n-1$ and $\rho(a_j) < \rho(a_0) \forall j = 1, \dots, n-1$. Then every non-trivial solution $f$ of $(\mathcal{E})$ satisfies $\rho(f) \ge \rho(a_0)$.

**Theorem 33.7.** Suppose $\mathbb{K}$ has residue characteristic 0. Then for every $f \in A^*(D)$, we have $\rho(f') = \rho(f)$, $\theta(f') = \theta(f)$, $\sigma(f') = \sigma(f)$ and $\psi(f') = \psi(f)$.

**Remark 3.** Theorem 33.7 does not hold in residue characteristic $p > 0$ because there exist functions $f \in A^*(D)$ such that $\rho(f) > 0$ and that $f'$ is bounded, as shown in the following example with $R = 1$: $g(x) = \sum_{m=0}^{\infty} \frac{x^{p^m}}{p^m}$. We can see that $g'(x) = \sum_{n=0}^{\infty} x^{p^{m-1}}$, hence $g'$ is bounded, and therefore $\rho(g') = 0$. However, consider the sequence $(r_m)_{m \in \mathbb{N}}$ defined as $r_m = 1 - \frac{1}{p^m}$. We can check that $|g|(r_m) \ge p^m(r_m)^{p^m}$, hence

$$ \mathrm{Log}(|g|(r_m)) \ge m + p^m \mathrm{Log}(r_m) = m + p^m \mathrm{Log} \left( 1 - \frac{1}{p^m} \right). $$
When $m$ is big enough, we have $\mathrm{Log}(1 - \frac{1}{p^m}) \ge \frac{2}{p^m}$, hence

$$ \mathrm{Log}(|g|(r_m)) \ge m - p^m \left(\frac{2}{p^m}\right) = m - 2. $$

Therefore, when $m$ is big enough, we have

$$ \frac{\mathrm{Log}(\mathrm{Log}(|g|(r_m)))}{-\mathrm{Log}(r_m)} \ge \frac{\mathrm{Log}(m-2)}{-\mathrm{Log}(1 - \frac{1}{p^m})} > \frac{\mathrm{Log}(m-2)}{\frac{2}{p^m}} = \frac{p^m}{2}\mathrm{Log}(m-2). $$

Thus, we have $\rho(g) = +\infty$.

**Remark 4.** Theorem 33.7 applies, for instance, to the complex Levi-Civita field whose residue characteristic is 0 [88].

**Theorem 33.8.** Let $f \in \mathcal{A}^*(E)$. Then, $\lambda(f) = \rho(f)$.

**Proof.** First, we prove that $\rho(f) \le \lambda(f)$. Obviously, we can assume that $\lambda(f) < +\infty$. Let $u$ be such that $\lim_{r \to R^{-}} \mathrm{Log}(|f|(r))(R-r)^u = 0$. Let us fix $\epsilon > 0$. For $r$ close enough to $R$, we have $\mathrm{Log}(|f|(r))(R-r)^u \le \epsilon$, hence $\mathrm{Log}(|f|(r)) \le \frac{\epsilon}{(R-r)^u}$, therefore $\mathrm{Log}(\mathrm{Log}(|f|(r))) \le \mathrm{Log}\epsilon - u\mathrm{Log}(R-r)$, hence

$$ \frac{\mathrm{Log}(\mathrm{Log}(|f|(r))))}{(-\mathrm{Log}(R-r))} \le \frac{\mathrm{Log}(\epsilon)}{(-\mathrm{Log}(R-r))} + u, $$

and hence

$$ \limsup_{r \to R^{-}} \frac{\mathrm{Log}(\mathrm{Log}(|f|(r)))}{(-\mathrm{Log}(R-r))} \le u $$

i.e. $\rho(f) \le u$. This is true for every $s$ such that $\lim_{r \to R^{-}} \mathrm{Log}(|f|(r))(R-r)^u = 0$ and hence $\rho(f) \le \lambda(f)$.

On the other hand, we note that, by definition of $\lambda(f)$, either $\lambda(f) = 0$ and then $\lambda(f) \le \rho(f)$, or

$$ \lambda(f) = \sup\{u \in ]0, +\infty[ \mid \limsup_{r \to R^{-}} \mathrm{Log}(|f|(r))(R-r)^u > 0\}. $$

Thus, suppose that $\lambda(f) > 0$. Let us take $u \in ]0, \lambda(f)][$. We have a number $b > 0$ such that

$$ \limsup_{r \to R^{-}} \mathrm{Log}(|f|(r)(R-r)^u) \ge b > 0. $$

Let us fix $\epsilon \in ]0, b[$. There exists a sequence $(r_n)_{n \in \mathbb{N}}$ in $]0, R[$ such that $\lim_{n \to +\infty} r_n = R$ and such that, when $n$ is big enough, we have
Growth of an Analytic Function in an Open Disk

315

$b - \epsilon \le \text{Log}(|f|(r_n))(R - r_n)^u$, hence $-u\text{Log}(R - r_n) + \text{Log}(b - \epsilon) < \text{Log}(\text{Log}(|f|(r_n)))$, therefore

$$u + \frac{\text{Log}(b - \epsilon)}{(-\text{Log}(R - r_n))} \le \frac{\text{Log}(\text{Log}(|f|(r_n)))}{(-\text{Log}(R - r_n))}.$$

Consequently, $\limsup_{n \to +\infty} \frac{\text{Log}(\text{Log}(|f|(r_n)))}{(-\text{Log}(R - r_n))} \ge u$, therefore $\rho(f) \ge u$. But this holds for every $u < \lambda(f)$. Thus, $\rho(f) \ge \lambda(f)$, and finally, $\rho(f) = \lambda(f)$. $\square$

**Theorem 33.9.** Let $f, g \in A^*(E)$. Then, $\sigma(fg) \le \sigma(f) + \sigma(g)$. If $\rho(f) \ge \rho(g)$, then $\sigma(f) \le \sigma(fg)$. If $\rho(f) = \rho(g)$, then $\max(\sigma(f), \sigma(g)) \le \sigma(fg)$.

If $\rho(f) = \rho(g)$ and $\sigma(f) > \sigma(g)$, then $\sigma(f+g) \ge \sigma(f)$. If $\rho(f+g) = \rho(f) \ge \rho(g)$, then $\sigma(f+g) \le \max(\sigma(f), \sigma(g))$.

**Proof.** Let $u = \rho(f)$, $t = \rho(g)$ and suppose $u \ge t$. When $r$ is close enough to $R$, we have $\max(\text{Log}(|f|(r)), \text{Log}(|g|(r)) \le \text{Log}(|f.g|(r)) = \text{Log}(|f|(r)) + \text{Log}(|g|(r))$, and by Theorem 1, we have $\rho(fg) = u$. Therefore,

$$\begin{aligned} \sigma(fg) &= \limsup_{r \to R^{-}} (\text{Log}(|f.g|(r))(R-r)^u) \\ &\le \limsup_{r \to R^{-}} (\text{Log}(|f|(r))(R-r)^u) + \limsup_{r \to R^{-}} (\text{Log}(|g|(r))(R-r)^t) \\ &= \sigma(f) + \sigma(g). \end{aligned}$$

On the other hand,

$$\sigma(f) = \limsup_{r \to R^{-}} \text{Log}(|f|(r))(R-r)^u \le \limsup_{r \to R^{-}} (\text{Log}(|fg|(r))(R-r)^u.$$

But $\rho(fg) = u$, hence $\sigma(f) \le \sigma(fg)$. Particularly, if $\rho(f) = \rho(g)$, then $\max(\sigma(f), \sigma(g)) \le \sigma(fg)$.

Now, suppose again that $\rho(f) = \rho(g) = u$ and suppose $\sigma(f) > \sigma(g)$. Let $u = \rho(f)$, $b = \sigma(f)$. Then, $b > 0$. Let $(r_n)_{n \in \mathbb{N}}$ be a sequence such that $\lim_{n \to +\infty} r_n = R$ and $\lim_{n \to +\infty}(\text{Log}(|f|(r_n))(R - r_n)^u) = b$. Since $\sigma(g) < \sigma(f)$, we note that when $n$ is big enough, we have $|g|(r_n) < |f|(r_n)$. Consequently,
when $n$ is big enough, we have $|f + g|(r_n) = |f|(r_n)$, and hence

$$ (1) \qquad \lim_{n \to +\infty} (\mathrm{Log}(|f+g|(r_n))(R-r_n)^u) = b. $$

By definition of $\sigma$, we have $\sigma(f+g) \ge \lim_{n \to +\infty}(\mathrm{Log}(|f+g|(r_n))(R-r_n)^\rho)$.

$$ \begin{aligned} \sigma(f+g) &\ge \lim_{n \to +\infty} (\mathrm{Log}(|f+g|(r_n))(R-r_n)^\rho) \\ &\ge \lim_{n \to +\infty} (\mathrm{Log}(|f+g|(r_n))(R-r_n)^u \\ &= \lim_{n \to +\infty} \mathrm{Log}(|f|(r_n))(R-r_n)^u = \sigma(f), \end{aligned} $$

therefore by (1), $\sigma(f+g) \ge \sigma(f)$.

Finally, suppose now that $\rho(f+g) = \rho(f) \ge \rho(g)$. Let $u = \rho(f)$ and $t = \rho(g)$. Then,

$$ \begin{aligned} \sigma(f+g) &= \limsup_{r \to R^{-}} (\mathrm{Log}(|f+g|(r))(R-r)^u \\ &\le \max \left( \limsup_{r \to R^{-}} (\mathrm{Log}(|f|(r)))(R-r)^u, \limsup_{r \to R^{-}} (\mathrm{Log}(|g|(r)))(R-r)^u \right) \\ &\le \max \left( \limsup_{r \to R^{-}} (\mathrm{Log}(|f|(r)))(R-r)^u, \limsup_{r \to R^{-}} (\mathrm{Log}(|g|(r)))(R-r)^t \right) \\ &= \max(\sigma(f), \sigma(g)) \end{aligned} $$

which ends the proof. □

**Corollary 33.10.** Let $f, g \in A^*(E)$ be such that $\rho(f) \ne \rho(g)$. Then,

$$ \sigma(f+g) \le \max(\sigma(f), \sigma(g)). $$

**Lemma 33.11.** Let $a \in [1, +\infty[$ and $b \in [0, +\infty[$. Then, $\mathrm{Log}(a+b) \le \mathrm{Log}(a) + \mathrm{Log}(b+1)$.

**Proof.** Indeed, since $a \ge 1$, we have $\mathrm{Log}(a+b) \le \mathrm{Log}(a(b+1)) = \mathrm{Log}(a) + \mathrm{Log}(b+1)$. □

**Theorem 33.12.** Let $f \in A^*(E)$. Then, $\theta(f) - 1 \le \rho(f) \le \theta(f)$. Moreover, if $0 < \psi(f) < \infty$, then $\rho(f) = \theta(f)$.
**Proof.** We denote by $|\cdot|_{\infty}$ the Archimedean absolute value of $\mathbb{R}$. Let us first choose $u > \theta(f)$. Then, $\lim_{r \to R^{-}} s(r, f)(R-r)^{u} = 0$. Now, since $\lim_{r \to R^{-}} |f|(r) = +\infty$, we can take $\ell \in [0, R]$ such that $|f|(\ell) > 1$. Then we can take $b > 0$ such that

$$s(r, f) \leq b(R - r)^{-u} \quad \forall r \in [\ell, R].$$

Now, taking $r \in [\ell, R[$, by Theorem 22.26, we have

$$\log(|f|(r)) \leq \log(|f|(\ell)) + s(r, f) \left(\log\left(\frac{r}{\ell}\right)\right)$$

which leads to

$$\log(|f|(r)) \leq \log(|f|(\ell)) + b(R - r)^{-u} \left(\log\left(\frac{r}{\ell}\right)\right),$$

hence

$$\log(\log(|f|(r))) \leq \log\left(\log(|f|(\ell)) + b(R - r)^{-u} \left(\log\left(\frac{r}{\ell}\right)\right)\right),$$

therefore, by Lemma 33.11, we can derive

(1)

$$\log(\log(|f|(r))) \leq \log(\log(|f|(\ell))) + \log(b(R-r)^{-u} \left(\log\left(\frac{r}{\ell}\right)\right) + 1).$$

Now, since $u > 0$, there obviously exists $h \in [\ell, R[$ such that $b(R - r)^{-u} \geq 1 \forall r \in [h, R^{-}]$, therefore by Lemma 33.11 again,

$$\log(\log(|f|(r))) \leq \log(\log(|f|(\ell))) + \log\left(b(R - r)^{-u}\left(\log\left(\frac{r}{\ell}\right)\right)\right) + \log(1 + 1)$$

i.e.

(2)

$$\log(\log(|f|(r))) \leq \log(\log(|f|(\ell))) + \log(b) - s\log(R - r) + \log(\left(\log(\frac{r}{\ell})\right) + \log(2))$$

Consequently, by (2), we obtain

$$\frac{\log(\log(|f|(r)))}{-\log(R-r)} \leq \frac{\log(\log(|f|(\ell)))}{-\log(R-r)} + \frac{\log(b)}{-\log(R-r)} + u + \frac{\log(\log(\frac{r}{\ell})) + \log(2)}{-\log(R-r)}.$$
We can check that

$$ \lim_{r \to R^{-}} \frac{\text{Log}(\text{Log}(|f|(\ell))) + \text{Log}(b)}{-\text{Log}(R-r)} = \lim_{r \to R^{-}} \frac{\text{Log}(\text{Log}(\frac{r}{\ell}) + \text{Log}(2))}{-\text{Log}(R-r)} = 0, $$

and hence $\limsup_{r \to R^{-}} \frac{\text{Log}(\text{Log}(|f|(r)))}{-\text{Log}(R-r)} \le u$. Consequently, choosing $\epsilon > 0$, there exists $u \in [\ell, 1[$ such that $\frac{\text{Log}(\text{Log}(|f|(r)))}{-\text{Log}(R-r)} \le u + \epsilon \forall r \in [u', R[$, and hence $\rho(f) \le u+\epsilon$. But since that holds for every $u > \theta(f)$ and for every $\epsilon > 0$, we have $\rho(f) \le u$ and hence $\rho(f) \le \theta(f)$.

Let us now show that $\rho(f) \ge \theta(f) - 1$. By Theorem 22.26, we have

$$ (3) \quad \begin{aligned} \text{Log}(|f|(r)) - \text{Log}(|f|(\frac{r^2}{R})) &\ge s(\frac{r^2}{R}, f)(\text{Log}(r) - \text{Log}(\frac{r^2}{R})) \\ &= (\frac{r^2}{R}, f)(\text{Log}(R) - \text{Log}(r)). \end{aligned} $$

Consider now a number $s < \theta(f)$ and a sequence $(r_n)_{n \in \mathbb{N}}$ of $]0, R[$ such that $\lim_{n \to +\infty} r_n = R$ and such that $\limsup_{n \to +\infty} s(r_n, f)$ $(R - r_n)^s \ge b > 0$. Then, by (3), we have

$$ \text{Log}(|f|(r_n)) \ge \frac{b(\text{Log}(R) - \text{Log}(r_n))}{(R - \frac{r_n^2}{R})^s}. $$

Consequently,

$$ \text{Log}(\text{Log}(|f|(r_n))) \ge \text{Log}(b) + \text{Log}(\text{Log}(R) - \text{Log}(r_n)) \\ - s(\text{Log}(R - r_n) + \text{Log}(R + r_n)) + 2u\text{Log}(R), $$

and therefore

$$ \frac{\text{Log}(\text{Log}(|f|(r_n)))}{-\text{Log}(R-r_n)} \ge \frac{\text{Log}(b)}{-\text{Log}(R-r_n)} + \frac{\text{Log}(\text{Log}(R) - \text{Log}(r_n))}{-\text{Log}(R-r_n)} \\ + u \left( 1 + \frac{\text{Log}(R+r_n) + 2\text{Log}(R)}{-\text{Log}(R-r_n)} \right). $$

Clearly,

$$ \lim_{n \to +\infty} \left( \frac{\text{Log}(b)}{\text{Log}(R - r_n)} \right) = \lim_{n \to +\infty} \frac{\text{Log}(R + r_n) + 2\text{Log}(R)}{\text{Log}(R - r_n)} = 0, $$
Growth of an Analytic Function in an Open Disk

319

and by elementary reasonings, we can check that

$$ \lim_{t \to R^{-}} \frac{\log(\log(R) - \log(t))}{\log(R - t)} = 1, $$

therefore

$$ \lim_{n \to +\infty} \frac{\log(\log(R) - \log(r_n))}{\log(R - r_n)} = 1. $$

Consequently,

$$ \limsup_{n \to +\infty} \frac{\log(\log(|f|(r_n)))}{-\log(R - r_n)} \geq u - 1, $$

and therefore

$$ \limsup_{r \to R^{-}} \frac{\log(\log(|f|(r))))}{-\log(R - r)} \geq u - 1. $$

That holds for every $u < \theta(f)$ and shows that if $\theta(f) < +\infty$, then $\rho(f) \geq \theta(f) - 1$. Next, if $\theta(f) = +\infty$, then we would have $\rho(f) = +\infty$, which is excluded by hypothesis since $f \in \mathcal{A}^*(E)$. Consequently, the inequality $\rho(f) \geq \theta(f) - 1$ is established.

Let us now show that $\rho(f) \geq \theta(f)$ when $\psi(f) < +\infty$. Suppose $\theta(f) > \rho(f)$ and let $u \in [\rho(f), \theta(f)]$. Then, by Remark 1, we have $\limsup_{r \to R^{-}} s(r, f)(R - r)^s = +\infty$, but then $\limsup_{r \to R^{-}} s(r, f)(R - r)^\rho(f) = +\infty$, i.e. $\psi(f) = +\infty$, a contradiction. Therefore, $\theta(f) \leq \rho(f)$, and hence whenever $\psi(f) < +\infty$, we have $\theta(f) = \rho(f)$. $\square$

Theorem 33.12 obviously suggests the following conjecture:

**Conjecture.** Let $f \in \mathcal{A}^*(E)$. Then, $\rho(f) = \theta(f)$.

Theorem 33.13 is much different from the relations concerning $\rho$, $\sigma$, $\psi$ obtained for entire functions.

**Theorem 33.13.** Let $f \in \mathcal{A}^*(E)$ be such that $\psi(f) < +\infty$. Then, $\sigma(f) = 0$.

**Proof.** Without loss of generality, we can assume that $f(0) \neq 0$. Let us fix $\epsilon > 0$ and let $R'$ be such that $\log(R) - \log(R') = \epsilon$. Let $(a_n)_{n \in \mathbb{N}}$ be the sequence of zeros of $f$, for each $n \in \mathbb{N}$, let $w_n$ be the
order of $a_n$, and let $r_n = |a_n|$. Now, let $u$ be the biggest integer $n$ such that $r_n < R'$ and for each $r > 0$, let $m(r)$ be the biggest integer $n$ such that $r_n \le r$.

Let $A_u = \sum_{n=0}^{u} w_n$ and let $B_u = \text{Log}(|f(0)|) + \sum_{n=0}^{u} w_n(\text{Log}(R') - \text{Log}(r_n))$. Let us take $r \in ]R', R[$. Now, we can write

$$ \frac{\sigma(f, r)}{\psi(f, r)} = \frac{B_u + \sum_{n=u+1}^{m(r)} w_n(\text{Log}(r) - \text{Log}(r_n))}{A_u + \sum_{n=u+1}^{m(r)} w_n} $$

But by hypothesis, $\text{Log}(r) - \text{Log}(r_n) \le \epsilon \ \forall n \ge u$, hence

$$ \sigma(f, r)\text{over}\psi(f, r) \le \frac{B_u + \epsilon \sum_{n=u+1}^{m(r)} w_n}{A_u + \sum_{n=u+1}^{m(r)} w_n} $$

Let us put $\phi(r) = \sum_{n=u+1}^{m(r)} w_n$. Thus,

$$ \frac{\sigma(f, r)}{\psi(f, r)} \le \frac{B_u + \epsilon\phi(r)}{A_u + \phi(r)} $$

But since $f$ belongs to $A^*(D)$, it has infinitely many zeros in $D$, hence $\phi(r)$ is an increasing unbounded function tending to $+\infty$ when $r$ tends to $R$. Consequently, it is obvious that

$$ \lim_{r \to R} \sigma(f, r)\text{over}\psi(f), r = 0. $$

Therefore, if $\limsup_{r \to R^-} \psi(f, r) < +\infty$, then $\sigma(f) = 0$. $\square$

<figure><img src="image_12.png" /></figure>
# Chapter 34

## The Set Mult($\mathcal{A}_b(d(0, R^{-}))$

Throughout this chapter, we fix $R \in ]0, +\infty[$, we denote by $D$ the disk $d(0, R^{-})$ and by $A$ the $\mathbb{K}$-algebra $\mathcal{A}_b(d(0, R^{-}))$.

In this chapter, we examine the continuous multiplicative norms and semi-norms on $A$. Following a work by J. Araujo [6], we show that there exist continuous multiplicative semi-norms whose kernel is a prime closed ideal that is neither null nor a maximal ideal. Theorem 34.1 is easily checked.

**Theorem 34.1.** Each circular filter of diameter $r \in ]0, R[$ secant with the disk $d(0, R^{-})$ defines a continuous multiplicative norm on $A$. The norm $\| \cdot \|_D$ is a multiplicative norm on $A$. Each ultrafilter $\mathcal{U}$ on $D$ defines an element $\varphi_{\mathcal{U}} \in \text{Mult}(A, \| \cdot \|)$ as $\varphi_{\mathcal{U}}(f) = \lim_{\mathcal{U}} |f(x)|$.

**Proof.** Let $\mathcal{F}$ be a circular filter of diameter $r \in ]0, R[$, secant with $D$. There exists a disk $D'$ of diameter $r' \in [r, R[$ such that $\mathcal{F}$ is secant with $D'$. And then, by Proposition 14.12, $A$ is included in $H(D')$. Consequently, $\varphi_{0,r'}$ belongs to $\text{Mult}(H(D'), \| \cdot \|_D)$ and therefore belongs to $\text{Mult}(A, \| \cdot \|)$. On the other hand, by Theorem 14.20, $\| \cdot \|_D$ is a multiplicative norm defined on $A$. $\square$

Thus, given an ultrafilter $\mathcal{U}$ on $D$, the function $|f(x)|$ from $D$ to $[0, \|f\|]$ has a limit $\varphi_{\mathcal{U}}(f)$ which clearly defines an element of $\text{Mult}(A, \| \cdot \|)$.

**Definition.** Two filters $\mathcal{F}$ and $\mathcal{G}$ on a subset $S$ of $\mathbb{K}$ are said to be contiguous if for every $F \in \mathcal{F}$ and $G \in \mathcal{G}$ we have $\delta(F, G) = 0$.
Proposition 34.2 is immediate due to the uniform continuity of functions in A:

**Proposition 34.2.** Let $\mathcal{U}$ and $\mathcal{V}$ be two contiguous ultrafilters on $D$. Then, $\varphi_{\mathcal{U}} = \varphi_{\mathcal{V}}$.

**Proof.** Let $f \in A$, and let $l = \varphi_{\mathcal{U}}|(f)$ and $m = \varphi_{\mathcal{V}}(f)$. Let $\epsilon > 0$. Since $f$ is uniformly continuous, we can find $\eta > 0$ such that $|f(x) - f(y)| \le \epsilon \ \forall x, y \in D$ such that $|x - y| \le \eta$. Let us take $F \in \mathcal{U}$ and $G \in \mathcal{V}$ such that $|f(x) - l|_{\infty} \le \epsilon \ \forall x \in F$ and $|f(x)| - m|_{\infty} \le \epsilon \ \forall y \in G$. But since $\mathcal{U}$ and $\mathcal{V}$ are contiguous, there exists $x \in F$ and $y \in G$ such that $|x - y| \le \eta$, hence $|l - m|_{\infty} \le 3\epsilon$. That holds for every $\epsilon > 0$, hence $l = m$. $\square$

**Remark.** It seems very hard to know whether two ultrafilters $\mathcal{U}$ and $\mathcal{V}$ on $D$ such that $\varphi_{\mathcal{U}} = \varphi_{\mathcal{V}}$ are contiguous.

**Definition.** Similar to the definition for analytic elements, an element $f \in A$ is said to be *quasi-invertible* if it factorizes in the form $Pg$ with $P \in \mathbb{K}[x]$ and $g$ an invertible element of $A$.

By Corollary 14.16 and Theorem 14.19, we have the following obvious lemma:

**Lemma 34.3.** An element $f \in A$ is quasi-invertible if and only if it has finitely many zeros.

**Notation.** Let $\mathcal{Y}$ be the filter on $d(0, R^{-})$ admitting for basis the family of sets of the form $d(0, R^{-}) \setminus \bigcup_{n=1}^{\infty} (\bigcup_{j=1}^{q_n} d(a_{n,j}, r_n^{-}))$ with $|a_{j,n}| = r_n < r_{n+1} < R \ \forall n \in \mathbb{N}$ and $\lim_{n \to \infty} r_n = R$.

**Theorem 34.4.** For every $f \in A$, $\|f\|_D = \lim_{y} |f(x)|$.

**Proof.** Let $f \in A$, let $(C(0, r_n))_{n \in \mathbb{N}}$ be the sequence of circles of center 0 containing zeros of $f$ and for every $n \in \mathbb{N}$, and let $a_{n,1}, \dots, a_{n,q_n}$ be the zeros of $f$ in $C(0, r_n)$. Then we can see that $|f(x)| = |f|(|x|) \ \forall x \notin d(0, R^{-}) \setminus \bigcup_{n=1}^{\infty} (\bigcup_{j=1}^{q_n} d(a_{n,j}, r_n^{-}))$. Next, by Theorem 14.20, we have $\|f\|_D = \lim_{r \to R} |f|(r)$ which ends the proof. $\square$

**Theorem 34.5.** Let $\phi \in \text{Mult}(A, \|\cdot\|_D)$ and assume that its restriction $\varphi_F$ to $H(D)$ is not $\|\cdot\|_D$. Then, $\phi(f) = \lim_F |f(x)| \ \forall f \in A$.
**Proof.** By hypothesis, $\mathcal{F}$ is a circular filter of diameter $l < R$. Suppose first $l = 0$. Then $\mathcal{F}$ is the filter of neighborhoods of a point $a \in D$. Let $r \in ]0, R[$. Then $f$ belongs to $H(d(a, r))$ and we show that $\phi(h) \le \varphi_{\mathcal{F}}(h) \forall h \in A$.

Suppose first $f$ is invertible. Since $f$ has no zero in $D$, by Theorem 22.16, $|f(x)|$ is a constant $b > 0$. Consequently, $\|f\|_D = b = \varphi_{\mathcal{F}}(f)$. Suppose $\phi(f) \ne b$. Then $\phi(f) < b$ because $b = \|f\|_D$. Now consider $h = \frac{1}{f}$. Since $\| \cdot \|_D$ is multiplicative, we see that $\phi(h) > \|h\|_D$, a contradiction. Consequently, $\phi(f) = \varphi_{\mathcal{F}}(f)$.

Suppose now $f$ is quasi-invertible. Then $f$ is of the form $Pg$ with $P \in \mathbb{K}[x]$ and $g$ invertible in $A$. Then, $\phi(f) = \phi(P)\phi(g) = \varphi_{\mathcal{F}}(P)\varphi_{\mathcal{F}}(g) = \varphi_{\mathcal{F}}(f)$.

We now suppose that $f$ is not quasi-invertible. By Lemma 34.3, $f$ has a sequence of zeros $(a_n)_{n \in \mathbb{N}}$ in $D$, each having a multiplicity order $u_n$. By Corollary 22.19, we have $\lim_{n \to +\infty} |a_n| = R$, so we can assume $|a_n| \le |a_{n+1}| \forall n \in \mathbb{N}$. Let $t = \phi(f)$ and $s = \lim_{\mathcal{F}} |f(x)|$.

First, we show that $t \le s$.

Suppose first that $\mathcal{F}$ is secant with a disk $d(a, r)$ which contains none of the $a_n$. By Corollary 22.29, we have $\frac{\|f\|_{d(a,r)}}{\|f\|_D} = \prod_{n=0}^{\infty} \left(\frac{|a_n - a|}{R}\right)^{u_n}$, hence inside the disk $d(a, r)$, $|f(x)|$ is a constant equal to $\|f\|_D \prod_{n=0}^{\infty} \left(\frac{|a_n - a|}{R}\right)^{u_n}$, and therefore,

$$ (1) \qquad s = \|f\|_D \prod_{n=0}^{\infty} \left( \frac{|a_n - a|}{R} \right)^{u_n}. $$

For each $q \in \mathbb{N}$, let $f_q = \frac{f}{\prod_{n=0}^{q} (x - a_n)^{u_n}}$ and let $l_q = \sum_{k=0}^{q} u_k$. So, clearly, $\|f_q\|_D = \frac{\|f\|_D}{R^{l_q}} \forall q \in \mathbb{N}$.

Now, since $\phi(P) = \varphi_{\mathcal{F}}(P) \forall P \in \mathbb{K}[x]$, we have $\phi(\prod_{n=0}^{q} (x - a_n)^{u_n}) = \prod_{n=0}^{q} |a_n - a|^{u_n}$, hence $\phi(f_q) = \frac{t}{\prod_{n=0}^{q} |a_n - a|^{u_n}}$. But since $\phi(f_q) \le \|f_q\|_D$, that yields

$$ \frac{t}{\prod_{n=0}^{q} |a_n - a|^{u_n}} \le \|f_q\|_D = \frac{\|f\|_D}{R^{l_q}} \quad \forall q \in \mathbb{N}, $$

hence

$$ \frac{t R^{l_q}}{\prod_{n=0}^{q} |a_n - a|^{u_n}} \le \|f\|_D \quad \forall q \in \mathbb{N}. $$
Since this is true for every $q \in \mathbb{N}$, we can derive

$$t \prod_{n=0}^{\infty} \left( \frac{R^{u_n}}{|a_n - a|^{u_n}} \right) \le \|f\|_D,$$

hence by (1), $\frac{t\|f\|_D}{s} \le \|f\|_D$ and therefore $t \le s$.

Now, consider the case when there exists no disk $d(a, r)$ such that none of the $a_n$ lie in $d(a, r)$ and such that $\mathcal{F}$ is secant with that disk. Since $\lim_{n \to \infty} |a_n| = R$, $\mathcal{F}$ is a filter admitting a center $\alpha$. Let $\rho$ be its diameter: of course $\rho < R$ because $\varphi_{\mathcal{F}}$ is not $\|\cdot\|_D$. Consequently, $d(\alpha, \rho)$ contains finitely many zeros of $f$. Suppose $|a_j - \alpha| < \rho$ whenever $j = 0, \dots, q$. Let $h = \frac{f}{\prod_{j=0}^q (x-a_j)^{u_j}}$. Then, $\varphi_{\mathcal{F}}(h) = \frac{s}{\prod_{j=0}^q |a_j - \alpha|^{u_j}}$ and $\phi(h) = \frac{t}{\prod_{j=0}^q |a_j - \alpha|^{u_j}}$. Thus, we are led to the same problem with $h$. Setting $s' = \frac{s}{\prod_{j=0}^q |a_j - \alpha|^{u_j}}$, $t' = \frac{t}{\prod_{j=0}^q |a_j - \alpha|^{u_j}}$, we have $t' \le s'$, hence $t \le s$ in all cases and therefore we have proven

$$(2) \qquad \phi(h) \le \varphi_{\mathcal{F}}(h) \quad \forall h \in A.$$

Suppose now that for some $f \in A$, we have $\phi(f) < \varphi_{\mathcal{F}}(f)$. Let $r \in ]l, R[$. The filter $\mathcal{F}$ is then secant with a unique disk $d(a, r)$. Without loss of generality, we can suppose that $0$ belongs to $d(a, r)$ and then we can take $a = 0$. Let $f(x) = \sum_{n=0}^{\infty} b_n x^n$. For every $q \in \mathbb{N}$, let $g_q(x) = \sum_{n=0}^q b_n x^n$. We note that when $q$ is big enough we have $\varphi_{\mathcal{F}}(g_q) = \sup_{n \in \mathbb{N}} |b_n| r^n$. Set $w = \sup_{n \in \mathbb{N}} |b_n| r^n$. Now $\varphi_{\mathcal{F}}(f - g_q) \le \sup_{n > q} |b_n| r^n$, therefore $\lim_{q \to +\infty} \varphi_{\mathcal{F}}(f - g_q) = 0$, and hence, by (2),

$$(3) \qquad \lim_{q \to +\infty} \phi(f - g_q) = 0.$$

So, we can take $q$ such that $\varphi_{\mathcal{F}}(f - g_q) < \varphi_{\mathcal{F}}(f)$, and hence, by (2), $\phi(f - g_q) < \varphi_{\mathcal{F}}(f)$. But since $g_q$ is a polynomial, we have $\phi(g_q) = \varphi_{\mathcal{F}}(g_q)$, hence $\phi(g_q) > \phi(f)$. Consequently, $\phi(f - g_q) = \phi(g_q) = w$ when $q$ is big enough, a contradiction to (3). $\square$

By Corollary 22.29, we can derive Lemma 34.6:

**Lemma 34.6.** Let $f \in A$ be not quasi-invertible such that $f(0) = 1$ and let $(a_n)_{n \in \mathbb{N}}$ be the sequence of zeros with respective multiplicity $q_n$. Then the series $\sum_{n=0}^{\infty} q_n \log \left( \frac{R}{|a_n|} \right)$ converges to $\log \|f\|_D$.
**Lemma 34.7.** Let $\phi \in \mathrm{Mult}(A, \| \cdot \|_D)$ satisfy $\phi(P) = \|P\|_D \forall P \in \mathbb{K}[x]$. Every quasi-invertible element $f \in A$ also satisfies $\phi(f) = \|f\|_D$.

**Proof.** First, suppose $f \in A$, invertible in $A$. Then, $1 = \phi(f)\phi(f^{-1})$. But $\phi(f) \le \|f\|_D$, $\phi(f^{-1}) \le \|f^{-1}\|_D$, hence both inequalities must be equalities. Now, take $f = Pg \in A$ which is quasi-invertible, with $P \in \mathbb{K}[x]$ and $g \in A$, invertible in $A$. Then, $\phi(f) = \phi(P)\phi(g) = \|P\|_D\|g\|_D = \|Pg\|_D = \|f\|_D$. $\square$

**Theorem 34.8.** Let $\phi \in \mathrm{Mult}(A, \| \cdot \|_D)$ be a norm and have a restriction to $\mathbb{K}[x]$ equal to $\| \cdot \|_D$. Then, $\phi(f) = \|f\|_D \forall f \in A$.

**Proof.** Suppose the claim is wrong. Suppose that there exists $f \in A$ such that $\phi(f) \ne \|f\|_D$. By Lemma 10.5, we have $\phi(f) < \|f\|_D$. Actually, without loss of generality, we can choose $f \in A$ such that $\phi(f) < 1 \le \|f\|_D$ and that $f(0) \ne 0$. By Lemma 34.7, $f$ is not quasi-invertible, hence $f$ has a sequence of zeros $(a_n)_{n \in \mathbb{N}}$ in $D$, with $|a_n| \le |a_{n+1}|$ and $\lim_{n \to +\infty} |a_n| = R$. For each $n \in \mathbb{N}$, let $q_n$ be the multiplicity order of $a_n$. By Lemma 34.6, we know that $\sum_{n=0}^{\infty} q_n \log \left( \frac{R}{|a_n|} \right) < +\infty$. Consequently, by Lemma 34.6, there exists a sequence $t_n$ of strictly positive integers satisfying $t_n \le t_{n+1}$, $n \in \mathbb{N}$,

$$ \lim_{n \to \infty} t_n = +\infty, $$

$$ \sum_{n=0}^{\infty} t_n q_n \log \left( \frac{R}{|a_n|} \right) < +\infty. $$

By Theorem 27.14, there exists a function $g \in \mathcal{A}(d(0, R^-))$ admitting each $a_n$ as a zero of order $s_n \ge t_n q_n$ such that $g(0) = f(0)$ and $|g|(|a_n|) \le 2 \left| \prod_{k=0}^{n} \left( \frac{a_n}{a_k} \right)^{t_n q_n} \right| \forall n \in \mathbb{N}$. Consequently, $\lim_{n \to +\infty} |g|(|a_n|) < +\infty$, and hence $g$ belongs to $A$.

Now, for each $n \in \mathbb{N}$ and for each $k = 0, \dots, n$, set $u_{n,k} = \max(0, t_n q_k - s_k)$, and let $P_n(x) = \prod_{k=0}^{n} \left(\frac{x}{a_k} - 1\right)^{u_{n,k}}$. Since the sequence $t_n$ is increasing, we can check that for each fixed $n \in \mathbb{N}$, each zero $a_k$ of $f^{t_n}$ is a zero of $P_n g$ of order $\ge t_n q_k$: indeed, this is obvious, by definition of $g$, for all $k \ge n$ and when $k < n$, the order of $a_k$ as a zero of $P_n g$ is at least $u_{n,k} + s_k = \max(s_k, t_n q_k)$. Consequently, by Theorem 27.1, in the ring $A$, we can write $P_n g$ in the form $f^{t_n} \sigma_n$, with $\sigma_n \in A$.
We have $\|\sigma_n\|_D \|f^{t_n}\|_D = \|P_n\|_D \|g\|_D$, hence, since $\|f\|_D \ge 1$, we
can see that

$$
(1) \qquad \|\sigma_n\|_D \le \|P_n\|_D \|g\|_D.
$$

On the other hand, since the restriction of $\phi$ to $\mathbb{K}[x]$ is $\|\cdot\|_D$, we
have $\phi(P_n) = \|P_n\|_D$, hence $\phi(P_n g) = \phi(P_n)\phi(g) = \|P_n\|_D\phi(g)$, and
therefore, by (1), we have

$$
\begin{align*}
\|P_n\|_D \phi(g) &= \phi(P_n g) = \phi(f^{t_n} \sigma_n) = (\phi(f))^{t_n} \phi(\sigma_n) \le (\phi(f))^{t_n} \|\sigma_n\|_D \\
&\le (\phi(f))^{t_n} \|P_n\|_D \|g\|_D,
\end{align*}
$$

and hence $\phi(g) \le (\phi(f))^{t_n} \|g\|_D$. Now, since $\phi(f) < 1$, we can see
that $g=0$, a contradiction. That ends the proof of Theorem 34.8. $\square$

Let us now look at the maximal spectrum of A.

**Theorem 34.9.** Let $\mathcal{M}$ be an ideal of $A$. Then $\mathcal{M}$ is a maximal ideal of codimension 1 if and only if it is of the form $(x-a)A$ with $a \in D$.

**Proof.** Given $a \in D$, the ideal $(x-a)A$ is obviously a maximal ideal of codimension 1. because the mapping $\chi_a$ from $A$ to $\mathbb{K}$ defined as $\chi_a(f) = f(a)$ maps $A$ onto $\mathbb{K}$.

Now, let $\mathcal{M}$ be a maximal ideal of codimension 1 and let $\tau$ be the
$\mathbb{K}$-algebra homomorphism from $A$ onto $\mathbb{K}$ admitting $\mathcal{M}$ for kernel.
Let $a = \tau(x)$. Since $\tau(x-a) = \tau(x) - a$ is not invertible, we have
$|a| < 1$ because if $|a| \ge 1$, then by Theorem 22.20, $x-a$ is invertible
in $A$. Thus, $a$ belongs to $D$. By Corollary 10.10, we know that $\tau$
is continuous, hence $\tau(f) = f(a) \forall f \in A$. Consequently, $\text{Ker}(\tau) =$
$(x-a)A$.
$\square$

**Notation.** We denote by $\text{Mult}_1(A, \|\cdot\|)$ the set of multiplicative semi-norms of $A$ whose kernel is a maximal ideal of codimension 1.

By Theorems 34.5 and 34.9, we have the following corollary:

**Corollary 34.10.** Let $\phi \in \text{Mult}(A, \|\cdot\|) \setminus \text{Mult}_1(A, \|\cdot\|)$. If $\phi$ is not a norm, its restriction to $\mathbb{K}[x]$ is $\|\cdot\|_D$.

**Remark.** A admits maximal ideals of infinite codimension. The following theorem shows such examples.
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327

**Theorem 34.11.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of $D$ such that $\lim_{n \to +\infty} |a_n| = 1$ and $\prod_{n \in \mathbb{N}} |a_n| > 0$ and let $\mathcal{I}$ be the ideal of the $f \in A$ such that $\lim_{n \to +\infty} f(a_n) = 0$. Then $\mathcal{I}$ is not null and is included in a maximal ideal of infinite codimension.

**Proof.** Let $T$ be the divisor of $D$ defined by the sequence $(a_n)_{n \in \mathbb{N}}$. By Theorem 27.14, there exists an analytic function $f$ in $D$ such that $f(a_n) = 0 \forall n \in \mathbb{N}$ and such that $|f|(r) \le |T|(r) + 1 \forall r \in ]0, R[$. But since $\prod_{n \in \mathbb{N}} |a_n| > 0$, we can check that $|T|(r)$ is bounded in $]0, R[$ because

$$ \lim_{r \to R} |T|(r) = \frac{1}{\prod_{n \in \mathbb{N}} |a_n|} $$

Consequently, $f$ is bounded in $D$ and hence $f \in A$. Therefore, $\mathcal{I}$ is not null. But clearly, it is not included in any maximal ideal of the form $(x - a)A$. Consequently, it is included in a maximal ideal of infinite codimension. $\square$
Chapter 35

The Corona Problem on $\mathcal{A}_b(d(0, 1^{-}))$

The Corona conjecture, stated by Kakutani in 1941 in the field $\mathbb{C}$, was solved by L. Carlesson in 1962 [28]. Consider the open unit disk $O$ in $\mathbb{C}$ and the Banach algebra $B$ of bounded holomorphic functions in $O$. Each point of $O$ obviously defines a maximal ideal of $B$. On the other hand, all maximal ideals are of codimension 1 and the Gelfand transform defines a topology on the maximal spectrum. The question was whether the set of maximal ideals defined by points of $O$ was dense inside the whole spectrum of maximal ideals, with respect to the Gelfand topology.

Consider now the situation in our field $\mathbb{K}$. Let us keep the notation introduced in Chapter 34 but put $D = d(0, 1^{-})$ and $A = \mathcal{A}_b(d(0, 1^{-}))$. The maximal ideals of $A$ of codimension 1 are defined by the points of $D$, but looking for a Gelfand topology on the maximal spectrum makes no sense because of maximal ideals of infinite codimension. However, the only link existing between the various kind of maximal ideals comes from continuous multiplicative semi-norms. For simplicity, we denote $\| \cdot \|_A$ the Gauss norm on $A$. Here, we have the topology of pointwise convergence on $\text{Mult}(A, \| \cdot \|_A)$ and we can ask whether the set of continuous multiplicative semi-norms whose kernel is a maximal ideal of codimension 1 is dense in the whole set of continuous multiplicative semi-norms whose kernel is a maximal ideal, or even whether it is dense in the whole set of all continuous multiplicative semi-norms. Both questions are very hard. Here we solve the first with the use of the ultrametric holomorphic functional calculus [40], [45], [49], [53]. A partial answer is given in
[48] when the field is spherically complete. We have generalized it by using Banachic processes.

**Notation.** Recall that we denote by $U$ the disk $d(0,1)$. Let $\mathcal{H}$ be the family of ideals $J$ of $U[x]$ such that $J \cap U \neq \{0\}$, and, given an integer $s \in \mathbb{N}^*$, let $\mathcal{H}_s$ be the set of $J \in \mathcal{H}$ generated by $s$ elements. For every ideal $J \in \mathcal{H}$, we put $t(J) = \sup\{|x| \mid x \in J \cap U\}$ and $\ell(J) = \inf\{\sup_{f \in J} |f(x)| \mid x \in V\}$ and we denote by $u(J)$ the number such that $t(J) = \ell(J)^{u(J)}$. Finally, we put $m(s) = \sup\{u(J) \mid J \in \mathcal{J}_s\}$. Henceforth, given $f_1, \dots, f_s \in H(U)$ such that $\|f_i\| < 1 \ \forall i = 1, \dots, s$, we set $w(f_1, \dots, f_s) = \inf\{\max_{1 \le i \le s} |f_i(x)| \mid x \in U\}$. Moreover, given $f_1, \dots, f_s \in A$, we set $\lambda(f_1, \dots, f_s) = \inf\{\max_{1 \le i \le s} |f_i(x)| \mid x \in D\}$.

**Proposition 35.2** is also called Corona Statement in dimension 1. However, on a non-archimedean field, it is not at all proven to be equivalent to a property of density for maximal Proposition 35.1 is proved in [91] and is indispensable for further results.

**Proposition 35.1.** Let $J$ be a finitely generated ideal of $U[x]$ such that $J \cap U \neq \{0\}$. Then, $m(s) = 2 \ \forall s \geq 2$.

**Proposition 35.2.** Let $s \in \mathbb{N}^*$. For any $f_1, \dots, f_s \in A$ satisfying $\|f_i\| < 1 \ (1 \le i \le s)$ and $\lambda(f_1, \dots, f_s) > 0$, there exist $g_1, \dots, g_s \in A$ satisfying $\sum_{i=1}^s f_i g_i = 1$ and $\|g_i\| < \lambda(f_1, \dots, f_s))^{-2}$.

**Proof.** As the first step, we prove that for any $f_1, \dots, f_s \in H(U)$ satisfying $\|f_i\| < 1 \ \forall i = 1, \dots, s$ and $w(f_1, \dots, f_s) > 0$, there exist $g_1, \dots, g_s \in H(d(0, R))$ satisfying $\sum_{i=1}^s f_i g_i = 1$ and $\|g_i\| < (w(f_1, \dots, f_s))^{-2} \ \forall i = 1, \dots, s$.

Since $\mathbb{K}[x]$ is dense in $H(U)$, we can find polynomials $P_1, \dots, P_s \in \mathbb{K}[x]$ such that $\|P_i - f_i\| \le (w(f_1, \dots, f_s))^2 \ \forall i = 1, \dots, s$. Since $w(f_1, \dots, f_s) > 0$, by Corollary 24.13, there exists no maximal ideal $\mathcal{M}$ of $H(U)$ containing the ideal generated by $f_1, \dots, f_s$. Consequently, there exist $g_1, \dots, g_s \in H(U)$ such that $\sum_{i=1}^s g_i f_i = 1$. Therefore, we can define $h_1, \dots, h_s \in H(U)$ such that $\|h_i\| \le 1 \ \forall i = 1, \dots, s$ and such that $\sum_{i=1}^s h_i f_i$ is an element $P_0 \in U$. Let $I$ be the ideal of $U[x]$ generated by $P_0, P_1, \dots, P_s$. Since $I \cap U \neq \{0\}$, we have $0 < t(I) < \ell(I)$. Consequently, by Proposition 35.1, we can find $Q_0, Q_1, \dots, Q_s \in U[x]$ such that $\sum_{i=0}^s Q_i P_i$ is an element $a \in U$.
satisfying $|a| \ge (w(f_1, \dots, f_s))^{m(s)} = (w(f_1, \dots, f_s))^2$. Then,

$$ \sum_{i=1}^{s} a^{-1}f_i + a^{-1}Q_0 \sum_{i=1}^{s} h_i f_i = 1 + \sum_{i=1}^{s} a^{-1}Q_i(f_i - P_i). $$

By construction, we have $\|\sum_{i=1}^{s} a^{-1}Q_i(f_i - P_i)\| < 1$, and hence $1 + \sum_{i=1}^{s} a^{-1}Q_i(f_i - P_i)$ is a unit $u$ in $H(U)$. So, $\sum_{i=1}^{s} u^{-1}(a^{-1}(Q_i + Q_0h_i))f_i = 1$, and hence for every $i = 1, \dots, s$, we have $\|u^{-1}a^{-1}(Q_i + Q_0h_i)\| \le (w(f_1, \dots, f_s))^{-2} \forall i = 1, \dots, s$. Thus, putting $g_i = u^{-1}(a^{-1}(Q_i+Q_0h_i))$, $i = 1, \dots, s$, we have proven that for any $f_1, \dots, f_s \in H(U)$ satisfying $\|f_i\| < 1 \forall i = 1, \dots, s$ and $w(f_1, \dots, f_s) > 0$, there exist $g_1, \dots, g_s \in H(U)$ satisfying $\sum_{i=1}^{s} f_i g_i = 1$ and $\|g_i\| < (w(f_1, \dots, f_s))^{-2} \forall i = 1, \dots, s$.

Now, let us prove the conclusion of Proposition 35.2. So, we take $f_1, \dots, f_s \in A$. Let $(u_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $0 < |u_n| < |u_{n+1}| < \dots < 1$ and $\lim_{n \to +\infty} |u_n|^n = 1$. For each $i = 1, \dots, s$, set $f_{i,n}(x) = f_i(u_n x)$. Since each $u_n$ belongs to $D$, each $f_{i,n}$ is a power series of radius $r > 1$ and hence belongs to $H(U)$. Then by the claim we have just proven, for each $n \in \mathbb{N}$, there are $g_{i,n} \in H(U)$ such that $\sum_{i=1}^{s} g_{i,n} f_{i,n} = 1$ and $\|g_{i,n}\| < (w(f_{1,n}, \dots, f_{s,n}))^{-2}$. Now, each $g_{i,n}$ is a power series $\sum_{k=0}^{+\infty} g_{i,n,k} x^k$. Let $h_{i,n}$ be the power series $\sum_{k=0}^{2n} g_{i,n,k} (u_n)^{-k} x^k$. Then we have $\|h_{i,n}\| < |u_n|^{-2n} (w(f_1, \dots, f_s))^{-2}$ and hence $\sum_{i=1}^{s} h_{i,n} f_i$ is of the form $1 + x^n t_n(x)$ with $t_n \in H(U)$.

We get to a conclusion thanks to a Banach process. Let $\mathcal{E}$ be the Banach space of bounded sequences in $\mathbb{K}$ provided with the classic norm $\|(a_n)_{n \in \mathbb{N}}\|_0 = \sup_{n \in \mathbb{N}} |a_n|$ and let $\mathcal{C}$ be the closed sub-space of converging sequences. For every sequence $(a_n)_{n \in \mathbb{N}} \in \mathcal{C}$, we put $\mathcal{L}((a_n)) = \lim_{n \to +\infty} a_n$.

Suppose first that $\mathbb{K}$ is spherically complete, there exists a linear map of norm 1 from $\mathcal{E}$ to $\mathbb{K}$ expanding $\mathcal{L}$ to $\mathcal{E}$. Let $\tilde{\mathcal{L}}$ be this continuation of $\mathcal{L}$. Set $h_{i,n} = \sum_{k=0}^{+\infty} x^k$ and let $l_{i,k}$ be the sequence $(h_{i,n,k})_{1 \le n}$. For each pair $(i, k)$, we can now define $g_{i,k} = \tilde{\mathcal{L}}(l_{i,k})$ and put $g_i = \sum_{k=0}^{+\infty} g_{i,k} x^k$. Since $\tilde{\mathcal{L}}$ is of norm 1, each sequence $g_{i,k}$ is bounded and hence $g_i$ belongs to $A$. Moreover, by construction, $g_i$ satisfies $\|g_i\| < (\lambda(f_1, \dots, f_s))^{-2} \forall i = 1, \dots, s$. Now, we have

$$ (1) \quad \sum_{i=1}^{s} g_i f_i = \sum_{k=0}^{+\infty} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} \tilde{\mathcal{L}}(l_{i,j}) f_{i,k-j} \right) \right) x^k $$
and for each fixed $k \in \mathbb{N}$,

$$ \sum_{i=1}^{s} \left( \sum_{j=0}^{k} \tilde{\mathcal{L}}(l_{i,j}) f_{i,k-j} \right) = \tilde{\mathcal{L}} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right). $$

Consequently, since

$$ \sum_{i=1}^{s} h_{i,n} f_i = 1 + x^n t_n(x), $$

we can check that

$$ \lim_{n \to +\infty} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right) = 1 \text{ whenever } k = 0 $$

and

$$ \lim_{n \to +\infty} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right) = 0 \text{ whenever } k \neq 0. $$

Consequently, since $\tilde{\mathcal{L}}$ extends to $\mathcal{E}$ the limit on $\mathcal{C}$, we have

$$ \tilde{\mathcal{L}} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right) = 1 \text{ whenever } k = 0 $$

and

$$ \tilde{\mathcal{L}} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right) = 0 \text{ whenever } k \neq 0. $$

Therefore, by (1), we obtain $\sum_{i=1}^{s} g_i f_i = 1$.

Consider now the general case when $\mathbb{K}$ is no longer supposed to be spherically complete. Consider a spherically complete algebraically closed extension $\tilde{\mathbb{K}}$ of $\mathbb{K}$. Set $B = \lambda(f_1, \dots, f_s)^{-2}$. Then by what forgoes, there exist $h_1, \dots, h_s \in \tilde{\mathbb{A}}$ such that $\sum_{j=1}^{s} h_j f_i = 1$ and $\max_{1 \le j \le s} \|h_j\| < B$.

Inside $\tilde{\mathbb{A}}$, let $E$ be the closed subspace of the $\mathbb{K}$-Banach space generated by 1 and all coefficients of all the $h_j$. Let us take $\epsilon > 0$
The Corona Problem on $A_b(d(0, 1^{-}))$

333

such that $(1+\epsilon) \max_{1\le j\le s} \|h_j\| \le B$. Since $E$ is a $\mathbb{K}$-Banach space of countable type, there exists a $\mathbb{K}$-linear map $\ell$ from $E$ to $\mathbb{K}$ satisfying $\ell(1) = 1$ with a norm $\|\ell\|$ satisfying $\|\ell\| \le 1 + \epsilon$. Let $T$ be the closed subspace of $\hat{A}$ consisting of the power series with coefficients in $E$. Then $T$ is a $A$-module and then we have an extension $\mathcal{L}$ of $\ell$ from $T$ to $A$ defined as $\mathcal{L}(\sum_{k=0}^{\infty} e_k x^k) = \sum_{k=0}^{\infty} \ell(e_k) x^k$ which is $A$-linear and its norm satisfies $\|\mathcal{L}\| \le 1 + \epsilon$. Now, putting $g_j = \mathcal{L}(h_j)$, $1 \le j \le s$, we have $\sum_{i=1}^s f_i g_i = 1$ and $\max_{1\le j\le s} \|g_j\| < B$. $\square$

**Notation.** Let $\mathcal{I}$ be an ideal of $A$. For each $f \in A$ and for each $\epsilon > 0$, we set $\mathcal{D}(f, \epsilon) = \{x \in D \mid |f(x)| \le \epsilon\}$.

Given $\phi \in \mathrm{Mult}(A, \| \cdot \|)$, $f_1, \dots, f_q \in A$ and $\epsilon > 0$, we denote by $\mathcal{V}(\phi, f_1, \dots, f_q, \epsilon)$ the set $\{\psi \in \mathrm{Mult}(A, \| \cdot \|) \mid |\phi(f_j) - \psi(f_j)|_\infty \le \epsilon, j = 1, \dots, q\}$.

**Corollary 35.3.** Let $\mathcal{I}$ be a proper ideal of $A$. The family $(\mathcal{D}(f, \epsilon), f \in \mathcal{I}, \epsilon > 0)$ generates a filter $\mathcal{F}$ on $D$ such that $\mathcal{I} \subset \mathcal{J}(\mathcal{F})$.

**Proof.** Let $(\mathcal{D}(f_j, \epsilon_j), 1 \le j \le n$ be such that $\bigcap_{j=1}^n \mathcal{D}(f, \epsilon_j) = \emptyset$. Let $\epsilon = \min_{1 \le j \le n}(\epsilon_j)$ and let $F_j = D \setminus \mathcal{D}(f_j, \epsilon)$, $1 \le j \le n$. Then, $\bigcup_{j=1}^n F_j = D$, hence by Theorem 35.2, $1 \in \mathcal{I}$, a contradiction. $\square$

By Theorem 34.9 and Corollary 35.3, we can derive Corollary 35.4:

**Corollary 35.4.** Let $\mathcal{M}$ be a maximal ideal of $A$. Then there exists an ultrafilter $\mathcal{U}$ on $D$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$. Moreover, if $\mathcal{U}$ converges, $\mathcal{M}$ is of codimension 1. If $\mathcal{U}$ does not converge, then $\mathcal{M}$ is of infinite codimension.

**Proof.** Indeed, let $\mathcal{M}$ be a maximal ideal of $A$. The family $(\mathcal{D}(f, \epsilon), f \in \mathcal{U}, \epsilon > 0)$ generates a filter $\mathcal{F}$ on $D$ such that $\mathcal{M} \subset \mathcal{J}(\mathcal{U})$. But since $\mathcal{M}$ is a maximal ideal, then $\mathcal{M} = \mathcal{J}(\mathcal{U})$. The conclusion on the codimension comes from Theorem 34.9. $\square$

Conversely, now we see that certain ultrafilters define maximal ideals. First, we need a few basic results.

**Definition and notation.** We denote by $\mathcal{W}$ the filter on $D$ admitting for basis the annuli $\Gamma(0, r, 1)$. A sequence $(a_n)_{n \in \mathbb{N}}$ is called a *regular sequence* if $\inf_{j \in \mathbb{N}} \prod_{n \neq j} |a_n - a_j| > 0$.
An ultrafilter $\mathcal{U}$ is said to be *regular* if it is thinner than a regular sequence.

We denote by $B(\mathbb{N}, \mathbb{K})$ the $\mathbb{K}$-algebra of bounded sequences of $\mathbb{K}$.
Let $S = (a_n)_{n \in \mathbb{N}}$ be a sequence in $D$ thinner than $W$. We denote
by $T_S$ the mapping from $A$ into $B(\mathbb{N}, \mathbb{K})$ which associates with each
$f \in A$ the sequence $(f(a_n))_{n \in \mathbb{N}}$, and we denote by $\Sigma(S)$ the set of
ultrafilters thinner than $S$ and by $\mathcal{I}(S)$ the ideal of the $f \in A$ such
that $f(a_n) = 0 \ \forall n \in \mathbb{N}$.

From Corollary 4.6 of [91], we have Theorem 35.5 which is purely
technical:

**Theorem 35.5.** Let $S$ be a sequence in $D$ thinner than $W$. Then $T_S$
is surjective on $B(\mathbb{N}, \mathbb{K})$ if and only if $S$ is regular.

**Notations.** Let $S$ be a regular sequence. Since $T_S$ is surjective,
there exists a $\mathbb{K}$-algebra isomorphism $\Lambda_S$ from $\frac{A}{\mathrm{Ker}(T_S)}$ onto $B(\mathbb{N}, \mathbb{K})$,
where $\mathrm{Ker}(T_S) = \mathcal{I}(S)$.

For every ultrafilter $\mathcal{G}$ on $\mathbb{N}$, we denote by $\Theta(\mathcal{G})$ the ideal of $B(\mathbb{N}, \mathbb{K})$ consisting of sequences $(a_n)_{n \in \mathbb{N}}$ such that $\lim_{\mathcal{G}} a_n = 0$.

Theorem 35.6 is also classical:

**Theorem 35.6.** $\Theta$ is a bijection from the set of ultrafilters on $\mathbb{N}$ onto $\mathrm{Max}(B(\mathbb{N}, \mathbb{K}))$. The restriction of $\Theta$ to the subset of non-principal ultrafilters on $\mathbb{N}$ is a bijection from this set onto the set of non-principal maximal ideals of $B(\mathbb{N}, \mathbb{K})$. Moreover, a maximal ideal of $B(\mathbb{N}, \mathbb{K})$ is principal if and only it is of codimension 1.

**Theorem 35.7.** Let $S$ be a regular sequence and let $\mathcal{M}$ be a maximal ideal of $A$. The following two statements are equivalent:

(i) $\mathcal{I}(S) \subset \mathcal{M}$.

(ii) There exists an ultrafilter $\mathcal{U}$ thinner than $S$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$.

Moreover, the mapping $G$ which associates with each ultrafilter $\mathcal{U}$ thinner than $S$ the ideal $\mathcal{J}(\mathcal{U})$ is a bijection from $\Sigma(S)$ onto the set of maximal ideals of $A$ containing $\mathcal{I}(S)$.

**Proof.** Obviously, (ii) implies (i). Thus, suppose (i) true. Let $S = (a_n)_{n \in \mathbb{N}}$. By Theorem 35.6, the isomorphism $\Lambda_S$ makes a bijection $G$ from the set of maximal ideals of $A$ containing $\mathcal{I}(S)$ to the set of maximal ideals of $B(\mathbb{N}, \mathbb{K})$, and more precisely, it makes a bijection
from the set of maximal ideals of $A$ of infinite codimension containing $\mathcal{I}(S)$ to the set of maximal ideals of $B(\mathbb{N}, \mathbb{K})$ of infinite codimension which actually are the non-principal maximal ideals of $B(\mathbb{N}, \mathbb{K})$. Let $\mathcal{N} = G(M)$. By Theorem 35.5, there exists an ultrafilter $\mathcal{U}$ on $\mathbb{N}$ such that $\mathcal{N}$ is the ideal of the bounded sequences tending to zero along $\mathcal{U}$. Now, let $\Xi$ be the natural bijection from the set of non-principal ultrafilters of $\mathbb{N}$ onto the set of ultrafilters thinner than $S$ and let $V = \Xi(\mathcal{U})$. Then, $\mathcal{N} = \Lambda_S(M)$, hence $M = \mathcal{I}(S)$. Moreover, in this way, we can see that $G \circ \Xi^{-1}$ is a bijection from $\Sigma(S)$ onto the set of maximal ideals of $A$ containing $\mathcal{I}(S)$.

**Corollary 35.8.** If $\mathcal{U}$ is a regular ultrafilter on $D$, $\mathcal{J}(\mathcal{U})$ is a maximal ideal of $A$.

**Proposition 35.9.** Let $(B, \| \cdot \|)$ be a unital commutative ultra-metric $\mathbb{K}$-Banach algebra. Suppose there exist $\ell \in B$, $\phi, \psi \in \mathrm{Mult}(B, \| \cdot \|)$ such that $\psi(\ell) < \phi(\ell)$, $\mathrm{sp}(\ell) \cap \Gamma(0, \psi(\ell), \phi(\ell)) = \emptyset$ and there exists $\epsilon \in ]0, \phi(\ell) - \psi(\ell)[$ satisfying further $\|(\ell - a)^{-1}\| \le M \quad \forall a \in \Gamma(0, \psi(\ell), \phi(\ell) - \epsilon)$. Then there exists $g \in B$ such that $\psi(g) = 1$, $\phi(g) = 0$.

**Proof.** By Theorem 10.6, we have $\| \ell \|_{si} = \sup\{\theta(\ell) \mid \theta \in \mathrm{Mult}(B, \| \cdot \|)\}$.

Let $s = \psi(\ell)$, $t = \phi(\ell) - \epsilon$. Consider now the construction made in Theorem 21.15 by taking $s = r_0$, $Q = \|\ell\|$, $R = t - \epsilon$, and $l = \frac{1}{M}$ and consider the sequence $(a_{n,j})_{n \in \mathbb{N}, 1 \le j \le q_n}$ and the set $E = d(0, Q^-) \setminus (\bigcup_{j=1}^{q_n} d(a_{n,j}, l^-))$ defined in Theorem 21.15. Then, in $H(E)$, we have

$$ (1) \qquad \left\| \frac{1}{x-b} \right\|_E \le l \quad \forall b \in \bigcup_{n \in \mathbb{N}} (\cup_{j=1}^{q_n} d(a_{n,j}, l^-)). $$

There exists a natural homomorphism $\sigma$ from $R(E)$ into $B$ such that $\sigma(x) = \ell$. And since $Q = \|\ell\|$ and $\|(\ell-b)^{-1}\| \le M \forall b \in \Gamma(0, s, t)$, by Theorem 15.1 and by (1), $\sigma$ is clearly continuous with respect to the norms $\| \cdot \|_E$ of $R(E)$ and $\| \cdot \|$ of $B$. Consequently, $\sigma$ has continuation to a continuous homomorphism from $H(E)$ to $B$.

Now, let $\psi' = \psi \circ \sigma$, $\phi' = \phi \circ \sigma$. Then both $\phi'$, $\psi'$ belong to $\mathrm{Mult}(H(E), \| \cdot \|)$ and satisfy $\psi'(x) = s$, $\phi'(x) = t - \epsilon$. So, $\psi'$ is of the form $\varphi_F$ with $F$ a circular on $E$ secant with $C(0, s)$ and $\phi'$ is of the form $\varphi_G$ with $G$ a circular on $E$ secant with $C(0, t)$.
Consider now the function $f$ constructed in Theorem 21.15 which, by construction, belongs to $H(E)$ and has no zero and no pole in $d(0, s^{-})$. Consequently, $|f(x)| = |f(0)| = 1 \ \forall x \in d(0, s^{-})$. Moreover, as showed in Theorem 21.15, we have $\lim_{\mathcal{G}} f(x) = 0$, hence $\phi'(f) = 0$. Putting $g = \sigma(f)$, we have $\psi(g) = \psi'(f) = 1$ and $\phi(g) = \phi'(f) = 0$, which ends the proof. □

**Definition and notation.** We denote by $\mathcal{W}$ the filter on $D$ admitting for basis the annuli $\Gamma(0, r, R)$, $r \in ]0, R[$. Given an ideal $\mathcal{I}$ of $A$, we denote by $\mathcal{G}_I$ the filter generated by the sets $\mathcal{D}(f, \epsilon)$, $f \in I, \epsilon > 0$. By definition, $\mathcal{G}_I$ is minimal, with respect to the relation of thinness, among the filters $\mathcal{H}$ such that $\lim_{\mathcal{H}} f(x) = 0 \ \forall f \in I$. A filter $\mathcal{F}$ on $D$ is said to be *coroner* if it is thinner than $\mathcal{W}$.

By Theorem 10.9, each maximal ideal of a Banach $\mathbb{K}$-algebra is the kernel of at least one element of $\mathrm{Mult}(A, \| \cdot \|)$. A maximal ideal $\mathcal{M}$ of $A$ is said to be *univalent* if there exists only one $\phi \in \mathrm{Mult}_1(A, \| \cdot \|)$ such that $\mathrm{Ker}(\phi) = \mathcal{M}$ and the algebra $A$ is said to be *multibijective* if every maximal ideal of $A$ is univalent.

**Theorem 35.10.** Let $\mathcal{M}$ be a non-principal maximal ideal of $A$ and let $\mathcal{U}$ be an ultrafilter thinner than $\mathcal{G}_{\mathcal{M}}$. Then $\varphi_{\mathcal{U}}$ belongs to the closure of $\mathrm{Mult}_1(A, \| \cdot \|)$ in $\mathrm{Mult}_m(A, \| \cdot \|)$.

**Proof.** Let $V$ be neighborhood of $\varphi_{\mathcal{U}}$ in $\mathrm{Mult}(A, \| \cdot \|)$. It contains some set of the form $\mathcal{V}(\varphi_{\mathcal{U}}, f_1, \dots, f_q, \epsilon)$. For each $j = 1, \dots, q$, there exists $E_j \in \mathcal{U}$ such that

$$ \left| |f_j(x)| - \varphi_{\mathcal{U}}(f_j) \right|_{\infty} \le \epsilon \quad \forall x \in E_j. $$

Let $E = \bigcap_{j=1}^q E_j$. Then,

$$ \left| |f_j(x)| - \varphi_{\mathcal{U}}(f_j) \right|_{\infty} \le \epsilon \quad \forall x \in E, \forall j = 1, \dots, q. $$

Consequently, $\varphi_a$ belongs to $\mathcal{V}(\varphi_{\mathcal{U}}, f_1, \dots, f_q, \epsilon)$ for all $a \in E$. □

**Corollary 35.11.** Let $\mathcal{M}$ be a univalent non-principal maximal ideal of $A$ and let $\phi \in \mathrm{Mult}_m(A, \| \cdot \|)$ satisfy $\mathrm{Ker}(\phi) = \mathcal{M}$. Then $\phi$ is of the form $\phi(f) = \lim_{\mathcal{U}} |f(x)|$ with $\mathcal{U}$ a coroner ultrafilter and such that $\mathcal{J}(\mathcal{U}) = \mathcal{M}$. Moreover, $\phi$ belongs to the closure of $\mathrm{Mult}_1(A, \| \cdot \|)$ in $\mathrm{Mult}_m(A, \| \cdot \|)$.
**Proposition 35.12.** Let $\mathcal{U}$ be a coroner ultrafilter on $D$, let $f \in A\backslash\mathcal{J}(\mathcal{U})$ be non-invertible in $A$, such that $\|f\| \le 1$, and let $g \in A$, $h \in \mathcal{J}(\mathcal{U})$ be such that $fg = 1+h$. Let $\tau = \varphi_{\mathcal{U}}(f)$, let $\epsilon \in ]0, \tau[$, and let $\Lambda = \{x \in D \mid |f(x)g(x)| - 1|_{\infty} < \epsilon, \mid |f(x)| - \tau|_{\infty} < \epsilon\}$.

Suppose that there exist a function $\tilde{h} \in A$ admitting for zeros in $D$ the zeros of $h$ in $D\backslash\Lambda$ and a function $\overline{h} \in A$ admitting for zeros the zeros of $h$ in $\Lambda$, each counting multiplicities, so that $h = \overline{h}\tilde{h}$. Then $|\tilde{h}(x)|$ has a strictly positive lower bound in $\Lambda$ and $\overline{h}$ belongs to $\mathcal{J}(\mathcal{U})$.

Moreover, there exists $\omega \in ]0, \tau[$ such that $\omega \le \inf\{\max(|f(x)|, |\overline{h}(x)|) \mid x \in D\}$. Further, for every $a \in d(0, (\tau - \epsilon))$, we have $\omega \le \inf\{\max(|f(x) - a|, |\overline{h}(x)|) \mid x \in D\}$.

**Proof.** Let $u \in \Lambda$ and let $s$ be the distance of $u$ from $K\backslash\Lambda$. So, the disk $d(u, s^{-})$ is included in $\Lambda$, hence $fg$ has no zero inside this disk. Consequently, $|f(x)g(x)|$ is a constant $b$ in $d(u, s^{-})$. Consider the family $F_u$ of radii of circles $C(u, r)$, containing at least one zero of $fg$. By Theorem 22.20, $F_u$ has no cluster point different from 1. Consequently, there exists $\rho \ge s$ such that $fg$ admits at least one zero in $C(u, \rho)$ and admits no zero in $d(u, \rho^{-})$. And then $|f(x)g(x)|$ is a constant $c$ in $d(u, \rho^{-})$. But then, at $u$, we see that $b = c$, and therefore, $d(u, \rho^{-})$ is included in $\Lambda$. Hence, $\rho = s$, and therefore, $fg$ admits at least one zero $\alpha$ in $C(u, s)$. Thus, at $\alpha$, we have $h(\alpha) = -1$. Therefore, in the disk $d(\alpha, s^{-})$, we can check that $\varphi_{\alpha,s}(h) \ge 1$. But by Lemma 13.4, $\varphi_{\alpha,s}(h) = \varphi_{u,s}(h)$, hence $\varphi_{u,s}(h) \ge 1$.

Now,

$$ \frac{\|h\|}{\varphi_{u,s}(h)} = \frac{\|\tilde{h}\|}{\varphi_{u,s}(\tilde{h})} \frac{\|\overline{h}\|}{\varphi_{u,s}(\overline{h})} \ge \frac{\|\tilde{h}\|}{\varphi_{u,s}(\tilde{h})}. $$

Therefore, since $\varphi_{u,s}(h) \ge 1$, we obtain

$$ (1) \qquad \frac{\|\tilde{h}\|}{\varphi_{u,s}(\tilde{h})} \le \|h\|. $$

But since by definition $d(u, s^{-})$ is included in $\Lambda$, $\tilde{h}$ has no zero in this disk, hence $|\tilde{h}(x)|$ is constant and equal to $\varphi_{u,s}(\tilde{h})$. Consequently, by
(1), we obtain $\frac{\|\tilde{h}\|}{|h(u)|} \le \|h\|$, and therefore,

$$|\tilde{h}(u)| \ge \frac{\|\tilde{h}\|}{\|h\|} \quad \forall u \in \Lambda.$$

This shows that $\tilde{h}$ does not belong to $\mathcal{J}(\mathcal{U})$, hence, by Theorem 35.1, $\varphi_{\mathcal{U}}(\tilde{h}) \ne 0$. Consequently, $\varphi_{\mathcal{U}}(\tilde{h}) = 0$.

Now, by hypothesis, we have $fg - \tilde{h}\tilde{h} = 1$. Since both $g, \tilde{h}$ belong to $A$ and therefore are bounded in $D$, it is obvious that $\inf\{\max(|f(x)|, |\tilde{h}(x)|) \mid x \in D\} > 0$. So, we may obviously choose $\omega \in ]0, \tau - \epsilon[$ such that

$$(2) \qquad \omega \le \inf\{\max(|f(x)|, |\tilde{h}(x)|) \mid x \in D\}.$$

Let us now show that for every $a \in d(0, (\tau - \epsilon))$, we have $\omega \le \inf\{\max(|f(x) - a|, |\tilde{h}(x)|) \mid x \in D\}$.

Let $\Lambda' = \{x \in D \mid |f(x)| \ge \tau - \epsilon\}$ and let $a \in d(0, (\tau - \epsilon)^-)$. When $\beta$ lies in $\Lambda'$, we have $|f(\beta)| > |a|$, hence by (2), $\max(|f(\beta) - a|, |\tilde{h}(\beta)|) \ge \omega$ because by (2), either $\omega \le |\tilde{h}(\beta)|$, or $\omega \le |f(\beta)| = |f(\beta) - a|$.

Now, let $\beta$ lie in $D\setminus\Lambda'$ and let $t$ be the distance from $\beta$ to $\Lambda'$. Since $D\setminus\Lambda'$ is open, $t$ is $> 0$. Consider $\varphi_{\beta,t}(f)$. Either there exists $\ell \in \Lambda'$ such that $|\beta - \ell| = t$ and then $\varphi_{\beta,t}(f) \ge |f(\ell)|$ or there exists a sequence $(x_n)_{n \in \mathbb{N}} \in \Lambda'$ such that $\lim_{n \to +\infty} |\beta - x_n| = t$. Now, let $\beta$ lie in $D\setminus\Lambda'$ and let $t$ be the distance from $\beta$ to $\Lambda'$. Since $D\setminus\Lambda'$ is open, $t$ is $> 0$. Consider $\varphi_{\beta,t}(f)$. Either there exists $\mu \in \Lambda'$ such that $|\beta - \mu| = t$ and then $\varphi_{\beta,t}(f) \ge |f(\mu)| \ge \tau - \epsilon$ or there exists a sequence $(x_n)_{n \in \mathbb{N}} \in \Lambda'$ such that $\lim_{n \to +\infty} |\beta - x_n| = t$ and $|x_n - \beta| > t$. Suppose that we are in the second case: there exists a sequence $(x_n)_{n \in \mathbb{N}} \in \Lambda'$ such that $\lim_{n \to +\infty} |\beta - x_n| = t$ and $|x_n - \beta| > t$. Then the sequence is thinner than the circular filter of center $\beta$ and diameter $t$, hence

$$\lim_{n \to +\infty} |f(x_n)| = \varphi_{\beta,t}(f),$$

hence $\varphi_{\beta,t}(f) \ge \tau - \epsilon$ again. If $f$ has no zero in $d(\beta, t^{-})$, then $|f(x)|$ is a constant in that disk, hence of course $\varphi_{\beta,t}(f) < \tau - \epsilon$, a contradiction. Consequently, $f$ must have a zero $\gamma$ in $d(\beta, t^{-})$. Therefore, due to (2), we have $|\tilde{h}(\gamma)| \ge \omega$. But since, by definition, $\Lambda \subset \Lambda'$, the zeros of $\tilde{h}$ belong to $\Lambda'$. And since $d(\beta, t^{-}) \cap \Lambda' = \emptyset$ actually $\tilde{h}$ has
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no zero in $d(\beta, t^{-})$. Consequently, $|\bar{h}(x)|$ is constant in $d(\beta, t^{-})$, and hence $|\bar{h}(\beta)| \ge \omega$, which completes the proof. □

The following basic lemma is easily checked:

**Lemma 35.13.** Let $S$ be a set and let $E$ be a subset. Let $\mathcal{F}$ be an ultrafilter on $E$. Then the filter $\tilde{\mathcal{F}}$ on $S$ with basis $\mathcal{F}$ is an ultrafilter inducing on $E$ the ultrafilter $\mathcal{F}$.

**Corollary 35.14.** Let $S$ be a set and let $E$ be a subset of $S$. Let $\mathcal{F}$ be an ultrafilter on $E$ and let $\tilde{\mathcal{F}} = \mathcal{G}$ be the ultrafilter on $S$ having $\mathcal{F}$ as a basis of filter. Let $f$ be a function defined on $S$ with values in a compact topological space $T$. Then, $\lim_{\mathcal{G}} f(x) = \lim_{\mathcal{F}} f(x)$.

**Proof.** Suppose that $f$ admits distinct limits on $\mathcal{F}$ and $\mathcal{G}$. Then $\mathcal{F}$ is a basis of a filter on $S$ that is not secant with $\mathcal{G}$, a contradiction since $\mathcal{F}$ is the ultrafilter induced by $\mathcal{G}$ on $E$. □

**Proposition 35.15.** Let $\mathcal{M}$ be a non-principal maximal ideal of $A$ and let $\mathcal{U}$ be an ultrafilter on $D$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$. Let $f \in A \setminus \mathcal{M}$ satisfy $\|f\| < 1$, let $\tau = \varphi_{\mathcal{U}}(f)$, and let $\epsilon \in ]0, \tau[$. There exists $c > 0$ such that, for every $a \in d(0, \tau - \epsilon)$, there exists $g_a \in A$ satisfying $(f-a)g_a - 1 \in \mathcal{M}$ and $\|g_a\| \le c$.

**Proof.** Suppose first that $f$ is invertible in $A$. By Theorem 22.20, $|f(x)|$ is a constant and hence is equal to $\tau$. Therefore, $|f(x) - a| = \tau \quad \forall a \in d(0, \tau - \epsilon)$. Consequently, $f - a$ is invertible and its inverse $g_a$ satisfies $\|g_a\| = \tau^{-1}$. Thus, we only have to show the claim when $f$ is not invertible.

Since $f$ does not belong to $\mathcal{M}$, we can find $g \in A$ and $h \in \mathcal{M}$ such that $fg = 1 + h$ with $h \in \mathcal{M}$.

Let $\tilde{\mathbb{K}}$ be an algebraically closed spherically complete extension of $\mathbb{K}$, and let $\tilde{D}$ be the disk $\{x \in \tilde{\mathbb{K}} \mid |x| < 1\}$. Let $\tilde{A}$ be the algebra of bounded power series converging in $\tilde{D}$ with coefficients in $\tilde{\mathbb{K}}$.

$\mathcal{U}$ makes a basis of a filter $\tilde{\mathcal{U}}$ on $\tilde{D}$, and by definition, $\tilde{\mathcal{U}}$ is the filter induced by $\tilde{\mathcal{U}}$ on $D$. By Lemma 35.13, $\tilde{\mathcal{U}}$ is an ultrafilter on $\tilde{D}$.

Consider now $f$ as an element of $\tilde{A}$. Then $\tilde{\mathcal{U}}$ defines an element $\psi$ of $\text{Mult}(\tilde{A}, \| \cdot \, \|)$ as $\psi(\ell) = \lim_{\tilde{A}} |\ell(x)|, \forall \ell \in \tilde{A}$. Consequently, by
Corollary 35.14, $\tau$ is equal to $\lim_{\mathcal{U}} |f(x)|$. Let

$$ \Lambda = \{x \in \hat{D} \mid |f(x)g(x)| - 1|_{\infty} < \epsilon, \mid |f(x)| - \tau|_{\infty} < \epsilon\}. $$

By Proposition 28.6, we can factorize $h$ in the form $\tilde{h}\overline{h}$, where $\tilde{h} \in \hat{A}$ is a function admitting for zeros in $\hat{D}$ the zeros of $h$ in $\hat{D} \setminus \Lambda$ and $\overline{h} \in \hat{A}$ is a function admitting for zeros the zeros of $h$ in $\Lambda$, each counting multiplicities. Moreover, we can choose $\overline{h}$ so that $\|\overline{h}\| < 1$.

Now, in the field $\mathbb{K}$, by Proposition 35.12, there exists $\omega > 0$ such that for every $a \in \hat{d}(0, (\tau - \epsilon))$, we have $\omega \le \inf\{\max(|f(x) - a|, |\overline{h}(x)|) \mid x \in \hat{D}\}$. This implies that $\inf\{\max(|f(x) - a|, |\overline{h}(x)|) \mid x \in \hat{D}\} \ge \omega \forall a \in \hat{d}(0, \tau - \epsilon)$. We note that $\|f - a\| < 1$ for every $a \in \hat{d}(0, \tau - \epsilon)$, so we may apply Theorem 35.1 and obtain a bound $b$ only depending on $f$ and $h$ and functions $\ell_a$, $h_a \in \hat{A}$ such that $(f - a)\ell_a + \overline{h}h_a = 1$, with

$$ (1) \qquad \|\ell_a\| < b, \quad \|h_a\| < b \quad \forall a \in \hat{d}(0, \tau - \epsilon). $$

By hypothesis, we have $\lim_{\mathcal{U}} h(x) = 0$. Hence, by Corollary 35.14, on $\hat{D}$ we have $\lim_{\hat{A}} h(x) = 0$. Then, by Proposition 35.12, we have $\lim_{\hat{A}} \overline{h}(x) = 0$, hence, on $D$,

$$ (2) \qquad \lim_{\mathcal{U}} \overline{h}h_a(x) = 0 \quad \forall a \in d(0, \tau - \epsilon). $$

Now, let us fix $a \in d(0, \tau - \epsilon)$. Let $G$ be the closed $\mathbb{K}$-vector subspace of $\mathbb{K}$ (considered as a $\mathbb{K}$-Banach space), linearly generated over $\mathbb{K}$ by 1 and all coefficients of $\ell_a$. Take $\eta > 0$ such that $(1 + \eta)\max(\|\ell_a\|, \|h_a\|) \le b$. We note that $G$ is a $\mathbb{K}$-Banach space of countable type, hence there exists a $\mathbb{K}$-linear mapping $\Xi$ from $G$ to $\mathbb{K}$ of norm $\le 1 + \eta$ such that $\Xi(1) = 1$ [89]. Let $F$ be the closed $\mathbb{K}$-vector subspace of $\hat{A}$ consisting of all power series with coefficients in $E$. Then $F$ is a $A$-module and $\Xi$ has continuation to a $A$-linear mapping $\Xi$ from $F$ to $A$ defined as $\Xi(\sum_{n=0}^{\infty} b_n x^n) = \sum_{n=0}^{\infty} \Xi(b_n)x^n$. This mapping $\Xi$ has a norm bounded by $1 + \eta$. Set $g_a = \Xi(\ell_a)$. Then, by (1), we have

$$ (3) \qquad \|g_a\| \le b(1 + \eta) \quad \forall a \in d(0, \tau - \epsilon). $$
On the other hand, by construction, for every $z \in G$, we have $|\Xi(z)| \le |z|(1 + \eta)$: that holds particularly for elements of $G \cap D$. Now, since $(f-a)(l_a) - \bar{h}h_a = 1$, for all $x \in D$, we have $l_a(x) \in G$, $f(x) - a \in K$ and hence $\bar{h}h_a(x)$ belongs to $G$. Therefore, the inequality applies and shows that $|\Xi(\bar{h}h_a)(x)| \le |(\bar{h}h_a)(x)|(1 + \eta)$, hence by (2), we can derive $\lim_{\mathcal{U}} |\Xi(\bar{h}h_a)(x)| = 0$ $\forall a \in d(0, \tau - \epsilon)$. But since $\Xi$ is a $A$-module linear mapping, we have $\Xi((f-a)h_a - 1) = (f-a)g_a - 1$. Consequently, $\lim_{\mathcal{U}} |(f(x)-a)g_a(x)-1| = 0$ $\forall a \in d(0, \tau - \epsilon)$ and hence $(f-a)g_a - 1$ belongs to $\mathcal{J}(\mathcal{U})$. Putting $c = b(1+\eta)$, by (3), we are done. $\square$

**Theorem 35.16.** *A is multibijective.*

**Proof.** Suppose $A$ is not multibijective and let $\mathcal{M}$ be a maximal ideal which is not univalent. Let $F$ be the quotient field $\textstyle\frac{\mathcal{A}}{\mathcal{M}}$, let $\theta$ be the canonical surjection from $A$ onto $F$, and let $\|\cdot\|_q$ be the $\mathbb{K}$-Banach algebra quotient norm of $F$. By Corollary 35.11, there exists an ultrafilter $\mathcal{U}$ on $D$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$. Thus, there exists $\psi \in \text{Mult}(A, \|\cdot\|)$ such that $\text{Ker}(\psi) = \mathcal{M}$ and $\psi \neq \varphi_{\mathcal{U}}$. Consequently, there exists $f \in A$ such that $\psi(f) \neq \varphi_{\mathcal{U}}(f)$, with $\psi(f) \neq 0$, $\varphi_{\mathcal{U}}(f) \neq 0$. We check that we may also assume $\psi(f) < \varphi_{\mathcal{U}}(f)$. Indeed, suppose $\psi(f) > \varphi_{\mathcal{U}}(f)$. Let $g \in A$ be such that $\theta(g) = \theta(f)^{-1}$. Then we can see that $\psi(g) = \psi(f)^{-1}$, $\varphi_{\mathcal{U}}(g) = (\varphi_{\mathcal{U}}(f))^{-1}$, therefore $\psi(g) < \varphi_{\mathcal{U}}(g)$. Thus, we may assume $\psi(f) < \varphi_{\mathcal{U}}(f)$ without loss of generality. Similarly, we may obviously assume that $\|f\| < 1$. By Lemma 35.7, we know that $f$ is not invertible.

By construction, $\varphi_{\mathcal{U}}$ factorizes in the form $\phi_1 \circ \theta$ and similarly, $\psi$ factorizes in the form $\phi_2 \circ \theta$ with $\phi_1, \phi_2 \in \text{Mult}(F, \|\cdot\|_q)$. So, on $F$ we have $\phi_1(\theta(f)) > \phi_2(\theta(f))$.

Let $\tau = \varphi_{\mathcal{U}}(f)$ and let $\epsilon \in ]0, \tau[$. By Proposition 35.15, there exists $c > 0$ such that, for every $a \in d(0, \tau - \epsilon)$, there exists $g_a \in A$ satisfying $(f-a)g_a - 1 \in \mathcal{M}$ and $\|g_a\| \le c$. Now, $\theta(g_a) = (\theta(f-a))^{-1}$. Thus, $\|(\theta(f-a))^{-1}\|_q \le c$ $\forall a \in d(0, \tau - \epsilon)$. Therefore, by applying Proposition 35.9 to the $\mathbb{K}$-Banach algebra $F$, we can see that there exists $y \in F$ such that $\phi_1(y) = 1, \phi_2(y) = 0$. Therefore, taking $g \in A$ such that $\theta(g) = y$, we get $\varphi_{\mathcal{U}}(g) = 0$, $\psi(g) = 1$, a contradiction to the hypothesis $\text{Ker}(\varphi_{\mathcal{U}}) = \text{Ker}(\psi)$. This finishes showing that $A$ is multibijective. $\square$
By Corollary 35.11, we have the following two corollaries:

**Corollary 35.17.** For every $\phi \in \text{Mult}_m(A, \|\cdot\|) \setminus \text{Mult}_1(A, \|\cdot\|)$, there exists a coroner ultrafilter $\mathcal{U}$ such that $\phi(f) = \lim_{\mathcal{U}} |f(x)|$ for every $f \in A$.

**Corollary 35.18.** $\text{Mult}_1(A, \|\cdot\|)$ is dense in $\text{Mult}_m(A, \|\cdot\|)$.

Another interesting question was whether certain elements of $\text{Mult}(A, \|\cdot\|)$ may have a kernel that is neither null nor a maximal ideal. The question was solved by Jesus Araujo thanks to this nice example [6].

**Theorem 35.19 (J. Araujo, [6]).** Let $h(x) = \sum_{n=0}^{\infty} a_n x^n$ and suppose that the sequence $(\frac{|a_n|}{|a_{n+1}|})_{n \in \mathbb{N}}$ is strictly increasing of limit 1. Then, $h$ belongs to $A$. Moreover, putting $r_n = \frac{|a_n|}{|a_{n+1}|}$, $n \in \mathbb{N}$, $h$ admits a unique zero on each circle $C(0, r_n)$ and has no other zero in $D$.

Let $\mathcal{N}$ be an ultrafilter on $\mathbb{N}$, and for every $f \in A$, let $\phi(f, n) = \|f\|_{d(\alpha_n, r_n)}$. Let $\phi_r(f) = \lim_{\mathcal{N}} \phi(f, n)$.

Then $\phi_r$ belongs to $\text{Mult}(A, \|\cdot\|)$ and $\text{Ker}(\phi_r)$ is neither null nor a maximal ideal of $A$. Moreover, $\text{Ker}(\phi_r)$ does not depend on $r \in ]0, 1[$.

However, each so defined semi-norm $\phi_r$ belongs to the closure of $\text{Mult}_1(A, \|\cdot\|)$ in $\text{Mult}(A, \|\cdot\|)$.

**Proof.** Let $\mathcal{M}$ be the ideal of the function $f \in A$ such that $\lim_{\mathcal{N}} |f(\alpha_n)| = 0$. Of course, $h$ belongs to $\mathcal{M}$ and $\text{Ker}(\phi_r)$ is strictly included in $\mathcal{M}$. Indeed, since $h$ admits a unique zero in the disk $d(\alpha_n, r)$, it satisfies $\|h\|_{d(\alpha_n, r)} = |h|(r_n)^{\frac{r_n}{r}}$ and therefore $\lim_{\mathcal{N}} \|h\|_{d(\alpha_n, r)} = \frac{R}{r}$, which proves that $h$ does not belong to $\text{Ker}(\phi_r)$.

On the other hand, we prove that $\text{Ker}(\phi_r)$ is not null. Let $(q_n)_{n \in \mathbb{N}^*}$ be a sequence of positive integers satisfying $q_n \le q_{n+1}$ for all $n \in \mathbb{N}^*$, $\lim_{n \to +\infty} q_n = +\infty$ and such that the series $\sum_{n=1}^{+\infty} q_n \log(R/q_n)$ converges: we can easily find the sequence $(q_n)$ since $\lim_{n \to +\infty} q_n = R$. Now, consider the divisor $(\alpha_n, q_n)_{n \in \mathbb{N}}$ of the disk $d(0, R^{-})$. By Theorem 27.14, there exists $g \in \mathcal{A}(d(0, R^{-}))$ admitting each $\alpha_n$ as a zero of order $t_n \ge q_n$ and such that $|g|(r_n) \le |T|(r_n) + 1$ for all $n \in \mathbb{N}^*$. Consequently, $g$ is bounded in $d(0, R^{-})$ and hence belongs to $A$. Next, for
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every $n \in \mathbb{N}^*$, by Corollary 22.30, we have

$$ \|g\|_{d(\alpha_n, r)} \le |g|(r_n) \left(\frac{r}{r_n}\right)^{t_n} \le \|g\| \left(\frac{r}{r_n}\right)^{q_n}. $$

Since the sequence $(q_n)_{n \in \mathbb{N}^*}$ tends to $+\infty$ and the sequence $(r_n)$ is increasing, we have $\lim_{n \to +\infty} \|g\|_{d(\alpha_n, r)} = 0$, which proves that $g$ belongs to $\mathrm{Ker}(\varphi_r)$. Let $f \in \mathrm{Ker}(\varphi_r)$ and let $s \in ]0, R[$. If $s < r$, it is obvious that $f$ belongs to $\mathrm{Ker}(\varphi_s)$. Now, suppose $s > r$. Consider an element $L$ of $\mathcal{N}$ such that $\inf_{n \in L} \|f\|_{d(\alpha_n, r)} = 0$. We prove that $\inf_{n \in L} \|f\|_{d(\alpha_n, s)} = 0$. For each $n \in \mathbb{N}^*$, by Corollary 22.32, we have

$$ (1) \qquad \log(\|f\|_{d(\alpha, s)}) - \log(\|f\|_{d(\alpha, r)}) $$

$$ \le \left( \log(\|f\|_{d(\alpha, r_n)}) - \log(\|f\|_{d(\alpha, s)}) \right) \left( \frac{\log(s) - \log(r)}{\log(r_n) - \log(s)} \right). $$

Suppose that the sequence $\|f\|_{d(\alpha_n, s)}$ does not tend to 0. There exists a sequence $(u_m)_{m \in \mathbb{N}}$ of $\mathbb{N}^*$ such that $\|f\|_{d(\alpha_{u_m}, s)} > b \ \forall m \in \mathbb{N}$ with $b > 0$. But then we get to a contradiction with (1). Consequently, $\inf_{n \in L} \|f\|_{d(\alpha_n, s)} = 0$ and therefore $f$ belongs to $\mathrm{Ker}(\varphi_s)$, which proves that $\mathrm{Ker}(\varphi_s) = \mathrm{Ker}(\varphi_r)$ and finishes the proof of Theorem 35.19. $\square$

**Question.** The characterization of all multiplicative semi-norms is not yet complete: does $\phi \in \mathrm{Mult}(A, \| \cdot \|)$ exist different from those we have studied?
# Chapter 36

## Meromorphic Functions in $K$

In this chapter, we define and examine the basic properties of meromorphic functions: relations with poles of analytic elements, absolute values on fields of meromorphic functions defined by circular filters, value of the derivative on a circular filter, development in a Laurent series in an annulus, and existence of primitives [19].

**Definition and notation.** We denote by $M(\mathbb{K})$ the field of fractions of $\mathcal{A}(\mathbb{K})$. The elements of $M(\mathbb{K})$ are called meromorphic functions in $\mathbb{K}$.

In the same way, given $a \in \mathbb{K}$ and $r > 0$, we denote by $M(d(a, r^{-}))$ (respectively $M_b(d(a, r^{-})))$, respectively $M_u(d(a, r^{-})))$ the field of fractions of $\mathcal{A}(d(a, r^{-}))$ (respectively the field of fractions of $A_b(d(a, r^{-})))$, respectively the set $M(d(a, r^{-})) \setminus M_b(d(a, r^{-})))$.

The topology defined on $\mathcal{A}(\mathbb{K})$ (respectively $\mathcal{A}(d(a, R^{-})))$ has expansion to $M(\mathbb{K})$ (respectively to $M(d(a, R^{-})))$. The neighborhoods of a function $f \in M(\mathbb{K})$ (respectively $M(d(a, R^{-})))$ are the sets $W(f, r, \epsilon) = \{h \in \mathcal{M}(\mathbb{K}) \mid |f - h|(r) \le \epsilon\}$, with $r > 0$ (respectively $W(f, r, \epsilon) = \{h \in \mathcal{M}(d(a, R^{-})) \mid |f - h|(r) \le \epsilon\}$, with $r > 0$).

Let $b \in \mathbb{K}$ (respectively $b \in d(a, R^{-}))$ and let $r \in \mathbb{R}_+^*$ (respectively $r \in ]0, R[$). The absolute value $\varphi_{b,r}$ defined on $\mathcal{A}(\mathbb{K})$ (respectively on $\mathcal{A}(d(a, R^{-})))$ has an immediate continuation to $M(\mathbb{K})$ (respectively to $M(d(a, R^{-})))$ that we denote again by $\varphi_{b,r}$. In the same way, $\varphi_{0,r}$ is denoted by $|\cdot|(r)$ on $M(\mathbb{K})$ and on $M(d(0, R^{-}))$. Similarly,
the function $\Psi(\cdot, \mu)$ defined on $\mathcal{A}(\mathbb{K})$ and on $\mathcal{A}(d(0, R^{-}))$ has an
immediate continuation to $\mathcal{M}(\mathbb{K})$ and to $\mathcal{M}(d(0, R^{-}))$ as $\Psi(\frac{h}{l}, \mu) =$
$\Psi(h, \mu) - \Psi(l, \mu)$, with $h, l \in \mathcal{A}(\mathbb{K})$ (respectively $h, l \in \mathcal{A}(d(0, R^{-}))$).

Let $f = \frac{h}{l} \in \mathcal{M}(\mathbb{K})$ (respectively $f = \frac{h}{l} \in \mathcal{M}(d(a, R^{-}))$). For each $\alpha \in \mathbb{K}$ (respectively $\alpha \in d(a, R^{-})$), the number $\omega_{\alpha}(h) - \omega_{\alpha}(l)$ does not depend on the functions $h, l$ chosen to make $f = \frac{h}{l}$. Thus, we can generalize the notation by setting $\omega_{\alpha}(f) = \omega_{\alpha}(h) - \omega_{\alpha}(l)$. If $\omega_{\alpha}(f)$ is an integer $q > 0$, $\alpha$ is called a zero of $f$ of order $q$. If $\omega_{\alpha}(f)$ is an integer $q < 0$, $\alpha$ is called a pole of $f$ of order $-q$. If $\omega_{\alpha}(f) \ge 0$, $f$ will be said to be holomorphic at $\alpha$.

Similarly, as for $\mathcal{A}(\mathbb{K})$, given $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^{-}))$), we can define the divisor $\mathcal{D}(f)$ on $\mathbb{K}$ (respectively of $d(a, R^{-})$) as $\mathcal{D}(f)(\alpha) = 0$ whenever $f(\alpha) \neq 0$ and $\mathcal{D}(f)(\alpha) = s$ when $f$ has a zero of order $s$ at $\alpha$.

**Lemma 36.1.** Let $r, R \in \mathbb{R}_+^*$ with $0 < r < R$ and let $f \in \mathcal{M}(d(a, R^{-}))$. Then $f$ has finitely many poles $a_1, \dots, a_q$ in $d(a, r^{-})$. Let $E = d(a, r)\{a_1, \dots, a_q\}$. Then $f$ belongs to $H(E)$. If there exists $s \in \mathbb{N}^*$ such that $f^s$ is a constant, then so is $f$.

**Proof.** Without loss of generality, we can assume $a=0$. Let $f = \frac{h}{l}$ with $h, l \in \mathcal{A}(d(0, R))$. Since $l$ belongs to $\mathcal{A}(d(0, R))$, by Corollary 22.19, $l$ has finitely many zeros in $d(0, r)$, hence $f$ has finitely many poles $a_1, \dots, a_q$ in $d(0, r)$. Suppose first $f$ is of the form $\frac{1}{l}$ with $l \in \mathcal{A}(d(0, R^{-}))$. By Corollary 14.16, $l$ factorizes in the form $P(x)u(x)$ with $P \in \mathbb{K}[x]$ a polynomial whose zeros in $d(a, r)$ are $a_1, \dots, a_q$ and $u \in \mathcal{A}(d(0, R^{-}))$ is invertible in $H(d(0, r))$. On the other hand, $\frac{1}{P(x)}$ obviously belongs to $R(E)$. And by Proposition 17.3, $E$ belongs to Alg. Consequently, $l$ is invertible in $H(E)$. Consider now the general case $f = \frac{h}{l}$ with $h, l \in \mathcal{A}(d(0, R))$. Then both $h, \frac{1}{l}$ belong to $H(E)$), hence by Proposition 17.3, so does $f$.

Suppose now that $f^s$ is a constant. Since $\mathbb{K}$ is algebraically closed and since $\mathcal{M}(d(a, R^{-}))$ is a field extension of $\mathbb{K}$, $f$ belongs to $\mathbb{K}$. $\square$

**Corollary 36.2.** Let $f \in \mathcal{M}(d(a, R^{-}))$, let $r \in ]0, R[$, let $\alpha_j, 1 \le j \le q$ be the poles of $f$ in $d(a, r)$, let $\rho \in ]0, \min_{i \neq j} |\alpha_i - \alpha_j|[$, and for each $j = 1, \dots, q$, let $\rho_j \in ]0, \rho[$ and let $T_j = d(\alpha_j, \rho_j^-)$. Let $D = d(a, r)\(\bigcup_{j=1}^q T_j)$. Then $f$ belongs to $H(D)$.
**Lemma 36.3.** Let $f \in \mathcal{M}(\mathbb{K})$. There exists $h \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(h) = \mathcal{D}(f)$ and then the function $l = \frac{h}{f}$ belongs to $\mathcal{A}(\mathbb{K})$. Then $\mathcal{D}(\frac{1}{f}) = \mathcal{D}(l)$ and we can write $f$ in the form $\frac{h}{l}$ with $h, l \in \mathcal{A}(\mathbb{K})$, having no common zero.

**Proof.** Indeed, by Theorem 27.4, there exists $h \in \mathcal{A}(\mathbb{K})$ such that $\mathcal{D}(h) = \mathcal{D}(f)$ and hence conclusion follows. $\square$

**Remark.** Let $f \in \mathcal{M}(d(a, R^{-}))$, let $r \in ]0, R[$, and let $\alpha_j, 1 \le j \le n$ be the poles of $f$ in $d(a, r)$, of respective order $q_j$. By Lemma 36.1, $f$ belongs to $H(d(a, r)\setminus\{\alpha_1, \dots, \alpha_n\})$. Now, according to the definition of poles for analytic elements (see Chapter 11), $f$ also admits each $\alpha_j$ as a pole of order $q_j$, considered as an element of $H(d(a, r)\setminus\{\alpha_1, \dots, \alpha_n\})$.

By Theorem 27.4, we have already seen that if $f \in \mathcal{M}(\mathbb{K})$ has no zero and no pole in $\mathbb{K}$, then it is a constant. Here we can generalize that with functions inside a disk.

**Theorem 36.4.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^{-}))$) have no pole in $\mathbb{K}$ (respectively in $d(a, R^{-})$). Then $f$ belongs to $\mathcal{A}(\mathbb{K})$ (respectively to $\mathcal{A}(d(a, R^{-}))$).

**Proof.** Suppose $f \in \mathcal{M}(\mathbb{K})$ has no pole in $\mathbb{K}$. By Lemma 36.3, we can write $f$ in the form $\frac{h}{l}$ with $\mathcal{D}(f) = \mathcal{D}(h)$. Since $f$ has no pole in $\mathbb{K}$, $l$ has no zero and hence is a constant, which ends the proof when $f$ belongs to $\mathcal{M}(\mathbb{K})$.

Suppose now that $f$ belongs to $\mathcal{M}(d(a, R^{-}))$ and has no pole in $d(a, R^{-})$. By Proposition 14.12, it is sufficient to show that for each $\rho \in ]0, R[$, $f$ belongs to $H(d(a, \rho))$. Let $f = \frac{h}{l}$, with $h, l \in \mathcal{A}(d(a, R^{-}))$. By Proposition 14.12, both $h, l$ belong to $H(a, \rho)$.

By hypothesis, each zero $\alpha$ of $l$ is a zero of $h$ such that $\omega_{\alpha}(h) \ge \omega_{\alpha}(l)$. Let $P$ be the polynomial admitting for zeros the zeros of $l$ inside $d(a, \rho)$ with the same multiplicity and no other zero. Then $P$ divides $h$ and $l$ in $\mathcal{A}(d(a, R^{-}))$, say $h = P\phi$, $l = P\psi$. So, $\psi$ is a power series that has no zero in $d(a, \rho)$, hence by Theorem 16.9, it is an invertible element of $H(d(a, \rho))$, which ends the proof. $\square$

**Corollary 36.5.** Let $f, g \in \mathcal{A}(\mathbb{K})$ (respectively $f, g \in \mathcal{A}(d(a, R^{-}))$) be such that $\mathcal{D}(g) \le \mathcal{D}(f)$. There exists $h \in \mathcal{A}(\mathbb{K})$ (respectively $h \in \mathcal{A}(d(a, R^{-}))$) such that $f = gh$.
**Proof.** Indeed, $\frac{f}{g}$ belongs to $M(\mathbb{K})$ (respectively to $M(d(a, R^{-}))$) and has no pole. $\square$

**Corollary 36.6.** Let $f \in M(\mathbb{K})$ (respectively $f \in M(d(a, R^{-}))$ have no zero and no pole in $\mathbb{K}$ (respectively in $d(a, R^{-})$). Then it is a constant (respectively an invertible element of $A_b(d(a, R^{-}))$).

**Corollary 36.7.** Let $f, g \in A(\mathbb{K})$ (respectively $f, g \in A(d(a, R^{-}))$ satisfy $D(f) = D(g)$. Then $\frac{f}{g}$ belongs to $\mathbb{K}$ (respectively is invertible in $A_b(d(a, R^{-}))$).

**Corollary 36.8.** Let $f \in A(\mathbb{K})$ be such that $D(f) = (a_n, q_n)_{n \in \mathbb{N}}$ with $a_n \neq 0 \forall n \in \mathbb{N}$, $\lim_{n \to +\infty} |a_n| = +\infty$. Then $f(x)$ is of the form $\lambda \prod_{n=0}^{\infty} \left(1 - \frac{x}{a_n}\right)^{q_n}$ with $\lambda \in \mathbb{K}$.

**Theorem 36.9.** Let $f$ be a function defined in $\mathbb{K}$ such that for every $r > 0$, the restriction of $f$ to $d(0, r^{-})$ belongs to $M(d(0, r^{-}))$. Then, $f \in M(\mathbb{K})$.

**Proof.** Let $(a_n, q_n)_{n \in \mathbb{N}}$ be the sequence of the poles of $f$ in $\mathbb{K}$ of respective order $q_n$, with $|a_n| \le |a_{n+1}|$. Since the sequence $(|a_n|)_{n \in \mathbb{N}}$ tends to $+\infty$, the product $\prod_{n=1}^{\infty} \left(1 - \frac{x}{a_n}\right)^{q_n}$ converges in $\mathbb{K}$ and the function $g(x) = \prod_{n=1}^{\infty} \left(1 - \frac{x}{a_n}\right)^{q_n}$ belongs to $A(d(0, r^{-}))$ for every $r > 0$, hence it is a power series converging in all $\mathbb{K}$, therefore it lies in $A(\mathbb{K})$. Consequently, $f$ lies in $M(\mathbb{K})$. $\square$

By Theorem 28.6, Lemma 36.9 and Corollary 36.10 are immediate:

**Lemma 36.10.** Let $\mathbb{K}$ be spherically complete, let $a \in \mathbb{K}$, $r \in \mathbb{R}_+^*$, and let $\mathcal{B}, \mathcal{C}$ be divisors on $d(a, R^{-})$. There exists $f \in M(d(a, R^{-}))$ such that $D(f) = \mathcal{B}$ and $D(\frac{1}{f}) = \mathcal{C}$.

**Corollary 36.11.** Let $\mathbb{K}$ be spherically complete, let $a \in \mathbb{K}$, $r \in \mathbb{R}_+^*$, and let $f \in M(d(a, R^{-}))$. There exist $g, h \in A(d(a, R^{-}))$, having no common zero, such that $f = \frac{g}{h}$.

**Remark.** If $\mathbb{K}$ is not spherically complete, in the general case, as shown in Theorem 28.1, we cannot find an analytic function $h \in A(d(a, R^{-}))$ such that $D(h) = D(f)$. Consequently, in a field such
as $\mathbb{C}_p$, we can't write $f$ in the form $f = \frac{h}{l}$ with $h, l \in \mathcal{A}(d(a, R^{-}))$, having no common zero (this gap was forgotten in several works).

However, by Theorem 8.4, we can take an algebraically closed spherically complete extension $\tilde{\mathbb{K}}$ of $\mathbb{K}$ and consider $f$ as an analytic function on the disk $\tilde{d}(a, R^{-})$ in the field $\tilde{\mathbb{K}}$: then $f$ may be written in the form $f = \frac{\hat{h}}{\hat{l}}$ with $\hat{h}, \hat{l} \in \mathcal{A}(\hat{d}(a, R^{-}))$, with $\hat{h}, \hat{l}$ having no common zero.

**Theorem 36.12.** Let $r \in \mathbb{R}_+^*$, let $f(x) = \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, r^{-}))$, let $S$ be the set of zeros and poles of $f$ in $\mathbb{K}$ (respectively in $d(a, r^{-})$), let $t$ be the g.c.d. of $\{\omega_\alpha(f) \mid \alpha \in S\}$, and let $n \in \mathbb{N}^*$. If there exists $g \in \mathcal{M}(\mathbb{K})$ (respectively $g \in \mathcal{M}(d(a, r^{-}))$) such that $g^q = f$, then $q$ divides $t$. Conversely, if $q$ divides $t$, then there exists $g \in \mathcal{M}(\mathbb{K})$ such that $g^q = f$ (respectively if $p$ is prime to $q$ and if $q$ divides $t$ then there exists $g \in \mathcal{M}(d(a, R^{-}))$ such that $g^q = f$).

**Proof.** If there exists $g \in \mathcal{M}(d(a, r^{-}))$ (respectively $g \in \mathcal{M}(\mathbb{K})$) such that $g^q = f$, then of course, $\omega_\alpha(g)$ divides $\omega_\alpha(f)$ for every $\alpha \in S$ and hence it divides $t$. Now suppose $q$ divides $t$ and set $t = lq$. For each $\alpha \in S$, $\omega_\alpha(f)$ is of the form $ts_\alpha = qls_\alpha$.

Suppose first $f \in \mathcal{M}(\mathbb{K})$. By Lemma 36.3, in $\mathcal{M}(\mathbb{K})$, there exists $g \in \mathcal{M}(\mathbb{K})$ admitting each zero $\alpha$ of $f$ as a zero of order $ls_\alpha$ and each pole $\alpha$ of $f$ as a pole of order $-ls_\alpha$. Then $\frac{f}{g^q}$ has no zero and no pole in $\mathbb{K}$, hence, by Corollary 36.6, it is a constant $\lambda$. Let $v$ be a $q$-th root of $\lambda$. Then, $f = (vg)^q$.

Now, suppose that $q$ is prime to $p$ and suppose $f \in \mathcal{M}(d(a, R^{-}))$. Suppose first that $\mathbb{K}$ is spherically complete. By Lemma 36.10 there exists $g \in \mathcal{M}(d(a, R^{-}))$ admitting each zero $\alpha$ of $f$ as a zero of order $ls_\alpha$ and each pole $\alpha$ of $f$ as a pole of order $-ls_\alpha$. Then $\frac{f}{g^q}$ has no zero and no pole in $d(a, R^{-})$, hence, it belongs to $\mathcal{A}(d(a, r^{-}))$. But since it has no zero, by Corollary 23.2, it satisfies $|h(x) - h(a)| < |h(a)| \quad \forall x \in d(a, r^{-})$. Let $\psi(x) = \frac{h(x)}{h(a)}$. Then we have $|\psi(x) - 1| < 1 \quad \forall x \in d(a, r^{-})$ and then, since $q$ is prime to $p$, by Theorem 29.23, we can apply the function $\sqrt[q]{\psi(x)}$ to $\psi(x)$ in order to get a function $\sqrt[q]{\psi(x)} \in \mathcal{A}(d(a, r^{-}))$. Now, let $v$ be a $q$-th root of $h(a)$. We have $f(x) = h(a)\psi(x)(g(x))^q = (v^{\frac{n}{q}}\sqrt[q]{\psi(x)}g(x))^q$ which ends the proof when $\mathbb{K}$ is spherically complete.

Consider now the general case, when $\mathbb{K}$ is no longer supposed to be spherically complete. Let $\tilde{\mathbb{K}}$ be a spherically complete algebraically
closed extension of $\mathbb{K}$. The function $f$ has continuation to a function $\hat{f}$ which belongs to $\mathcal{A}(\hat{d}(\alpha, R^{-}))$ and hence there exists a function $g \in \mathcal{A}(\hat{d}(\alpha, R^{-}))$ such that $g^q = f$. Then, by Lemma 14.5, $g$ is a power series that has all coefficients in $\mathbb{K}$ and hence belongs to $\mathcal{A}(d(a, R^{-}))$. $\square$

<figure><img src="image_13.png" /></figure>

**Corollary 36.13.** Let $f(x) \in \mathcal{M}(\mathbb{K})$, let $S$ be the set of zeros and poles of $f$ in $\mathbb{K}$, and let $t$ be the g.c.d. of $\{\omega_\alpha(f) \mid \alpha \in S\}$. Then $t$ is the greatest of the integers $n$ such that there exists $g \in \mathcal{M}(\mathbb{K})$ satisfying $g^n = f$.

**Theorem 36.14.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^{-}))$, respectively $f \in \mathcal{M}(D)$) be constant inside a disk included in $\mathbb{K}$ (respectively in $d(a, R^{-})$, respectively in $D$). Then $f$ is constant in $\mathbb{K}$ (respectively in $d(a, R^{-})$, respectively in $D$).

**Proof.** For a non-identically zero meromorphic function, the zeros and the poles of $f$ are isolated. Consequently, if $f(x)$ is equal to a constant inside a disk, it is constant in the set of definition. $\square$

<figure><img src="image_14.png" /></figure>

**Definition.** Given $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^{-}))$, respectively $f \in \mathcal{M}(D)$), we call divisor of the poles of $f$ on $\mathbb{K}$ (respectively on $d(a, R^{-})$) the divisor of $\frac{1}{f}$ on $\mathbb{K}$ (respectively on $d(a, R^{-})$, respectively on $D$).

**Lemma 36.15.** Let $f \in \mathcal{M}(\mathbb{K})\mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^{-}))\mathcal{A}(d(0, R^{-}))$) and suppose $0$ is not a pole of $f$. Let $r$ be the minimal distance of the poles of $f$ to $0$. Then $f$ belongs to $\mathcal{A}(d(0, r^{-}))$ and its radius of convergence is $r$.

**Proof.** Consider the divisor $T$ of the poles of $f$ on $d(0, R^{-})$. If $f \in \mathcal{M}(\mathbb{K})\mathcal{A}(\mathbb{K})$, there is no problem to write $f$ in the form $\frac{h}{l}$ with $h, l \in \mathcal{A}(\mathbb{K})$, where $l$ has no zero in $d(0, r^{-})$. Consequently, by Corollary 23.2, the restriction of $l$ to $d(0, r^{-})$ is invertible in $\mathcal{A}(d(0, r^{-}))$. Therefore, $\frac{h}{l}$ belongs to $\mathcal{A}(d(0, r^{-}))$ and hence its radius of convergence is $\ge r$. Conversely, since $f$ has a pole in $C(0, r)$, it is not equal to a power series in $x$ in $d(0, r)$ and hence, the radius of convergence is $r$.

Now, suppose $f \in \mathcal{M}(d(0, R^{-}))\mathcal{A}(d(0, R^{-}))$. By Theorem 27.15, we can find a function $l \in \mathcal{A}(d(0, R^{-}))$ such that $\mathcal{D}(l) \ge T$ and such that none of the zeros of $l$ lie in $d(0, R^{-})$. Next, we set $h = fl$ and