The Nevanlinna second main theorem is based on the following theorem:

**Theorem 46.13.** Let $f \in \mathcal{M}(D)$ and let $a_1, \dots, a_q \in \mathbb{K}$ be distinct. Then,

$$ (q-1)T_R(r, f) \le \max_{1 \le k \le q} \left( \sum_{j=1, j \ne k}^{q} Z_R(r, f - a_j) \right) + O(\log(r)) \quad (r \in I). $$

**Proof.** Suppose Theorem 46.13 is wrong. Thus, there exists $f \in \mathcal{M}(D)$ and $a_1, \dots, a_q \in \mathbb{K}$ such that $(q-1)T_R(r, f) - \max_{1 \le k \le q} \left( \sum_{j=1, j \ne k}^{q} Z_R(r, f - a_j) \right)$ admits no superior bound in $]0, +\infty[$. So, there exists a sequence of intervals $J_s = [w_s, y_s]$ such that $w_s < y_s < w_{s+1}$, $\lim_{s \to +\infty} w_s = +\infty$ and two distinct indices $m \le q$ and $t \le q$ such that

$$ \lim_{s \to +\infty} \left[ \inf_{r \in J_s} \frac{(T_R(r, f) - Z_R(r, f - a_m))}{\log(r)} \right] = +\infty $$

and

$$ \lim_{s \to +\infty} \left[ \inf_{r \in J_s} \frac{(T_R(r, f) - Z_R(r, f - a_t))}{\log(r)} \right] = +\infty. $$

But by Lemma 46.12, that is impossible.

<figure><img src="image_20.png" /></figure>

We can now state and prove the second main theorem for $\mathcal{M}(D)$.

**Theorem 46.14.** Let $\mathbb{K}$ be of characteristic 0. Let $f \in \mathcal{M}(D)$, let $\alpha_1, \dots, \alpha_q \in \mathbb{K}$, with $q \ge 2$, and let $W = \{\alpha_1, \dots, \alpha_q\}$. Then,

$$ (q-1)T_R(r, f) \le \sum_{j=1}^{q} \bar{Z}_R(r, f - \alpha_j) + Z_R(r, f') - Z_R^W(r, f') + O(\log(r)) \quad (r \in I). $$

Moreover, if $f$ belongs to $\mathcal{A}(D)$, then

$$ qT_R(r, f) \le \sum_{j=1}^{q} \bar{Z}_R(r, f - \alpha_j) + Z_R(r, f') - Z_R^W(r, f') + O(\log(r)) \quad (r \in I). $$
**Theorem 46.15 (Second Main Theorem).** Let $K$ be of characteristic 0. Let $f \in \mathcal{M}(D)$, let $\alpha_1, \dots, \alpha_q \in K$, $q \ge 2$, and let $W = \{\alpha_1, \dots, \alpha_q\}$. Then,

$$ (q-1)T_R(r, f) \le \sum_{j=1}^{q} \overline{Z}_R(r, f - \alpha_j) + \overline{N}_R(r, f) - Z_R^W(r, f') + O(\log(r)) \quad (r \in I). $$

**Proof.** (Proof of Theorems 46.14 and 46.15) By Theorem 46.13, there exists a constant $B > 0$, and for each $r > R$, there exists $k(r) \in \mathbb{N}$, $k(r) \le q$, such that

$$ (q-1)T_R(r, f) \le \sum_{j=1, j \ne k(r)}^{q} Z_R(r, f - a_j) + B \log(r). $$

Now, since $K$ is of characteristic 0, we have

$$ \sum_{j=1}^{q} Z_{R}(r, f-a_{j}) = \sum_{j=1}^{q} \overline{Z}_{R}(r, f-a_{j}) + Z_{R}(r, f') - Z_{R}^{W}(r, f') + B \log(r). $$

Consequently,

$$ (9) \qquad (q-1)T_R(r, f) \le \sum_{j=1}^{q} \overline{Z}_R(r, f - a_j) + Z_R(r, f') - Z_R^W(r, f') \\ \qquad \qquad \qquad \qquad - Z_R(r, f - a_{k(r)}) + O(\log(r)) $$

and this proves the first claim of Theorem 46.14.

Particularly, if $f \in \mathcal{A}(D)$, then we have $Z_R(r, f - a_j) = T_R(r, f - a_j) = T_R(r, f) + O(\log(r)) \quad \forall j = 1, \dots, q$, hence $Z_R(r, f - a_{k(r)}) = T_R(r, f) + O(\log(r))$, and therefore,

$$ qT_R(r, f) \le \sum_{j=1}^{q} \overline{Z}_R(r, f - a_j) + Z_R(r, f') - Z_R^W(r, f') + O(\log(r)), $$

which ends the proof of Theorem 46.14.

Consider now the situation in Theorem 46.15. By Theorem 46.9, for each $j = 1, \dots, q$, there exists a constant $B_j > 0$ such that $Z_R(r, f') \le Z_R(r, f - a_j) + \overline{N}_R(r, f - a_j) + B_j \log(r)$. Consequently, there exists a constant $C > 0$ such that $Z_R(r, f') \le Z_R(r, f - a_{k(r)}) + \overline{N}_R(r, f - a_{k(r)}) + C \log(r) \quad \forall r > R$.
Therefore, by Relation (9), that remains true in Theorem 46.15, we can derive

$$ (q-1)T_R(r, f) \le \sum_{j=1}^{q} \overline{Z}_R(r, f - \alpha_j) + \overline{N}_R(r, f) - Z_R^W(r, f') + O(\log(r)) \quad \forall r \in I. \quad \square $$

**Corollary 46.16.** Let $f \in \mathcal{M}(\mathbb{K})$ and let $a_1, \dots, a_q \in \mathbb{K}$ be distinct. Then, $(q-1)T_R(r, f) \le \sum_{j=1}^q Z_R(r, f - a_j) + O(\log(r)) \quad (r \in I)$.

**Corollary 46.17.** Let $\mathbb{K}$ be of characteristic 0. Let $f \in \mathcal{M}(D)$, let $\alpha_1, \dots, \alpha_q \in \mathbb{K}$, with $q \ge 2$, and let $W = \{\alpha_1, \dots, \alpha_q\}$. Then

$$ \sum_{j=1}^{q} (Z_R(r, f - \alpha_j) - \overline{Z}_R(r, f - \alpha_j)) \le T_R(r, f) + \overline{N}_R(r, f) - Z_R^W(r, f') + O(\log(r)) \quad (r \in I). $$
# Chapter 47

## Immediate Applications of the
Nevanlinna Theory

Throughout this chapter, the field $\mathbb{K}$ is supposed to be of characteristic 0.

**Notation.** In all the chapters, the field $\mathbb{K}$ is supposed to be of characteristic 0. As in Chapter 41, we denote by $D$ the set $\mathbb{K}\backslash d(0, R^{-})$ with $R$ a positive number. The definitions of $\mathcal{A}(D)$, $\mathcal{A}^{c}(D)$, $\mathcal{M}(D)$, and $\mathcal{M}^{c}(D)$ are those given in Chapter 46.

As immediate applications of the second main theorem, we can note Theorems 47.1–47.4.

**Theorem 47.1.** Let $a_1, a_2 \in \mathbb{K}(a_1 \neq a_2)$ and let $f, g \in \mathcal{A}(\mathbb{K})$ satisfy $f^{-1}(\{a_i\}) = g^{-1}(\{a_i\})(i=1,2)$. Then, $f=g$.

**Remark.** Theorem 47.1 does not hold in complex analysis. Indeed, let $f(z) = e^z$, $g(z) = e^{-z}$, and let $a_1 = 1, a_2 = -1$. Then, $f^{-1}(\{a_i\}) = g^{-1}(\{a_i\}) (i = 1, 2)$, though $f \neq g$.

**Theorem 47.2.** Let $a_1, a_2, a_3 \in \mathbb{K}$ ($a_i \neq a_j \forall i \neq j$) and let $f, g \in \mathcal{A}_u(d(a, R^{-}))$ (respectively, $g \in \mathcal{A}^c(D)$) satisfy $f^{-1}(\{a_i\}) = g^{-1}(\{a_i\})(i=1,2,3)$. Then, $f=g$.

**Theorem 47.3.** Let $a_1, a_2, a_3, a_4 \in \mathbb{K}(a_i \neq a_j \forall i \neq j)$ and let $f, g \in \mathcal{M}(\mathbb{K})$ satisfy $f^{-1}(\{a_i\}) = g^{-1}(\{a_i\})(i=1, 2, 3, 4)$. Then, $f=g$.
**Theorem 47.4.** Let $a_1, a_2, a_3, a_4, a_5 \in \mathbb{K}(a_i \neq a_j \forall i \neq j)$ and let $f, g \in \mathcal{M}_u(d(a, R^{-}))$ (respectively $f, g \in \mathcal{M}^c(D)$) satisfy $f^{-1}(\{a_i\}) = g^{-1}(\{a_i\})$ ($i = 1, 2, 3, 4, 5$). Then, $f = g$.

**Remark.** Let $f(x) = \frac{x}{3x-1}$, $g(x) = \frac{x^2}{x^2+2x-1}$. Let $a_0 = 0$, $a_1 = 1$, $a_2 = \frac{1}{2}$. Then we can check that $f^{-1}(\{a_i\}) = g^{-1}(\{a_i\})$, $i = 1, 2, 3$. So, Theorem 47.3 is sharp.

**Proof.** (Theorems 47.1–47.4) Let $I =]0, +\infty[$ in Theorems 47.1 and 47.3 and let $I =]0, R[$ in Theorems 47.2 and 47.4. For each $j = 1, \dots, n$, let $S_j$ be the set of all zeros of $f - a_j$ (without taking multiplicities into account). Since $a_i \neq a_j \forall i \neq j$, we have $S_i \cap S_j = \emptyset \forall i \neq j$. Next, we note that $f(x) = a_j$ implies $f(x) - g(x) = 0$. Consequently, we check the following:

$$ (1) \sum_{j=1}^{n} \bar{Z}(r, f - a_j) \leq \bar{Z}(r, f - g). $$

Suppose first that $f$ and $g$ either belong to $\mathcal{A}(K)$ or belong to $\mathcal{A}(d(0, R^{-}))$.

By applying Theorem 45.7 to $f$, we obtain

$$ \begin{aligned} (n-1)T(r, f) &\leq \sum_{j=1}^{n} \bar{Z}(r, f - a_j) + \bar{N}(r, f) - \log(r) + O(1) \\ &\leq n\bar{Z}(r, f - g) + \bar{N}(r, f) - \log(r) + O(1) \quad (r \in I), \end{aligned} $$

hence by (1),

$$ (n-1)T(r, f) \leq T(r, f - g) + \bar{N}(r, f) - \log(r) + O(1) \quad (r \in I), $$

and finally

$$ (n-1)T(r, f) \leq T(r, f - g) + N(r, f) - \log(r) + O(1) \quad (r \in I). $$

Similarly,

$$ (n-1)T(r, g) \leq T(r, f - g) + N(r, g) - \log(r) + O(1) \quad (r \in I), $$

therefore we obtain

$$ (2) \quad (n-1) \max(T(r, f), T(r, g)) \leq T(r, f - g) + \max(N(r, f), N(r, g)) - \log(r) + O(1) \quad (r \in I). $$
Assume we are in the hypothesis of Theorem 47.1. We have $N(r, f) = N(r, g) = 0$, and by Theorem 43.3, $T(r, f - g) \le \max(T(r, f), T(r, g)) + O(1)$. Consequently, by (2),

$$ (n-1)\max(T(r, f), T(r, g)) \\ \le \max(T(r, f), T(r, g)) - \log(r) + O(1) \quad (r \in I). $$

Since $r$ is not bounded, we can see that the inequality does not hold with $n = 2$, when $r$ goes to $+\infty$.

Now, assume the hypothesis of Theorem 47.2. Again, we have $N(r, f) = N(r, g) = 0$, and by Theorem 43.3, $T(r, f - g) \le \max(T(r, f), T(r, g)) + O(1)$, hence by (2),

$$ (n-1)\max(T(r, f), T(r, g)) \le \max(T(r, f), T(r, g)) + O(1) \quad (r \in I). $$

Since $f, g$ are unbounded, by Theorem 43.10, so are $T(r, f)$, $T(r, g)$ in intervals $]r_0, R[$, hence the inequality does not hold with $n = 3$.

Suppose now that $f$ and $g$ belong to $A^c(D)$. We then obtain

$$ (n-1)T_R(r, f) \le T_R(r, f-g) + O(\log(r)) \quad (r > R) \\ \le \max(T_R(r, f), T_R(r, g)) + O(\log(r)), $$

and similarly,

$$ (n-1)T_R(r, g) \le T_R(r, f-g) + O(\log(r)) \quad (r > R) \\ \le \max(T_R(r, f), T_R(r, g)) + O(\log(r)), $$

therefore

$$ (n-1)\max(R_R(r, f), T(r, g)) \le \max(T_R(r, f), T_R(r, g)) + O(\log(r)), $$

and hence $n \le 2$, which proves the conclusion whenever $n \ge 3$.

Assume now the hypothesis of Theorem 47.3. Since

$$ \max(N(r, f), N(r, g)) \le \max(T(r, f), T(r, g)), $$

by (2) and Theorem 43.3, we have

$$ (n-1)\max(T(r, f), T(r, g)) \\ \le 3\max(T(r, f), T(r, g)) - \log(r) + O(1) \quad (r \in I). $$

Since $r$ is not bounded, the inequality does not hold with $n = 4$, when $r$ goes to $+\infty$.
Finally, assume we are in the hypothesis of Theorem 47.4. Suppose first that $f$ and $g$ belong to $M_u(d(0, R^{-}))$. By (2) and Theorem 43.3, we have

$$ (n-1) \max(T(r, f), T(r, g)) \leq 3 \max(T(r, f), T(r, g)) + O(1) \quad (r \in I). $$

Since $T(r, f)$, $T(r, g)$ are not bounded, the inequality does not hold with $n=5$.

And now, suppose that $f$ and $g$ belong to $M^c(D)$. We then obtain

$$ \begin{aligned} (n-1)T_R(r, f) &\leq T_R(r, f-g) + O(\log(r)) \quad (r > R) \\ &\leq 3(T_R(r, f) + T_R(r, g)) + O(\log(r)), \end{aligned} $$

and similarly,

$$ \begin{aligned} (n-1)T_R(r, g) &\leq T_R(r, f-g) + O(\log(r)) \quad (r > R) \\ &\leq 3(T_R(r, f) + T_R(r, g)) + O(\log(r)), \end{aligned} $$

therefore

$$ (n-1)(T_R(r, f) + T_R(r, g)) \leq 3(T_R(r, f) + T_R(r, g)) + O(\log(r)), $$

and hence $n \leq 4$, which proves the conclusion whenever $n \geq 5$. That finishes the proof of Theorem 47.4. □

**Definitions.** Let $f \in \mathcal{M}(\mathbb{K})$. The function $f$ is called a function of uniqueness (respectively a function of strong uniqueness) for a family $\mathcal{F}$ of functions defined in a suitable subset of $\mathbb{K}$ if given any two functions $f, g \in \mathcal{F}$ satisfying $h \circ f = h \circ g$ (respectively $h \circ f = b(h \circ g)$ with $b \in \mathbb{K}^*$), then $f$ and $g$ are identical.

Similarly, we consider the same question in the purely algebraic context. Let $E$ be an algebraically closed field, let $h \in E(x)$, and let $\mathcal{F}$ be a subset of $E(x)$. Then $h$ is called a function of uniqueness for $\mathcal{F}$ (respectively a function of strong uniqueness for $\mathcal{F}$) if given any two functions $f, g \in \mathcal{F}$ satisfying $h \circ f = h \circ g$ (respectively $h \circ f = b(h \circ g)$, with $b \in E^*$), $f$ and $g$ are identical.

Particularly in each case, if $h$ is a polynomial, it is called a polynomial of uniqueness for the family $\mathcal{F}$ (respectively a polynomial of strong uniqueness for the family $\mathcal{F}$).
In Theorem 47.6, we need the following basic lemma [59]:

**Lemma 47.5.** Let E be an algebraically closed field of characteristic 0 and let $P(x) = (n-1)^2(x^n-1)-n(n-2)(x^{n-1}-1)^2 \in E[x]$. Then P admits 1 as a zero of order 4 and all other zeros $u_j$ ($1 \le j \le 2n-6$) are simple.

**Theorem 47.6.** Let

$$Q(x) = b((n+2)(n+1)x^{n+3} - 2(n+3)(n+1)x^{n+2} + (n+3)(n+2)x^{n+1})$$

with $b \in \mathbb{K}^*$. Let $R \in ]0, +\infty[$. Then $Q$ is a polynomial of uniqueness for $M(\mathbb{K})$ for every $n \ge 2$ and is a polynomial of uniqueness for $M_u(d(0, R^-))$ and for $M^c(D)$ for every $n \ge 3$.

**Proof.** Suppose $f, g \in M(\mathbb{K})$ (respectively $f, g \in M_u(d(0, R^-)$), respectively $f, g \in M^c(D)$) and suppose that $Q(f) = Q(g)$. Let $h = \frac{f}{g}$. We can derive

$$(n+2)(n+1)(h^{n+3}-1)g^2 - 2(n+3)(n+1)(h^{n+2}-1)g + (n+3)(n+2)(h^{n+1}-1) = 0.$$

If $h$ is a constant, it is 1, a contradiction. So, we suppose $h$ is not constant. If $g$ lies in $M(\mathbb{K})$, so does $h$. Now, if $g$ belongs to $M_u(d(0, R^-))$ or to $M^c(D)$, so does $h$, respectively. Indeed, suppose that $h \in M_b(d(0, R^-))$. Then clearly we have $T(r, (n+2)(n+1)(h^{n+2})g^2) \ge 2T(r, g) + O(1)$, while $T(r, -2(n+3)(n+1)(h^{n+2}-1)g + (n+3)(n+2)(h^{n+1}-1)) \le T(r, g) + O(1)$, a contradiction. Similarly, if $h$ has finitely many zeros and poles, we have the same contradiction.

Let $P(x) = (n+2)^2(x^{n+3})-1-(n+3)(n+1)(x^{n+2}-1)^2 \in \mathbb{K}[x]$. By Lemma 47.5, $P$ admits 1 as a zero of order 4 and all other zeros $u_j$ ($1 \le j \le 2n$) are simple. By change of variable, we can obviously assume that $h - u_j$ has no zero and no pole at 0. Consequently, we check that

$$\left(g - \left(\frac{n+3}{n+2}\right) \left(\frac{h^{n+2}-1}{h^{n+3}-1}\right)\right)^2 = \frac{(n+3)(h-1)^4 \prod_{j=1}^{2n}(h-u_j)}{(n+2)^2(n+1)(h^{n+3}-1)^2}.$$

Since $\frac{(n+3)(h-1)^4 \prod_{j=1}^{2n}(h-u_j)}{(n+2)^2(n-1)(h^{n+3}-1)^2}$ is equal to a square, clearly each zero of $h-u_j$ ($1 \le j \le 2n$) has order at least 2. Let $J = ]0, +\infty[$ (respectively
$J = ]0, R[$, respectively $J = [R, +\infty[$). Consequently,

$$ \sum_{j=1}^{2n} \bar{Z}(r, h - u_j) \le \frac{1}{2} \sum_{j=1}^{2n} Z(r, h - u_j) \le \frac{1}{2}(2n)T(r, h) + O(1) \quad (r \in J). $$

Suppose first that $f$ and $g$ belong to $\mathcal{M}(\mathbb{K})$ or to $\mathcal{M}(d(0, R^{-}))$. Then, applying Theorem 45.7 to $h$ at the points $u_j$ ($1 \le j \le 2n$), we obtain

$$ \begin{aligned} (2n - 1)T(r, h) &\le \sum_{j=1}^{2n} \bar{Z}(r, h - u_j) + \bar{N}(r, h) - \log(r) + O(1) \\ &\le \frac{1}{2} \sum_{j=1}^{2n} Z(r, h - u_j) + \bar{N}(r, h) + O(1) \\ &\le \frac{1}{2}(2n)T(r, h) + \bar{N}(r, h) - \log(r) + O(1) \quad (r \in J), \end{aligned} $$

and therefore $(2n - 1)T(r, h) \le nT(r, h) + T(r, h) - \log(r) + O(1)$. If $f, g$ belong to $\mathcal{M}(\mathbb{K})$, we conclude that $n \le 1$. And if $f, g$ belong to $\mathcal{M}_u(d(0, R^{-}))$, we conclude that $n \le 2$.

Suppose now that $f$ and $g$ belong to $\mathcal{M}^c(D)$. Then we can apply Theorem 46.13 and we have $(2n - 1)T_R(r, h) \le nT_R(r, h) + T_R(r, h) + O(\log(r))$, therefore we have again $n \le 2$, which ends the proof. $\square$

**Corollary 47.7.** Let $P(x) \in \mathbb{K}[x]$ have a derivative of the form $c(x - a)^n(x - b)^2$. Then $P$ is a polynomial of uniqueness for $\mathcal{M}(\mathbb{K})$ $\forall n \ge 2$ and is a polynomial of uniqueness for $\mathcal{M}_u(d(\alpha, R^{-}))$ and for $\mathcal{M}^c(D)$ for every $n \ge 3$.

**Theorem 47.8.** Let $Q(x) = x^{n+1} - x^n$. Let $a \in \mathbb{K}$ and $R \in ]0, +\infty[$. Then $Q$ is a polynomial of uniqueness for $\mathcal{A}(\mathbb{K})$, for $\mathcal{A}_u(d(0, R^{-}))$, and for $\mathcal{A}^c(D)$ for every $n \ge 2$.

**Proof.** Let $f, g \in \mathcal{A}(\mathbb{K})$ (respectively $f, g \in \mathcal{A}_u(d(0, R^{-}))$, respectively $f, g \in \mathcal{A}^c(D)$). Suppose $f^n(f - 1) = g^n(g - 1)$. Let $h = \frac{1}{g}$ and suppose $h$ is not the constant 1. Then we have $g = (\frac{h^n - 1}{h^{n+1} - 1})$. Consequently, if $h$ belongs to $\mathbb{K}$ (respectively to $\mathcal{M}_b(d(0, R^{-}))$, respectively to $\mathcal{M}_0(D)$), so does $g$, a contradiction. Thus, $h$ belongs to $\mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ (respectively to $\mathcal{M}_u(d(0, R^{-}))$, respectively to $\mathcal{M}^c(D)$).
Now, since $n \ge 2$, by Corollary 36.23, $h$ has to take at least one of the $(n+1)$th roots of 1 other than 1 and such an $(n+1)$th root of 1 cannot be an $n$th root of 1. Consequently, $g$ admits a pole, a contradiction. Therefore, $h$ is identically equal to 1 and hence $f = g$. $\square$

Now, we must examine the situation in $\mathcal{M}(D)$ in order to obtain a result similar to Theorem 47.8

**Theorem 47.9.** Let $Q(x) = x^{n+1} - x^n$. Let $a \in \mathbb{K}$ and $R \in [0, +\infty[$. Then, $Q$ is a polynomial of uniqueness for $\mathcal{A}^c(D)$ for every $n \ge 2$.

**Proof.** Let $f, g \in \mathcal{A}^c(D)$. Suppose $f^n(f-1) = g^n(g-1)$. Let $h = \frac{f}{g}$ and suppose $h$ is not the constant 1. Then, we have $g = \left(\frac{h^n-1}{h^{n+1}-1}\right)$. Consequently, if $h$ does not belong to $\mathcal{M}^c(D)$, neither does $g$, a contradiction. Thus, $h$ belongs to $\mathcal{M}^c(D)$).

Now, since $n \ge 2$, by Theorem 39.4, $h$ has to take at least one of the $(n+1)$th roots of 1 other than 1 and such an $(n+1)$th root of 1 cannot be an $n$th root of 1. Consequently, $g$ admits a pole, a contradiction. Therefore, $h$ is identically equal to 1, and hence, $f = g$. $\square$

We must also note Theorems 47.10 and 47.11:

**Theorem 47.10.** Let $f, g \in \mathcal{M}(\mathbb{K})$ satisfy $g^m + f^n = 1$, with $\min(m, n) \ge 2$, $\max(m, n) \ge 3$. Then $f$ and $g$ are constant. Moreover, if $f, g \in \mathcal{A}(\mathbb{K})$ and satisfy $g^m + f^n = 1$, with $\min(m, n) \ge 2$, then $f$ and $g$ are constant.

**Theorem 47.11.** Let $f, g \in \mathcal{M}(d(a, R^-))$ satisfy $g^m + f^n = 1$, with $\min(m, n) \ge 3$ and $\max(m, n) \ge 4$. Then $f$ and $g$ belong to $\mathcal{M}_b(d(a, R^-))$. Moreover, if $\mathbb{K}$ has characteristic different from 2, and if $f$ and $g$ belong to $\mathcal{A}(d(a, R^-))$ and satisfy $g^m + f^n = 1$, with $\min(m, n) \ge 2$, then $f$ and $g$ belong to $\mathcal{A}_b(d(a, R^-))$.

**Proof.** (Theorems 47.10 and 47.11) We first assume $\max(m, n) \ge 3$ and that $f, g$ belong to $\mathcal{M}(K)\backslash\mathbb{K}$ in Theorem 46.3 and belong to $\mathcal{M}_u(d(a, R^-))$ in Theorem 47.11.
Let $a_1, \dots, a_n$ be the $n$th roots of 1. Applying Theorem 45.7 to $f$ at the points $a_1, \dots, a_n$, we obtain the following:

$$ (1) \quad (n-1)T(r, f) \le \sum_{j=1}^{n} \overline{Z}(r, f-a_i) + \overline{N}(r, f) - \log r + O(1), \quad (r \in I). $$

Clearly, each zero of $f^n - 1$ has order $\ge m$. Hence, for each $j = 1, \dots, n$, each zero of $f - a_j$ has order $\ge m$. Consequently,

$$ (2) \quad \overline{Z}(r, f - a_i) \le \frac{1}{m} Z(r, f - a_i). $$

Therefore, in Theorem 47.10, by (1), we have

$$ (n-1)T(r, f) \le \left(\frac{1}{m}\right) \sum_{j=1}^{n} Z(r, f-a_i) + \overline{N}(r, f) - \log r + O(1) \quad (r \in I), $$

and therefore, we can see that $nm < n + m + 1$, hence $\max(n, m) \le 2$, which contradicts the general hypothesis of Theorem 47.10. Consequently, $f$ and $g$ are constants.

In the proof of Theorem 47.11, without loss of generality, we can obviously suppose that $a = 0$. Assuming now that $f, g$ belong to $M_u(d(0, R^-))$, Relation (1) holds again, whereas $r$ now is bounded and hence we now have $nm \le n + m + 1$, hence $\max(n, m) \le 3$ and $\min(m, n) \le 2$, a contradiction to the general hypothesis of Theorem 47.11, therefore $f$ and $g$ belong to $M_b(d(a, R^-))$.

Now, in Theorem 47.10, suppose that $f$ and $g$ belong to $A(\mathbb{K})$. Instead of (1), we obtain

$$ (3) \quad (n-1)T(r, f) \le \sum_{j=1}^{n} \overline{Z}(r, f - a_i) - \log r + O(1) \quad (r \in I), $$

and hence by (2), we derive $nm < n + m$, a contradiction as soon as $\min(m, n) \ge 2$. Hence, $f$ and $g$ are constants.

Finally, in Theorem 47.11, suppose that $f$ and $g$ belong to $A_u(d(0, R^-))$. Then, by (1) again, we have $nm \le n + m$, therefore $\max(m, n) \le 2$, hence $m = n = 2$. But then, by Theorem 14.22, we know that $f$ and $g$ are bounded in $d(0, R^-)$. This finishes the proof. $\square$
# Chapter 48

## Applications to Curves

Throughout this chapter, the field $K$ is supposed to be of characteristic 0.

We examine particular cases where curves are defined by their equations so that, for most of them, the $p$-adic Nevanlinna theory lets us find easy proofs. Most of results come from [24].

**Definition.** Let $F(x, y) \in K[x, y]$. A point $(a, b)$ of the algebraic curve of equation $F(x, y) = 0$ is called a singular point if $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = 0$. An algebraic curve is said to be degenerate if it admits a singular point. An algebraic curve of degree 2 (respectively 3) is called a conic (respectively an elliptic curve).

Throughout this chapter, the field $K$ is supposed to have characteristic 0.

**Remark.** Let $P, Q \in K[x]$. A point $(\alpha, \beta) \in K^2$ is a singular point of the curve of equation $P(x) = Q(y)$ if and only if $P(\alpha) = Q(\beta)$ and $P'(\alpha) = Q'(\beta) = 0$.

First, Theorem 48.1 is classical:

**Theorem 48.1.** Let $\Lambda$ be a non-degenerate conic with a center in $K$ and let $f, g \in \mathcal{A}(d(a, R^{-}))$ be such that $(f(u), g(u)) \in \Lambda \ \forall u \in D$. Then $f$ and $g$ belong to $\mathcal{A}_b(d(a, R^{-}))$. Moreover, if both $f$ and $g$ lie in $\mathcal{A}(K)$, then they are constant.

**Proof.** If the equation of $\Lambda$ is of the form $ax^2 + by^2 = t$, with $t \in K$, the proof is provided by Theorem 14.22. Now, we consider the
general case. Since $\Lambda$ has a center, by a suitable change of variable of the form $u = x - \alpha$, $w = y - \beta$, we may assume that the equation is of the form $au^2 + bw^2 + 2cuw = M$. Moreover, since $\Lambda$ is non-degenerate, we know that $M \neq 0$. We first note that if $a = b = 0$, then $f$ and $g$ are invertible in $\mathcal{A}(\mathbb{K})$ and therefore are constant. Thus, since $a$ and $b$ play the same role, we may assume, for instance, $b \neq 0$. In this way, $g$ is a solution (in $\mathcal{A}(\mathbb{K})$) of the equation $bY^2 + (2cf)Y + (af^2 - M) = 0$. Thereby, the reduced discriminant $f^2(c^2 - ab) + bM$ must be equal to $(bg - cf)^2$. Consequently, putting $h = bg - cf$, we check that $h^2 = f^2(c^2 - ab) + bM$. Thus, $f$ and $h$ satisfy an equation of the form $\lambda f^2 + \mu h^2 = l \in \mathbb{K}$. But since $bM \neq 0$, and since $h$ lies in $\mathcal{A}(\mathbb{K})$, we come back to the first case, proving that $f$ and $h$ are constant and therefore so is $g$. $\square$

**Remark.** The $p$-adic functions sin and cos are bounded inside $d(0, (p^{-\frac{1}{p-1}})^{-})$ when the residue characteristic is $p$ (respectively inside $d(0, 1^{-})$ when the residue characteristic is $0$) and satisfy $\sin^2 x + \cos^2 x = 1$. Throughout this chapter, we denote by $D$ an infinite bounded set included in a disk $d(a, r)$, for some $r < R$.

The second main theorem gives an easy proof of the impossibility to parametrize elliptic and hyperelliptic curves.

**Theorem 48.2.** Let $a_i \in \mathbb{K}, i = 1, 2, 3$, be pairwise distinct. There do not exist $f, g \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ and there do not exist $f, g \in \mathcal{A}_u(d(a, R^{-}))$ such that $(g(x))^2 = (f(x) - a_1)(f(x) - a_2)(f(x) - a_3)$.

**Proof.** Suppose two functions $f, g \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ satisfy $(g(x))^2 = (f(x) - a_1)(f(x) - a_2)(f(x) - a_3)$. Then each zero of $f - a_i, i = 1, 2, 3$, must be of order at least 2. And each pole of $(f - a_1)(f - a_2)(f - a_3)$ is a pole of $g^2$ and hence is of even order, and hence each pole of $f$ is at least of order 2. Consequently, $\bar{Z}(r, f) \le \frac{1}{2}Z(r, f)$ and $\bar{N}(r, f) \le \frac{1}{2}N(r, f)$. Now, applying the second main theorem, we have

$$2T(r, f) \le \sum_{i=1}^{3} \bar{Z}(r, f - a_i) + \bar{N}(r, f) - \log r + O(1),$$

and hence $2T(r, f) \le 2T(r, f) - \log r + O(1)$, a contradiction when $r$ goes to $+\infty$. Next, if $f, g$ belong to $\mathcal{A}_u(d(a, R^{-}))$, then assuming
$a = 0$, we have

$$2T(r, f) \leq \sum_{i=1}^{3} \bar{Z}(r, f - a_i) + O(1),$$

hence

$$2T(r, f) \leq \frac{\sum_{i=1}^{3} T(r, f - a_i)}{2} + O(1),$$

a contradiction.

□

**Theorem 48.3.** Let $P, Q$ be in $K[x]$ with $2 \leq \deg(P) \leq \deg(Q) = 3$. There exist non-constant meromorphic functions $f, g \in \mathcal{M}(K)$ such that $P(f) = Q(g)$ if and only if $P'$ has a zero $\alpha$ and $Q'$ has a zero $\beta$ satisfying $P(\alpha) = Q(\beta)$.

**Proof.** Let $\Lambda$ be the curve of equation $P(x) = Q(y)$. Since $\Lambda$ is a curve of degree 3, it is an elliptic curve, and hence by definition, it has a singular point if and only if there exists $(\alpha, \beta) \in \Lambda$ such that $P'(\alpha) = Q'(\beta) = 0$. By results in algebraic geometry, its genus is 0 if and only if the curve has a singular point. If the curve has genus zero, then there exist $f, g \in K(x)$ such that $P(f(t)) = Q(g(t))$. If the curve has no singular point, it is of genus 1, and then by classical results in algebraic geometry, it is birationally equivalent to a curve of equation $y^2 = G(f(t))$ with $G$ a polynomial of degree 3 with three distinct zeros. Consequently, by Theorem 48.2, there exist no non-constant meromorphic functions $f, g \in \mathcal{M}(K)$ such that $(g(t))^2 = G(f(t))$, and therefore, by birational equivalence, there exist no no-constant meromorphic functions $f, g \in \mathcal{M}(K)$ such that $P(f) = Q(g)$. □

**Theorem 48.4.** Let $a_i \in K$, $i = 1, \dots, q$, $q \geq 5$, be pairwise distinct, and let $R \in \mathbb{R}_+^*$. Let $m \in \mathbb{N}^*$ be $\geq 2$ and prime to $q$. There do not exist $f, g \in \mathcal{M}(K) \setminus K$ and there do not exist $f, g \in \mathcal{M}_u(d(0, R^-))$ such that $(g(x))^m = \prod_{i=1}^q (f(x) - a_i)$.

**Proof.** The proof is similar to that of Theorem 48.2. Suppose two functions $f, g \in \mathcal{M}(K) \setminus K$ satisfy $(g(x))^m = \prod_{i=1}^q (f(x) - a_i)$. Then each zero of $f - a_i$, $i = 1, \dots, q$, must be of order at least 2. And each pole of $\prod_{i=1}^q (f(x) - a_i)$ is a pole of $g^m$. But since $m$ is prime to $q$, each pole of $f$ is of order at least 2. Consequently, we have again
$\overline{Z}(r, f) \le \frac{1}{2}Z(r, f)$ and $\overline{N}(r, f) \le \frac{1}{2}N(r, f)$. Now, applying Theorem 45.7, we have

$$ (q-1)T(r, f) \le \sum_{i=1}^{q} \overline{Z}(r, f - a_i) + \overline{N}(r, f) - \log r + O(1), $$

and hence $(q-1)T(r, f) \le (\frac{q+1}{2})T(r, f) - \log r + O(1)$, a contradiction whenever $f, g \in \mathcal{M}(\mathbb{K})$ or whenever $f, g \in \mathcal{M}_u(d(0, R^-))$. $\square$

**Corollary 48.5.** Let $\Lambda$ be an algebraic curve on $\mathbb{K}$ of genus 1 and let $f, g \in \mathcal{M}(\mathbb{K})$ be such that $(f(u), g(u)) \in \Lambda \ \forall u \in D$. Then $f$ and $g$ are constants.

**Corollary 48.6.** Let $\Lambda$ be an algebraic curve on $\mathbb{K}$ of genus 1 or 2 on $\mathbb{K}$ and let $f, g \in \mathcal{A}(d(a, R^-))$ be such that $(f(u), g(u)) \in \Lambda \ \forall u \in D$. Then $f$ and $g$ are bounded.

**Proof.** (Corollaries 48.5 and 48.6) Indeed, every algebraic curve of genus 1 is birationally equivalent to an non-degenerate elliptic curve. So, we can apply Theorem 48.2 to prove Corollaries 48.5 and 48.6. $\square$

<figure><img src="image_21.png" /></figure>

Here we take this opportunity to recall that there exists no param- eterization of conics with a center, by entire functions, on the field $\mathbb{K}$. Such a result cannot be extended to bounded analytic functions as shown by the functions sin and cos defined in $d(0, (p^{-\frac{1}{p-1}})^{-})$ when the residue characteristic of $\mathbb{K}$ is $p$ (respectively $d(0, 1^{-})$ when the residue characteristic of $\mathbb{K}$ is 0).

We now consider more general algebraic equations of curves.

**Theorem 48.7.** Let $P, Q \in \mathbb{K}[X]$ be two relatively prime polyno- mials of degrees $s$ and $t$ respectively, let $n$ be the number of dis- tinct zeros of $Q$, let $m \in \mathbb{N}^*$, and let $g \in \mathcal{M}(\mathbb{K})$ (respectively $g \in \mathcal{M}(d(a, R^-)))$ be a non-constant function all poles of which have order $\ge m$. Suppose that there exists a function $f \in \mathcal{M}(\mathbb{K})$ (respec- tively $f \in \mathcal{M}(d(a, R^-)))$ satisfying $g(x)Q(f(x)) = P(f(x))$ for every $x \in D$ which is not a pole of $f$ or $g$:

(i) Assume that $f \in \mathcal{M}_u(d(a, R^-))$. Then $mn \le t + 2m$. Moreover, if $s > t$, then $mn \le \min(t + 2m, s + m)$.

(ii) Assume $f, g \in \mathcal{M}(\mathbb{K})$. Then $mn < t + 2m$. Moreover, if $s > t$, then $mn < \min(t + 2m, s + m)$.
**Proof.** If $n < 2$, the inequality $mn < t + 2m$ is trivial. So, we suppose $n \ge 2$. Let $Q(X) = \prod_{j=1}^n (X - b_j)^{t_j}$. Since $P$ and $Q$ have no common zeros, each zero $\alpha$ of $Q(f(t))$ is a pole of $g(t)$, and therefore, it is a zero of order at least $m$ of $Q(f(t))$. As a consequence, for each zero $\alpha$ of $f - b_j$, we have $t_j\omega_\alpha(f - b_j) \ge m$, hence

(1) $\overline{Z}(r, f - b_j) \le \frac{t_j}{m}Z(r, f - b_j)$.

But since $Z(r, f - b_j) \le T(r, f) + O(1) (r \in I)$, by (1), we have

(2) $\sum_{j=1}^n \overline{Z}(r, f - b_j) \le \frac{t}{m}T(r, f) + O(1) (r \in I)$.

Then, by Theorem 45.7, we obtain

(3) $(n-1)T(r, f) \le \frac{t}{m}T(r, f) + \overline{N}(r, f) - \log r + O(1) (r \in I)$.

In particular, this implies

(4) $(n-2)T(r, f) \le \frac{t}{m}T(r, f) - \log r + O(1) (r \in I)$.

Suppose that $f, g$ belong to $M(d(0, R^{-}))$. By Theorem 42.10, if $f$ does not lie in $M_b(d(0, R^{-}))$, we have $m(n-2) \le t$. And if $f, g$ lies in $M(\mathbb{K})$, then when $-\log r$ tends to $+\infty$, it is seen that inequality (4) implies $(n-2) < \frac{t}{m}$.

Now, suppose that $s > t$. Then each pole $\alpha$ of $f$ is a pole of $g$ and therefore satisfies $\omega_\alpha(f)(s-t) = \omega_\alpha(g)$, hence $\overline{N}(r, f) \le N(r, f)(\frac{s-t}{m})$. Consequently, Relation (3) becomes

(5) $(n-1)T(r, f) \le \frac{t}{m}T(r, f) + (\frac{s-t}{m})N(r, f) - \log r + O(1) (r \in I)$ and so we have $(n-1) \le \frac{t}{m} + \frac{s-t}{m}$, thereby $mn \le \min(s+m, t+2m)$. Finally, if in addition, $f$ and $g$ lie in $M(\mathbb{K})$, as $r$ tends to $+\infty$, inequality (5) becomes strict, and therefore, $mn < \min(s+m, t+2m)$, which finishes the proof. $\square$

**Examples.** (1) Let $\Lambda$ be the curve of equation $y^4(x - b')(x - b'') = (x - c)^3$ (with $b', b'', c$ all distinct) and let $f, g \in M(d(a, R^{-}))$ be such that $(f(u), g(u)) \in \Lambda$ for all $u \in D$. Then, by Theorem 48.7, $f, g \in M_b(d(a, R^{-}))$. (2) Let $\Lambda$ be the curve of equation $y^3(x - b')(x - b'') = (x - c)^3$ (with $b', b'', c$ all distinct) and let $f, g \in M(\mathbb{K})$ be such that $(f(u), g(u)) \in \Lambda$ for all $u \in D$. Then, by Theorem 48.7, $f, g$ are constant.

**Notation.** Henceforth, given two integers $m, n$, we denote by g.c.d.($m, n$) the greatest common divisor of $m$ and $n$.
In the proof of Theorem 48.9, we use the following arithmetical lemma:

**Lemma 48.8.** If $a, b, m, s \in \mathbb{N}^*$ satisfy $sa = mb$, then $a \ge \frac{m}{\text{g.c.d.}(s,m)}$.

**Theorem 48.9.** Let $P(X) = A \prod_{i=1}^k (X - a_i)^{s_i}$, $Q(X) = B \prod_{j=1}^n (X - b_j)^{t_j}$ be two relatively prime polynomials of $\mathbb{K}[X]$ of respective degrees $s$ and $t$. Let $m \in \mathbb{N}^*$ and let $f, g \in \mathcal{M}(\mathbb{K})$ (respectively $f, g \in \mathcal{M}(d(a, R^-))$) satisfy $(g(x))^m Q(f(x)) = P(f(x))$ for all $x \in D$.

(α) Suppose $f, g \in \mathcal{M}(d(a, R^-))$. If

$$k + n > 1 + \frac{1}{m} \left( g.c.d.(m, |s - t|_{\infty}) + \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right),$$

then both $f$ and $g$ lie in $\mathcal{M}_b(d(a, R^-))$.

Moreover, if $f$ lies in $\mathcal{A}(d(a, R^-))$, and if

$$k + n > 1 + \frac{1}{m} \left( \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right),$$

then $f \in \mathcal{A}_b(d(a, R^-))$ and $g \in \mathcal{M}_b(d(a, R^-))$.

(β) Suppose both $f, g$ lie in $\mathcal{M}(\mathbb{K})$. If

$$k + n \ge 1 + \frac{1}{m} \left( g.c.d.(m, |s - t|_{\infty}) + \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right),$$

then both $f$ and $g$ are constant.

Finally, if $f, g \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ and if

$$k + n \ge 1 + \frac{1}{m} \left( 1 + \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right),$$

then $f$ admits at least one pole of order $< m$.
**Proof.** When $f, g$ belong to $M(d(a, R^{-}))$, we can obviously assume $a = 0$. It is clear that if $f$ is constant, so is $g$. Suppose that $f$ is not constant. Then we have $s = \sum_{i=1}^{k} s_i, t = \sum_{j=1}^{n} t_j$. For each $i = 1, \dots, k$, every zero $\alpha$ of $f - a_i$ is a zero of $g$ and therefore is a zero of order $\omega_{\alpha}(f - a_i) = \frac{m\omega_{\alpha}(g)}{s_i}$. But by Lemma 48.8, we have $\omega_{\alpha}(f - a_i) \ge \frac{m}{g.c.d.(m, s_i)}$. In the same way, for each $j = 1, \dots, n$, every zero $\beta$ of $f - b_j$ is a pole of $g$ and therefore is a zero of order $\omega_{\beta}(f - b_j) = \frac{m\omega_{\beta}(g)}{t_j}$, and by Lemma 48.8, we have $\frac{m\omega_{\beta}(g)}{t_j} \ge \frac{m}{g.c.d.(m, t_j)}$. So, we have

$$ (6) \quad \bar{Z}(r, f - a_i) \le \frac{g.c.d.(m, s_i)}{m} T(r, f) + O(1) \quad (r \in I), $$

$$ (7) \quad \bar{Z}(r, f - b_j) \le \frac{g.c.d.(m, t_j)}{m} T(r, f) + O(1) \quad (r \in I). $$

Then, applying Theorem 45.7 to $f$ at the points $a_1, \dots, a_k, b_1, \dots, b_n$ and using (6) and (7), we obtain

$$ (8) \quad (k+n-1)T(r, f) \le \frac{1}{m} \left( \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right) T(r, f) + \bar{N}(r, f) - \log r + O(1) \quad (r \in I). $$

We now find an upper bound of $\bar{N}(r, f)$. Let $\gamma$ be a pole of $f$. We have

$$ (9) \quad (s-t)\omega_{\gamma}(f) = m\omega_{\gamma}(g). $$

In the same way, from (9), we obtain $|\omega_{\gamma}(f)|_{\infty} \ge \frac{m}{g.c.d.(m, |s-t|_{\infty})}$, and therefore,

$$ (10) \quad \begin{aligned} \bar{N}(r, f) &\le \frac{g.c.d.(m, |s-t|_{\infty})}{m} N(r, f) \\ &\le \frac{g.c.d.(m, |s-t|_{\infty})}{m} T(r, f). \end{aligned} $$
We prove (11):

$$ (11) \qquad (k+n-1)T(r,f) \le \frac{1}{m} \left( g.c.d.(m, |s-t|_{\infty}) + \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right) \times T(r,f) - \log r + O(1) (r \in I). $$

If $s = t$, we just have the inequality $\overline{N}(r, f) \le N(r, f) \le T(r, f)$. Then From (8), we obtain

$$ (k+n-1)T(r,f) \le \frac{1}{m} \left( m + \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right) \times T(r,f) - \log r + O(1) \quad (r \in I). $$

But in this case, we have $g.c.d.(m, |s-t|_{\infty}) = m$. So, we obtain (11). And if $s \ne t$, by (8), we obtain relation (11) again.

Thus, from relation (11), using Theorem 43.10, if $f$ does not lie in $M_b(d(0, R^{-}))$, we can get

$$ (12) \qquad (k+n-1) \le \frac{1}{m} \left( g.c.d.(m, |s-t|_{\infty}) + \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right), $$

a contradiction with the hypothesis. Hence, if $f$ belongs to $M(d(0, R^{-}))$, it must lie in $M_b(d(0, r^{-}))$ and so must $g$.

Now, suppose $f, g \in M(\mathbb{K})$. When $r$ tends to $+\infty$, we can see that the inequality (12) becomes strict. Therefore, if the inequality is not satisfied, $f$ and $g$ are constant. Conversely, suppose that $f$ is not constant and that all poles of $f$ have order $\ge m$. By (11), we have

$$ (k+n-1)T(r,f) \le \frac{1}{m} \left( \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right) \times T(r,f) + \frac{1}{m} T(r,f) - \log r + O(1) \quad (r \in I), $$
which implies $k + n - 1 < \frac{1}{m}(1 + \sum_{i=1}^{k} g.c.d.(m, s_i)) + \sum_{j=1}^{n} g.c.d.(m, t_j))$. Hence, if this inequality is not true, $f$ must admit at least one pole of order $< m$.

Finally, suppose $f \in \mathcal{A}(d(0, R^{-}))$. Then (8) becomes

$$ (13) \qquad (k + n - 1)T(r, f) \le \frac{1}{m} \left( \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right) \times T(r, f) - \log r + O(1) \quad (r \in I), $$

and then, if $f$ does not lie in $\mathcal{A}_b(d(0, R^{-}))$, by Theorem 43.10, we obtain

$$ k + n - 1 \le \frac{1}{m} \left( \sum_{i=1}^{k} g.c.d.(m, s_i) + \sum_{j=1}^{n} g.c.d.(m, t_j) \right), $$

hence $k + n - 1 \le \frac{1}{m}(s + t)$, a contradiction. That finishes the proof. □

**Examples.** (3) Let $c', c'' \in \mathbb{K}$ (with $c' \neq c''$) and let $\Lambda$ be the curve of equation $y^3 = (x - c')^2(x - c'')$. Let $f, g \in \mathcal{M}(\mathbb{K})$ be such that $(f(u), g(u)) \in \Lambda$ for all $u \in D$. If $f$ and $g$ are not constant, then $f$ admits at least one pole of order 1 or 2. Here the genus is clearly 0, therefore there exist $f, g \in \mathbb{K}(u)$ satisfying $g^3 = (f - c')^2(f - c'')$ and hence $f$ admits at least one pole of order 1 or 2.

(4) Let $\Lambda$ be the curve of equation $y^3(x - b)^2 = (x - c')^2(x - c'')$ (with $b, c', c''$ all distinct) and let $f, g \in \mathcal{M}(d(a, R^{-}))$ be such that $(f(u), g(u)) \in \Lambda$ for all $u \in D$. Then by Theorem 48.9, $f, g \in \mathcal{M}_b(d(a, R^{-}))$.

(5) Let $\Lambda$ be the curve of equation $y^2(x - b')(x - b'') = (x - c)$ (with $b', b'', c$ all distinct) and let $f \in \mathcal{A}(d(a, R^{-}))$ and let $g \in \mathcal{M}(d(a, R^{-}))$ be such that $(f(u), g(u)) \in \Lambda$ for all $u \in D$. Then by Theorem 48.9, $f \in \mathcal{A}_b(d(a, R^{-}))$ and $g \in \mathcal{M}_b(d(a, R^{-}))$.

**Remark.** In [24], it was proved that the equation $f^m + g^n = 1$ in $\mathcal{M}(\mathbb{K})$ leads to $f, g \in \mathbb{K}$ as soon as the least common multiple $q$ of $m$ and $n$ satisfies $\frac{1}{m} + \frac{1}{n} + \frac{1}{q} \ge 1$. Further, if $f, g \in \mathcal{A}(\mathbb{K})$, that leads to $f, g \in \mathbb{K}$ as soon as $\min(m, n) \ge 2$. Here we are now able to generalize these conclusions.
# Chapter 49

## Branched Values

Throughout this chapter, the field $K$ is supposed to be of characteristic 0.

In complex functions theory, a notion closely linked to Picard's exceptional values was introduced: the notion of “perfectly branched value” [29]. Here we consider the same notion on $M(K)$, on $M(d(a, R^{-}))$, and on $M(D)$. Most of results come from [51].

**Definition.** Let $f \in M(K)$ (respectively let $f \in M(d(a, R^{-}))$, respectively let $f \in M(D)$) and let $b \in K$. The value $b$ is said to be a perfectly branched value for $f$ if all zeros of $f-b$ are multiple zeros, except finitely many. And in this book, $b$ is said to be a totally branched value for $f$ if all zeros of $f-b$ are multiple zeros, without exception. Similarly, $\infty$ is called a perfectly branched value for $f$ if all poles of $f$ are multiple but finitely many and it is called a totally branched value for $f$ if all poles of $f$ are multiple, without exception.

In $\mathbb{C}$, it is known that a transcendental meromorphic function admits at most four perfectly branched values and an entire function admits at most two perfectly branched values. As explained by K. S. Charak in [29], these numbers, respectively 4 and 2, are sharp. The Weierstrass function $P$ has four totally branched values (considering $\infty$ as a value), and of course, sine and cosine functions admit two totally branched values: 1 and $-1$.

Here we do a similar study on $p$-adic functions and obtain sometimes certain better results. Particularly, an entire function admits at most one perfectly branched value.
Lemma 49.1 is immediate:

**Lemma 49.1.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}_u(d(0, R^{-}))$, respectively let $f \in \mathcal{M}^c(D)$), admitting a perfectly branched value $b \neq 0$. Then $\frac{1}{f}$ admits $\frac{1}{b}$ as a perfectly branched value. If $f$ admits $0$ as a perfectly branched value, $\frac{1}{f}$ admits $\infty$ as a perfectly branched value. If $f$ admits $\infty$ as a perfectly branched value, $\frac{1}{f}$ admits $0$ as a perfectly branched value.

We have an immediate application of the definition with meromorphic functions whose denominator is a small function with respect to the numerator or vice versa.

**Theorem 49.2.** Let $f, g \in \mathcal{A}(\mathbb{K}) \setminus \mathbb{K}[x]$ (respectively $f, g \in \mathcal{A}_u(d(0, R^{-}))$, respectively $f, g \in \mathcal{A}^c(D)$) be such that $\limsup_{r \to +\infty} \frac{T(r,f)}{T(r,g)} > 2$ (respectively $\limsup_{r \to R^{-}} \frac{T(r,f)}{T(r,g)} > 2$, respectively $\limsup_{r \to +\infty} \frac{T_R(r,f)}{T_R(r,g)} > 2$). Then both $\frac{f}{g}$ and $\frac{g}{f}$ have at most two perfectly branched values.

**Proof.** Set $\phi = \frac{f}{g}$. Without loss of generality, we can suppose that $f$ and $g$ have no common zero. Indeed, suppose first that $f, g \in \mathcal{A}(\mathbb{K}) \setminus \mathbb{K}[x]$ or $f, g \in \mathcal{A}^c(D)$. By Lemma 36.3, we can write $f$ and $g$ in the form $f = \tilde{f} \cdot h$ and $g = \tilde{g} \cdot h$, where $\tilde{f}$ and $\tilde{g}$ have no common zero and then $Z(r, f) = Z(r, \tilde{f}) + Z(r, h)$, $Z(r, g) = Z(r, \tilde{g}) + Z(r, h)$, and so much the more, we have $\frac{T(r, \tilde{f})}{T(r, \tilde{g})} > 2$.

Now, if $f, g \in \mathcal{A}_u(d(0, r)^{-})$, we can place ourselves in an algebraically closed spherically complete extension to obtain the same conclusion because the Nevanlinna functions are the same in such an extension. Therefore, we assume that $f$ and $g$ have no common zero.

Suppose first that $f$ and $g$ belong to $\mathcal{A}(\mathbb{K}) \setminus \mathbb{K}[x]$ or to $\mathcal{A}_u(d(0, R^{-}))$, Since $f, g$ have no common zero, we have $Z(r, \phi) = Z(r, f)$ and $N(r, \phi) = Z(r, g)$, hence $T(r, \phi) = \max(Z(r, f), Z(r, g)) + O(1)$.

Now, by hypothesis, there exists $\lambda < \frac{1}{2}$ and a sequence $(r_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} r_n = +\infty$ (respectively $\lim_{n \to +\infty} r_n = R$) and such that

$$ (1) \qquad T(r_n, g) \le \lambda T(r_n, f) \quad \forall n \in \mathbb{N}. $$
Suppose that $\phi$ has three perfectly branched values $b_j$, $j = 1, 2, 3$. Applying Theorem 45.7, we have

$$ (2) \qquad 2T(r, \phi) \le \sum_{j=1}^{3} \bar{Z}(r, \phi - b_j) + \bar{N}(r, \phi) - \log r + O(1). $$

But here, for each $j = 1, 2, 3$, we note that $\bar{Z}(r, \phi - b_j) \le \frac{Z(r, \phi - b_j)}{2} + q_j \log(r)$ with $q_j \in \mathbb{N}$ and $Z(r, \phi - b_j) = Z(r, f - b_j g) \le \max(T(r, f), T(r, g))$. But since $T(r_n, f) > T(r_n, g)$, we have $T(r_n, \phi - b_j) \le T(r_n, f) + O(1)$, hence $\bar{Z}(r, \phi - b_j) \le \frac{T(r_n, f)}{2} + q_j \log(r_n) + O(1)$. Now, putting $q = q_1 + q_2 + q_3$, by (2), we obtain

$$ 2T(r_n, f) \le \frac{3T(r_n, f)}{2} + T(r_n, g) + 2q \log(r_n) + O(1), $$

hence

$$ T(r_n, f) \le 2T(r_n, g) + q \log(r_n) + O(1), $$

a contradiction to (1).

Similarly, if $f$ and $g$ belong to $A^c(D)$, we can make the same reasoning by replacing $T$ by $T_R$, $Z$ by $Z_R$, and $N$ by $N_R$. $\square$

**Theorem 49.3.** Let $f, g \in A(\mathbb{K})$ be such that $+\infty > \rho(f) > \rho(g)$. Then,

$$ \liminf_{r \to +\infty} \frac{T(r, g)}{T(r, f)} = 0. $$

**Proof.** Suppose first $f, g \in A(\mathbb{K})$. Let $\gamma = \frac{\rho(g)}{\rho(f)}$ and let $(r_n)_{n \in \mathbb{N}}$ be a sequence in $]0, +\infty[$ such that $\lim_{n \to +\infty} r_n = +\infty$ and $\lim_{n \to +\infty} \frac{\log(\log(|f(r_n))))}{-\log(R)} = \rho(f)$. By hypothesis, we have

$$ \lim_{n \to +\infty} \frac{\log(\log(|g|(r_n))))}{\log(\log(|f|(r_n))))} \le \gamma, $$

hence

$$ \lim_{n \to +\infty} \frac{\log(T(r_n, g))}{\log(T(r_n, f))} \le \gamma. $$
Take $\beta \in ]\gamma, 1[$. Then when $n$ is big enough, we can get

$$ \frac{T(r_n, g)}{T(r_n, f)} \le (T(r_n, f))^{\beta-1}. $$

But since $\beta < 1$ and since $\lim_{n \to \infty} T(r_n, f) = +\infty$, one sees that $\lim_{n \to \infty} (T(r_n, f))^{\beta-1} = 0$, which ends the proof when $f$ and $g$ belong to $\mathcal{A}(\mathbb{K})$. $\square$

**Corollary 49.4.** Let $f, g \in \mathcal{A}(\mathbb{K})$ be such that $\rho(f) \ne \rho(g)$. Then both $\frac{f}{g}$ and $\frac{g}{f}$ have at most two perfectly branched values.

The proof of the following theorems require several basic lemmas.

**Lemma 49.6.** Let $(\alpha_i)_{1 \le i \le t}$, $(\beta_i)_{1 \le i \le t}$ be two finite sequences of $\mathbb{K}$ such that $|\alpha_i| < R$, $|\beta_i| < R \ \forall i = 1, \dots, t$. Let $\Theta(x) = \prod_{i=1}^t \left( \frac{1-\frac{\beta_i}{x}}{1-\frac{\alpha_i}{x}} \right)$. Then the function $\sqrt{\Theta(x)}$ is defined and belongs to $\mathcal{A}(\mathbb{K} \setminus d(0, 4R))$. Moreover, if $p \ne 2$, it belongs to $\mathcal{A}(\mathbb{K} \setminus d(0, R))$.

**Proof.** By Theorem 29.23, there exists a unique function $\ell \in \mathcal{A}(d(1, (\frac{1}{4})^{-}))$ with value in $d(1, 1^{-})$ such that $(\ell(u))^2 = u \ \forall u \in d(1, (\frac{1}{4})^{-})$. Moreover, if $p \ne 2$, then $\ell$ has continuation to a function $\bar{\ell} \in \mathcal{A}(d(1, 1^{-}))$ with value in $d(1, 1^{-})$ again and such that $(\bar{\ell}(u))^2 = u \ \forall u \in d(1, 1)$. Here, we put $u = \prod_{i=1}^t \left( \frac{1-\frac{\beta_i}{x}}{1-\frac{\alpha_i}{x}} \right)$. $\square$

**Lemma 49.7.** Let $g$, $h \in \mathcal{A}(\mathbb{K})$ with $\frac{g}{h}$ transcendental and let $\Theta(x) = \prod_{i=1}^t \left( \frac{1-\frac{\beta_i}{x}}{1-\frac{\alpha_i}{x}} \right)$ with $|\alpha_i| < R$ and $|\beta_i| < R \ \forall i = 1, \dots, t$. Then the function $g(x)^2 - h(x)^2 \Theta(x)$ belongs to $\mathcal{A}(\mathbb{K} \setminus d(0, 4R))$ and satisfies $\lim_{r \to +\infty} \frac{|g^2 - h^2 \Theta(r)|}{r^m} = +\infty \ \forall m \in \mathbb{N}$.

**Proof.** We first note that $g^2 - h^2 \Theta$ obviously belongs to $\mathcal{A}(\mathbb{K} \setminus d(0, R))$. Let us fix $m \in \mathbb{N}$. By Lemma 49.6, $\sqrt{\Theta}$ is defined in $\mathbb{K} \setminus d(0, R)$ and belongs to $\mathcal{A}(\mathbb{K} \setminus d(0, 4R))$. Let $\ell = \sqrt{\Theta}$. Then can write $g^2 - h^2 \Theta = (g - h\ell)(g + h\ell)$.

Since $\frac{g}{h}$ is transcendental, $g^2 - h^2 \Theta$ is not identically zero. So, by Theorem 22.22, there exists $a > 0$ and $q \in \mathbb{Z}$ such that $|g - h\ell|(r) \ge ar^q \ \forall r > R$.
Suppose first $h$ is transcendental. Since $h$ is entire and since $|\ell|(r) = 1 \forall r > R$, by Theorem 22.22, we have

$$ (1) \qquad \forall m \in \mathbb{N}, \lim_{r \to \infty} \frac{|h\ell|(r)}{r^m} = +\infty. $$

Consequently, $|h\ell|(r) > |g - h\ell|(r)$, and hence $|h\ell|(r) = |g + h\ell|(r)$. Thus,

$$ |g - h\ell|(r)|g + h\ell|(r) \ge ar^q|h\ell|(r). $$

Then the conclusion comes from (1). Suppose now $h$ is not transcendental, hence it is a polynomial and then $g$ is transcendental. Consequently, when $r$ is big enough, we have $|g - h\ell|(r) = |g|(r) = |g + h\ell|(r)$, and hence $|g^2 - h^2\Theta|(r) = (|g|(r))^2$, which yields the same conclusion. $\square$

**Theorem 49.8.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental (respectively let $f \in \mathcal{M}_u(d(a, R^{-}))$, respectively let $f \in \mathcal{M}^c(D)$). Then $f$ has at most four perfectly branched values. Moreover, any function $g \in \mathcal{M}(\mathbb{K})$ has at most three totally branched values.

**Remark.** Let $f \in \mathcal{M}(\mathbb{K})$. If $a \in \mathbb{K}$ is a perfectly branched value for $f$, then for every $b \in \mathbb{K}$, $a + b$ is a perfectly branched value for $f + b$. Moreover, if $a \neq 0$, then $\frac{1}{a}$ is a perfectly branched value for $\frac{1}{f}$. So, we are going to construct a function $f \in \mathcal{M}(\mathbb{K})$ admitting three distinct totally branched values. Let $\ell = \prod_{j=1}^{\infty} \left(1 - \frac{x}{a_j}\right) \in \mathcal{A}(\mathbb{K})$ with $\lim_{j \to +\infty} |a_j| = +\infty$ and $a_j \neq a_k \forall j \neq k$. Let $u = \prod_{k=1}^{\infty} \left(1 - \frac{x}{a_{2k}}\right)$ and let $w = \prod_{k=1}^{\infty} \left(1 - \frac{x}{a_{2k-1}}\right)$. So, both $u$ and $w$ belong to $\mathcal{A}(\mathbb{K})$ and satisfy $uw = \ell$. Now, let $\phi = \frac{u^2+w^2}{2}$ and $\theta = \frac{w^2-u^2}{2}$. Then, $\phi^2 - \theta^2 = \ell^2$. Now, let $g = \left(\frac{\phi}{\ell}\right)^2$. Note that $g$ admits 0 and 1 as totally branched values. Consequently, $g+1$ admits 1 and 2 as totally branched values and hence the function $f = \frac{1}{g+1}$ admits 1 and $\frac{1}{2}$ as totally branched values. But, on the other hand, all poles of $g$ are multiple, hence so are those of $g+1$. Consequently, $f$ also admits 0 as a totally branched value. Thus, Theorem 49.8 is sharp as far as totally branched values are concerned for meromorphic functions. One can only ask whether
there exist meromorphic functions admitting four perfectly branched
values where some of them are not totally branched values.

**Theorem 49.9.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental and have finitely many poles. Then $f$ has at most one perfectly branched value.

**Corollary 49.10.** Let $f \in \mathcal{A}(\mathbb{K})$ be transcendental. Then $f$ has at most one perfectly branched value.

**Remark.** However, a polynomial can admit two values looking like “perfectly branched values”. Yet, the definition of a perfectly branched value does not really apply to a polynomial or a rational function.

**Example.** Let $P(x) = x^3 - x^2 + \frac{4}{27}$. Then $0$ and $-\frac{4}{27}$ are two perfectly branched values that are not totally branched. Indeed, on one hand, $-\frac{4}{27}$ is perfectly but not totally branched since $P(x) - \frac{4}{27} = x^2(x-1)$. On the other hand, we can check that $P(x) = \left(x - \frac{2}{3}\right)^2\left(x + \frac{1}{3}\right)$.

**Theorem 49.11.** Let $f \in \mathcal{M}_u(d(a, R^-))$ have finitely many poles. Then $f$ has at most two perfectly branched values. Let $f \in \mathcal{M}^c(D)$ have finitely many poles in $D$. Then $f$ has at most two perfectly branched values.

**Corollary 49.12.** Let $f \in \mathcal{A}_u(d(a, R^-))$ (respectively let $f \in \mathcal{A}^c(D)$). Then $f$ has at most two perfectly branched values.

Theorem 49.8 gives an easy proof of the impossibility to parametrize elliptic and hyperelliptic curves by meromorphic functions in all $\mathbb{K}$.

**Corollary 49.13.** Let $a_i \in \mathbb{K}, i = 1, 2, 3$, be pairwise distinct. There do not exist $f, g \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$, there do not exist $f, g \in \mathcal{A}_u(d(a, R^-))$, and there do not exist $f, g \in \mathcal{A}^c(D)$ such that $(g(x))^2 = (f(x) - a_1)(f(x) - a_2)(f(x) - a_3)$.

**Proof.** Suppose two functions $f, g \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ satisfy $(g(x))^2 = (f(x)-a_1)(f(x)-a_2)(f(x)-a_3)$. Then each zero of $f-a_i, i = 1, 2, 3$, must be of order at least 2. And each pole of $(f-a_1)(f-a_2)(f-a_3)$ is a pole of $g^2$ and hence is of even order, and hence each pole of $f$ is at least of order 2. $f$ admits four totally branched values: $a_1, a_2, a_3, \infty$, what is impossible by Theorem 49.8. $\square$
**Remark.** We don't know whether there exists a function $f \in \mathcal{A}_u(d(a, R^{-}))$ admitting two perfectly branched values. The only case when we can improve Theorem 49.11 is the case when $\mathbb{K}$ has residue characteristic 0.

In the proofs of Theorems 49.8 and 49.9, without loss of generality, we can obviously assume that all supposed perfectly branched values of the functions we consider are finite, what we do for simplicity.

**Proof.** (Theorems 49.8, 49.9, and 49.11.) If $f$ lies in $M_u(d(a, R^{-}))$, we assume that $a = 0$. Suppose $f$ has $q$ perfectly branched values $b_j$ with $j = 1, \dots, q$. For each $j$, let $s_j$ be the number of simple zeros of $f - b_j$ and let $s = \sum_{j=1}^q s_j$. Applying Theorem 45.7, we have

$$ (1) \quad (q-1)T(r, f) \le \sum_{j=1}^{q} \overline{Z}(r, f-b_j) + \overline{N}(r, f) - \log r + O(1). $$

But since $f - b_j$ has $s_j$ simple zeros, we have

$$ \overline{Z}(r, f - b_j) \le \frac{Z(r, f - b_j) + s_j \log r}{2} + O(1) \le \frac{T(r, f) + s_j \log r}{2} + O(1) \quad \forall j = 1, \dots, q, $$

hence

$$ (2) \quad (q-1)T(r, f) \le \frac{qT(r, f)}{2} + T(r, f) + \left(\frac{s}{2} - 1\right) \log r + O(1). $$

By (2), clearly, we have $q \le 4$ in all cases, which shows the first statement of Theorem 49.8 whenever $f \in \mathcal{M}(\mathbb{K})$ or $f \in \mathcal{M}_u(d(0, R^{-}))$. Now, when $f \in \mathcal{M}^c(D)$, then (2) becomes $(q-1)T_R(r, f) \le \frac{qT_R(r, f)}{2} + T_R(r, f) + (O(\log(r)))$ and we have the same conclusion.

Further, suppose that $f$ lies in $\mathcal{M}(\mathbb{K})$ or in $\mathbb{K}(x)$ and that $b_1, \dots, b_q$ are totally branched values. Then, $s = 0$, hence by (2), we have

$$ (3) \quad \left(\frac{q}{2} - 2\right) T(r, f) \le -\log r + O(1). $$

In Theorem 49.8, since $f$ is transcendental, we have $\log r = o(T(r, f))$, hence $q \le 3$.

Consider now the hypotheses of Theorems 49.9 and 49.11. Let $t$ be the number of poles of $f$, taking multiplicity into account. We have
$N(r, f) \le t \log r + O(1)$, hence by (1), we obtain $(q-1)T(r, f) \le \frac{q}{2}T(r, f) + O(\log r)$, and hence $q \le 2$. When $f \in \mathcal{M}^c(D)$, we just replace $N$ by $N_R$.

Thus, Theorem 49.11 is proved.

For the proof of Theorem 49.9, without loss of generality, we may assume that these perfectly branched values are 0 and $b$. Suppose first that $f$ has infinitely many zeros of order $\ge 3$. Then $Z(r, f) - 2\bar{Z}(r, f)$ is a function $\theta(r)$ such that

$$ (4) \qquad \lim_{r \to +\infty} \frac{\theta(r)}{\log r} = +\infty, $$

therefore

$$ (5) \qquad \bar{Z}(r, f) \le \frac{T(r, f) - \theta(r)}{2}. $$

On the other hand, by (1), with $q = 2$, we have

$$ T(r, f) \le T(r, f) - \frac{\theta(r)}{2} + O(\log r), $$

and then, by (4), we can see a contradiction proving that $f$ cannot admit 0 and $b$ as branched values.

Suppose now that all zeros of both $f$ and $f-b$ are of order 2 except finitely many. So, there exists $S > 0$ satisfying the following properties:

(i) All poles of $f$ lie in $d(0, S)$.

(ii) $|f|(r) > |b| \forall r > S$.

(iii) All zeros of $f$ and of $f-b$ in $\mathbb{K} \setminus d(0, S)$ are of order 2 exactly.

We can then write $f$ in the form $\frac{Pg^2}{V}$ with $P, V \in \mathbb{K}[x]$, $g \in \mathcal{A}(\mathbb{K})$ and $\deg(P) = k$. Similarly, $f-b$ is in the form $\frac{Qh^2}{V}$ with $Q \in \mathbb{K}[x]$, $h \in \mathcal{A}(\mathbb{K})$, where all zeros of $P, Q, V$ lie in $d(0, S)$ and all zeros of $g, h$ lie in $\mathbb{K} \setminus d(0, S)$ and are simple zeros. Set $\deg(V) = t$. We note that $g$ is transcendental.

By (ii), we have $|f|(r) = |f - b|(r) \forall r > S$. Consequently, by Lemma 36.20, $f$ and $f-b$ have the same number of zeros in $d(0, S)$, and hence, $\deg(P) = \deg(Q)$. Let $P(x) = \prod_{i=1}^k (x - \alpha_i)$ and $Q(x) = \prod_{i=1}^k (x - \beta_i)$. Let $\Theta(x) = \prod_{i=1}^k \left( \frac{1 - \beta_i}{1 - \alpha_i} \right)$ and let $\Xi(x) = \frac{V(x)}{P(x)}$. Of course, $\lim_{r \to +\infty} \frac{|\Xi(r)|}{r^{t+1}} = 0$.
Now, we have

$$g(x)^2 = h(x)^2 \Theta(x) + b\Xi(x).$$

By Lemma 49.7, we can derive

$$\lim_{r \to +\infty} \frac{(|g^2 - h^2 \Theta|(r))}{r^m} = +\infty \quad \forall m \in \mathbb{N},$$

a contradiction to $\lim_{r \to +\infty} \frac{|\Xi|(r)}{r^{t+1}} = 0$. This completes the proof of Theorem 49.9. □

**Remark.** Corollary 49.12 may suggest that Theorem 28.4 is wrong. Indeed, by Theorem 28.4, given sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$, $(c_n)_{n \in \mathbb{N}}$ in $d(0, R^{-})$ such that

$$\lim_{n \to +\infty} |a_n| = \lim_{n \to +\infty} |b_n| = \lim_{n \to +\infty} |c_n| = R,$$

together with $\prod_{n=0}^{\infty} \frac{R}{|a_n|} = +\infty$, there exists functions $f \in \mathcal{A}(d(0, R^{-}))$ admitting each $a_n$ as a zero of order 2 and such that $f-1$ admits each $b_n$ as a zero of order 2 and such that $f-2$ admits each $c_n$ as a zero of order 2. Moreover, since $\prod_{n=0}^{\infty} \frac{R}{|a_n|} = +\infty$, we can check that $f \in \mathcal{A}_u(d(0, R^{-}))$.

The explanation is that, given such a function $f$, either $f$ or $f-1$ or $f-2$ has infinitely many other zeros of order 1, and therefore, at least one of these three values (0, 1, 2) is not perfectly branched.
# Chapter 50

## Exceptional Values of Functions and Derivatives

Throughout this chapter, the field $K$ is supposed to be of characteristic 0.

This chapter is aimed at studying various properties of derivatives of meromorphic functions, particularly their sets of zeros. Many important results are due to Jean-Paul Bézivin [10], [11], [12].

We first note a general property concerning quasi-exceptional values of meromorphic functions and derivatives.

**Notation.** Let $f \in \mathcal{M}(K)$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$) and let $\mathcal{T}$ be a property satisfied by $f$ at certain points. Let $r \in ]0, R[$. Assume that $f(0) \neq 0, \infty$. We denote by $Z(r, f | \mathcal{T})$ the counting function of zeros of $f$ in $K$ (respectively in $d(0, r)$) at the points where $f$ satisfies $\mathcal{T}$ (counting multiplicity) i.e. if $(a_n)$ is the finite or infinite sequence of zeros of $f$ in $d(0, R^{-})$ with respective multiplicity order $s_n$, where $\mathcal{T}$ is satisfied, we put $Z(r, f) = \sum_{|a_n| \le r, \mathcal{T}} s_n(\log r - \log |a_n|)$.

And we denote by $\bar{Z}(r, f | \mathcal{T})$ the counting function of zeros of $f$ in $K$ (respectively in $d(0, r)$) at the points where $f$ satisfies $\mathcal{T}$, (ignoring multiplicity).

Similarly, we denote by $N(r, f | \mathcal{T})$ the counting function of poles of $f$ in $K$ (respectively in $d(0, r)$) at the points where $f$ satisfies $\mathcal{T}$ (counting multiplicity) and we denote by $\bar{N}(r, f | \mathcal{T})$ the counting function of poles of $f$ in $K$ (respectively in $d(0, r)$) at the points where $f$ satisfies $\mathcal{T}$ (ignoring multiplicity).
Given two meromorphic functions $f, g \in \mathcal{M}(\mathbb{K})$ or $f, g \in \mathcal{M}(d(a, R^{-}))$ ($a \in \mathbb{K}, R > 0$), we denote by $W(f, g)$ the Wronskian of $f$ and $g$: $f'g - fg'$.

**Theorem 50.1.** Let $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}(x)$ (respectively let $f \in \mathcal{M}_u(d(\alpha, R^{-})))$. If $f$ admits a quasi-exceptional value, then $f'$ has no quasi-exceptional value different from 0.

**Proof.** Without loss of generality, we may assume $\alpha = 0$ and that $f$ has no zero and no pole at 0. Let $b \in \mathbb{K}$ and suppose that $b$ is a quasi-exceptional value of $f$. There exist $P \in \mathbb{K}[x]$ and $l \in \mathcal{A}(\mathbb{K}) \setminus \mathbb{K}[x]$ (respectively and $l \in \mathcal{A}_u(d(0, R^{-})))$ without common zeros, with $P$, such that $f = b + \frac{P}{l}$.

Let $c \in \mathbb{K}^*$. Remark that $f' - c = \frac{P'l - Pl' - cl^2}{l^2}$. Let $a \in \mathbb{K}$ (respectively let $a \in d(0, R^{-}))$. If $a$ is a pole of $f$, it is a pole of $f' - c$ and we can check that

$$ (1) \quad \omega_a(P'l - Pl' - cl^2) = \omega_a(l') = \omega_a(l) - 1 $$

because $a$ is not a zero of $P$.

Now, suppose that $a$ is not a pole of $f$. Then,

$$ (2) \quad \omega_a(f' - c) = \omega_a(P'l - Pl' - cl^2) $$

Consequently, $Z(r, f' - c) = Z(r, (P'l - Pl' - cl^2) | l(x) \neq 0)$. But, by (1), we have

$$ (3) \quad Z(r, (P'l - Pl' - cl^2) | l(x) = 0) < Z(r, l), $$

and therefore by (2) and (3), we obtain

$$ (4) \quad Z(r, f' - c) = Z(r, (P'l - Pl' - cl^2) | l(x) \neq 0) > Z(r, P'l - Pl' - cl^2) - Z(r, l). $$

Now, let us examine $Z(r, P'l - Pl' - cl^2)$. Let $r \in ]0, +\infty[$ (respectively let $r \in ]0, R[$). Since $l \in \mathcal{A}(\mathbb{K})$ is transcendental (respectively since $l \in \mathcal{A}_u(d(0, R^{-})))$, we can check that when $r$ is big enough, we have $|Pl'|(r) < |c|(l^2)$ and $|Pl|(r) < |c|(l^2)$, hence clearly $|P'l - Pl'|(r) < |c|(l^2)$, and hence
$|P'l - Pl' - cl^2|(r) = |c|(|l|(r))^2$. Consequently, when $r$ is big enough, by Theorem 43.2, we have $Z(r, P'l - Pl' - cl^2) = Z(r, l^2) + O(1)$. But $Z(r, l^2) = 2Z(r, l)$, hence $Z(r, P'l - Pl' - cl^2) = 2Z(r, l) + O(1)$, and therefore by (4), we check that when $r$ is big enough,

$$ (5) \quad Z(r, f' - c) > Z(r, l). $$

Now, if $l \in \mathcal{A}(\mathbb{K})$, since $l$ is transcendental, by (5), for every $q \in \mathbb{N}$, we have $Z(r, f' - c) > Z(r, l) > q \log r$, when $r$ is big enough, hence $f' - c$ has infinitely many zeros in $\mathbb{K}$. And similarly if $l \in \mathcal{A}_u(d(0, R^{-}))$, then by (5), $Z(r, f' - c)$ is unbounded when $r$ tends to $R$, hence $f' - c$ has infinitely many zeros in $d(0, R^{-})$. $\square$

We now note a property of differential equations of the form $y^{(n)} - \theta y = 0$ that is almost classical.

The problem of a constant Wronskian is involved in several questions.

**Theorem 50.2.** Let $h$, $l \in \mathcal{A}(\mathbb{K})$ (respectively $h$, $l \in \mathcal{A}(d(\alpha, R^{-}))$) and satisfy $h'l - hl' = c \in \mathbb{K}$, with $h$ non-affine. If $h$, $l$ belong to $\mathcal{A}(\mathbb{K})$, then $c = 0$ and $\frac{h''}{l}$ is a constant. If $c \neq 0$ and if $h$, $l \in \mathcal{A}(d(\alpha, R^{-}))$, there exists $\phi \in \mathcal{A}(d(\alpha, R^{-}))$ such that $h'' = \phi h$, $l'' = \phi l$.

**Proof.** Suppose $c \neq 0$. If $h(a) = 0$, then $l(a) \neq 0$. Next, $h$ and $l$ satisfy

$$ (1) \quad \frac{h''}{h} = \frac{l''}{l}. $$

Remark first that since $h$ is not affine, $h''$ is not identically zero. Next, every zero of $h$ or $l$ of order $\geq 2$ is a trivial zero of $h'l - hl'$, which contradicts $c \neq 0$. So, we can assume that all zeros of $h$ and $l$ are of order 1.

Now suppose that a zero $a$ of $h$ is not a zero of $h''$. Since $a$ is a zero of $h$ of order 1, $\frac{h''}{h}$ has a pole of order 1 at $a$ and so does $\frac{l''}{l}$, hence $l(a) = 0$, a contradiction. Consequently, each zero of $h$ is a zero of order 1 of $h$ and is a zero of $h''$, and hence, $\frac{h''}{h}$ is an element $\phi$ of $\mathcal{M}(\mathbb{K})$ (respectively of $\mathcal{M}(d(\alpha, R^{-}))$) that has no pole in $\mathbb{K}$ (respectively in $d(\alpha, R^{-})$). Therefore, $\phi$ lies in $\mathcal{A}(\mathbb{K})$ (respectively in $\mathcal{A}(d(\alpha, R^{-}))$).
The same holds for $l$, and so, $l''$ is of the form $\theta l$ with $\theta \in \mathcal{A}(\mathbb{K})$ (respectively in $\mathcal{A}(d(\alpha, R^-))$). But since $\frac{h''}{h} = \frac{l''}{l}$, we have $\phi = \theta$.

Now, suppose $h, l$ belong to $\mathcal{A}(\mathbb{K})$. Since $h''$ is of the form $\phi h$ with $\phi \in \mathcal{A}(\mathbb{K})$, we have $|h''|(r) = |\phi|(r)|h|(r)$. But by Theorem 37.10, we know that $|h''|(r) \le \frac{1}{r^2}|h|(r)$, a contradiction when $r$ tends to $+\infty$. Consequently, $c = 0$. But then $h'l - hl' = 0$ implies that the derivative of $\frac{h}{l}$ is identically zero, hence $\frac{h}{l}$ is constant. $\square$

**Corollary 50.3.** Let $h, l \in \mathcal{A}(\mathbb{K})$ with coefficients in $\mathbb{Q}$ also be entire functions in $\mathbb{C}$, with $h$ non-affine. If $h'l - hl'$ is a constant $c$, then $c = 0$.

**Theorem 50.4.** Let $\theta \in \mathcal{M}(\mathbb{K})$ (respectively let $\theta \in \mathcal{M}(d(\alpha, R^-))$) and let $(\mathcal{E})$ be the differential equations $y^{(n)} - \theta y = 0$. Let $E$ be the subvector space of $\mathcal{M}(\mathbb{K})$ (respectively of $\mathcal{M}(d(\alpha, R^-))$) of the solutions of $(\mathcal{E})$:

(i) If $n = 1$, then the dimension of $E$ is at most 1.

(ii) If $\theta$ belongs to $\mathcal{A}(\mathbb{K})$ (respectively if $\theta$ belongs to $\mathcal{A}_u(d(\alpha, R^-))$, then $E = \{0\}$.

(iii) If $\theta$ belongs to $\mathcal{A}_b(d(\alpha, R^-))$ and satisfies $\|\theta\|_{d(\alpha, R^-)} > \frac{1}{R^n}$, then $E = \{0\}$ again.

**Proof.** In each case, we assume that $(\mathcal{E})$ admits a non-identically zero solution $h$. Then $h^{(n)}$ may not be identically zero.

Suppose first that $n = 1$. Suppose that $g \in E$. Let $u = \frac{h}{g}$. Since $h' = \theta h$, we have $u'g + ug' = \theta ug$, therefore $u\frac{g'}{g} = u\theta = u' + u\frac{g'}{g}$, and hence $u' = 0$ i.e. $u$ is a constant. Consequently, $E$ is at most of dimension 1.

Suppose now that $\theta$ lies in $\mathcal{A}(\mathbb{K})$. Then $|\theta|(r) = \frac{|h^{(n)}|(r)}{|h|(r)}$ is an increasing function in $r$ in $]0, +\infty[$, a contradiction to the inequality $\frac{|h^{(n)}|(r)}{|h|(r)} \le \frac{1}{r^n}$ coming from Theorem 37.10.

Now, suppose that $\theta \in \mathcal{A}_u(d(\alpha, R^-))$. Without loss of generality, we may assume $\alpha = 0$. Similarly, $|\theta|(r)$ is unbounded in $]0, R^-[$, a contradiction to $\frac{|h^{(n)}|(r)}{|h|(r)} \le \frac{1}{r^n}$ again.
In the same way, if $\theta$ belongs to $\mathcal{A}_b(d(0, R^{-}))$ and satisfies
$\|\theta\|_{d(0, R^{-})} > \frac{1}{R^n}$, the inequality $\frac{|h^{(n)}(r)|}{|h|(r)} \le \frac{1}{r^n}$ is then violated when $r$
tends to $R$. $\square$

**Remark.** The hypothesis $\theta$ unbounded in $d(\alpha, R^{-})$ is indispensable to show that the space $E$ is of dimension $0$ or $1$, as shown by the example given again by the $p$-adic hyperbolic functions $h(x) = \cosh(x)$ and $l(x) = \sinh(x)$. The radius of convergence of both $h, l$ is $p^{\frac{1}{p-1}}$ when $\mathbb{K}$ has residue characteristic $p$ and is $1$ when $\mathbb{K}$ has residue characteristic $0$. Of course, both functions are solutions of $y'' - y = 0$, but they are bounded.

Theorem 50.5 given in [10] is an improvement of Theorem 50.2.

**Theorem 50.5 (J.P. Bézivin).** Let $f, g \in \mathcal{A}(\mathbb{K})$ be such that $W(f,g)$ is a non-identically zero polynomial. Then both $f, g$ are polynomials.

**Proof.** First, by Theorem 50.2, we check that the claim is satisfied when $W(f,g)$ is a polynomial of degree $0$. Now, suppose the claim holds when $W(f,g)$ is a polynomial of certain degree $n$. We show it for $n+1$. Let $f, g \in \mathcal{A}(\mathbb{K})$ be such that $W(f,g)$ is a non-identically zero polynomial $P$ of degree $n+1$.

Thus, by hypothesis, we have $f'g - fg' = P$, hence $f''g - fg'' = P'$.
We can extract $g'$ and get $g' = \frac{f'g - P}{f}$. Now consider the function
$Q = f''g' - f'g''$ and replace $g'$ by what we just found: we can get
$Q = f'\left(\frac{f''g - fg'}{f}\right) - \frac{Pf''}{f}$.

Now, we can replace $f''g - fg''$ by $P'$ and obtain $Q = \frac{(f'P' - Pf'')}{f}$.
Thus, in that expression of $Q$, we can write $|Q|(R) \le \frac{|f|(R)|P|(R)}{R^2|f|(R)}$,
hence $|Q|(R) \le \frac{|P|(R)}{R^2} \quad \forall R > 0$. But by definition, $Q$ belongs to $\mathcal{A}(\mathbb{K})$.
Consequently, $Q$ is a polynomial of degree $t \le n-1$.

Now, suppose $Q$ is not identically zero. Since $Q = W(f', g')$ and since $\deg(Q) < n$, by the induction, hypothesis $f'$ and $g'$ are polynomials and so are $f, g$. Finally, suppose $Q = 0$ meromorphic functions in $\mathbb{K}: (\mathcal{E})$ $y' = \theta y$ with $\theta = \frac{P'}{P}$, whereas $y$ belongs to $\mathcal{A}(\mathbb{K})$. By Theorem 50.4, the space of solutions of $(\mathcal{E})$ is of dimension $0$ or $1$.
Consequently, there exists $\lambda \in \mathbb{K}$ such that $f' = \lambda P$, hence $f$ is a polynomial. The same holds for $g$. □

Then, $P'f' - Pf'' = 0$, and therefore, $f'$, $P$ are two solutions of the differential equation of order 1 for the following:

**Remark.** In Archimedean analysis, Theorem 50.5 does not hold. For example, take $f(x) = e^x$, $g(x) = e^{-x}$. Then, $W(f,g) = 2$. We can also consider $f(x) = xe^x$, $g(x) = e^{-x}$. Then, $W(f,g) = 2x + 1$.

**Lemma 50.6.** Let $U, V \in \mathcal{A}(\mathbb{K})$ have no common zero and let $f = \frac{U}{V}$. If $f'$ has finitely many zeros, there exists a polynomial $P \in \mathbb{K}[x]$ such that $U'V - UV' = P\tilde{V}$.

**Proof.** If $V$ is a constant, the statement is obvious. So, we assume that $V$ is not a constant. Now $\tilde{V}$ divides $V'$ and hence $V'$ factorizes in the way $V' = \tilde{V}Y$ with $Y \in \mathcal{A}(\mathbb{K})$. Then no zero of $Y$ can be a zero of $V$. Consequently, we have

$$f'(x) = \frac{U'V - UV'}{V^2} = \frac{U'\overline{V} - UY}{\overline{V}^2\tilde{V}}.$$

The two functions $U'\overline{V} - UY$ and $\overline{V}^2\tilde{V}$ have no common zero since neither have $U$ and $V$. So, the zeros of $f'$ are those of $U'\overline{V} - UY$ which therefore has finitely many zeros and consequently is a polynomial $P$, hence $U'V - UV' = P\tilde{V}$. □

**Theorem 50.7.** Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$ have finitely many poles of order $\ge 3$ and admit primitives. Then $f$ has no quasi-exceptional value.

**Proof.** Suppose that $f$ admits a quasi-exceptional value. Without loss of generality, we can assume that this value is 0. Let $F$ be a primitive of $f$ and let $F = \frac{U}{V}$, with $U, V \in \mathcal{A}(\mathbb{K})$, having no common zero. By Lemma 50.7, there exists a polynomial $P$ such that $U'V - UV' = P\tilde{V}$. But since $f$ has finitely many poles of order $\ge 3$, $F$ has finitely many poles of order $\ge 2$, hence $\tilde{V}$ has finitely many zeros, hence it is a polynomial. But then $P\tilde{V}$ is a polynomial and then, by Theorem 50.5, both $U, V$ are polynomials, therefore $F \in \mathbb{K}(x)$ a contradiction. □
**Theorem 50.8.** Let $f \in \mathcal{M}(\mathbb{K})$ have finitely many multiple poles such that for certain $b \in \mathbb{K}$, $f' - b$ has finitely many zeros. Then, $f$ belongs to $\mathbb{K}(x)$.

**Proof.** Suppose first $b = 0$. Let us write $f = \frac{U}{V}$ with $U, V \in \mathcal{A}(\mathbb{K})$, having no common zeros. By Lemma 50.7, there exists a polynomial $P \in \mathbb{K}[x]$ such that $U'V - UV' = P\tilde{V}$. Since $f$ has finitely many multiple poles, $\tilde{V}$ is a polynomial, hence so is $U'V - UV'$. But then by Theorem 50.5, both $U, V$ are polynomials, which ends the proof when $b = 0$. Consider now the general case. $f' - b$ is the derivative of $f - bx$ that satisfies the same hypothesis, so the conclusion is immediate. $\square$

**Theorem 50.9.** Let $f \in \mathcal{M}(\mathbb{K})$ admit primitives. If $f$ has two perfectly branched values, it has no quasi-exceptional value. Moreover, if $f$ has one totally branched value, then it has no exceptional value.

**Proof.** Suppose that $f$ has two perfectly branched values $a$ and $b$ and a quasi-exceptional value $c$. Since $f$ admits primitives, $N(r, f)$ satisfies $\overline{N}(r, f) \le \frac{N(r, f)}{2} + o(T(r, f))$, hence by Theorem 45.7, we have

$$2T(r, f) \le \frac{(Z(r, f - a) + Z(r, f - b) + N(r, f))}{2} + o(T(r, f)),$$

hence $2T(r, f) \le \frac{3T(r, f)}{2} + o(T(r, f))$, a contradiction.

Suppose now that $f$ has one totally branched and an exceptional value $c$. Then $N(r, f)$ satisfies

$$2T(r, f) \le \frac{(Z(r, f - a) + N(r, f))}{2} - \log(r),$$

a contradiction. $\square$

**Notation.** For each $n \in \mathbb{N}^*$, we set $\lambda_n = \max\{\frac{1}{|k|}, 1 \le k \le n\}$. Given positive integers $n, q$, we denote by $C_n^q$ the combination $\frac{n!}{q!(n-q)!}$.

For convenience, in this chapter, log is the Neperian logarithm and we denote by $e$ the number such that $\log(e) = 1$ and Exp is the real exponential function.
**Remark.** For every $n \in \mathbb{N}^*$, we have $\lambda_n \le n$ because $k|k| \ge 1 \forall k \in \mathbb{N}$. The equality holds for all $n$ of the form $p^h$.

Lemmas 50.10 and 50.11 are due to Jean-Paul Bézivin [10]:

**Lemma 50.10.** Let $U, V \in \mathcal{A}(d(0, R^{-}))$. Then for all $r \in ]0, R[$ and $n \ge 1$, we have

$$|U^{(n)}V - UV^{(n)}|(r) \le |n!|\lambda_n \frac{|U'V - UV'|(r)}{r^{n-1}}.$$

More generally, given $j, l \in \mathbb{N}$, we have

$$|U^{(j)}V^{(l)} - U^{(l)}V^{(j)}|(r) \\ \le |(j!)(l!)|\lambda_{j+l} \frac{|U'V - UV'|(r)}{r^{j+l-1}}.$$

**Proof.** Set $g = \frac{U}{V}$ and $f = g'$. Applying Theorem 37.10 to $f$ for $k-1$, we obtain

$$|g^{(k)}|(r) = |f^{(k-1)}|(r) \le |(k-1)!| \frac{|f|(r)}{r^{k-1}} \\ = |(k-1)!| \frac{|U'V - UV'|(r)}{|V^2|(r)r^{k-1}}.$$

As in the proof of Theorem 37.10, we set $U = V(\frac{U}{V})$. By Leibniz formula again, now we can obtain

$$U^{(n)} = \sum_{q=1}^{n} C_{n}^{q} V^{(n-q)} \left(\frac{U}{V}\right)^{(q)} + V^{(n)} \left(\frac{U}{V}\right),$$

hence

$$(1) \qquad U^{(n)} - V^{(n)} \left(\frac{U}{V}\right) = \sum_{q=1}^{n} C_{n}^{q} V^{(n-q)} \left(\frac{U}{V}\right)^{(q)}.$$

Now, we have

$$\left|\left(\frac{U}{V}\right)^{(q)}\right|(r) = |g^{(q)}|(r) \le |(q-1)!| \frac{|U'V - UV'|(r)}{|V^2|(r)r^{q-1}}$$

and

$$|V^{(n-q)}|(r) \le |(n-q)!| \frac{|V|(r)}{r^{n-q}}.$$
Consequently, the general term in (1) is upper bounded as

$$ \begin{aligned} \left| C_n^q V^{(n-q)} \left( \frac{U}{V} \right)^{(q)} \right| (r) &\le \frac{|(n!)((n-q)!)((q-1)!)|}{|(q!)((n-q)!)|} \frac{|U'V - UV'|(r)}{|V|(r)r^{n-1}} \\ &\le \lambda_n \frac{|n!| |U'V - UV'|(r)}{|V|(r)r^{n-1}}. \end{aligned} $$

Therefore, by (1), we obtain

$$ \left| U^{(n)} - V^{(n)} \left( \frac{U}{V} \right) \right| (r) \le |n!| \lambda_n \frac{|U'V - UV'|(r)}{|V|(r)r^{n-1}}, $$

and finally

$$ \left| U^{(n)}V - V^{(n)}U \right|(r) \le |n!| \lambda_n \frac{|U'V - UV'|(r)}{r^{n-1}}. $$

We can now generalize the first statement. Set $P_j = U^{(j)}V - UV^{(j)}$. By induction, we can show the following equality that already holds for $l \le j$:

$$ U^{(j)}V^{(l)} - U^{(l)}V^{(j)} = \sum_{h=0}^{l} C_l^h (-1)^h P_{j+h}^{(l-h)}. $$

Then, the second statement gets just an application of the first. □

**Lemma 50.11.** Let $U, V \in \mathcal{A}(\mathbb{K})$ and let $r$, $R \in ]0, +\infty[$ satisfy $r < R$. For all $x, y \in \mathbb{K}$ with $|x| \le R$ and $|y| \le r$, we have the following inequality:

$$ |U(x + y)V(x) - U(x)V(x + y)| \le \frac{R|U'V - UV'|(R)}{e(\log R - \log r)} $$

**Proof.** By Taylor's formula at the point $x$, we have

$$ U(x + y)V(x) - U(x)V(x + y) = \sum_{n \ge 0} \frac{U^{(n)}(x)V(x) - U(x)V^{(n)}(x)}{n!}y^n. $$
Now, by Lemma 50.9, we have

$$ \left| \frac{U^{(n)}(x)V(x) - U(x)V^{(n)}(x)}{n!} y^n \right| \le \lambda_n \frac{|U'V - UV'|(R)}{R^{n-1}} r^n \\ = \lambda_n R |U'V - UV'|(R) \left(\frac{r}{R}\right)^n . $$

Consequently, $\lim_{n \to +\infty} \left| \frac{U^{(n)}(x)V(x) - U(x)V^{(n)}(x)}{n!} y^n \right| = 0$, therefore we can define $B = \max_{n \ge 1} \{\lambda_n (\frac{r}{R})^n\} R |U'V - UV'|(R)$ and we have $|U(x+y)V(x) - U(x)V(x+y)| \le B$. Now, as remarked above, we have $\lambda_n \le n$. We can check that the function $h$ defined in $]0, +\infty[$ as $h(t) = t(\frac{r}{R})^t$ reaches its maximum at the point $u = \frac{1}{e(\log R - \log r)}$. Consequently, $B \le \frac{1}{e(\log R - \log r)}$, and therefore,

$$ |U(x + y)V(x) - U(x)V(x + y)| \le \frac{R|U'V - UV'|(R)}{e(\log R - \log r)} \quad \square $$

**Notation.** Let $f \in \mathcal{M}(d(0, R^{-}))$. For each $r \in ]0, R[$, we denote by $s(r, f)$ the number of zeros of $f$ in $d(0, r)$, each counted with its multiplicity and we set $t(r, f) = s(r, \frac{1}{f})$. Similarly, we denote by $\beta(r, f)$ the number of multiple zeros of $f$ in $d(0, r)$, each counted with its multiplicity.

**Theorem 50.12.** Let $f \in \mathcal{M}(\mathbb{K})$ be such that for some $c, q \in ]0, +\infty[$, $t(r, f)$ satisfies $t(r, f) \le cr^q$ in $[1, +\infty[$. If $f'$ has finitely many zeros, then $f \in \mathbb{K}(x)$.

**Proof.** Suppose $f'$ has finitely many zeros and set $f = \frac{U}{V}$. If $V$ is a constant, the statement is immediate. So, we suppose $V$ is not a constant and hence it admits at least one zero $a$. By Lemma 50.7, there exists a polynomial $P \in \mathbb{K}[x]$ such that $U'V - UV' = P\tilde{V}$. Next, we take $r, R \in [1, +\infty[$ such that $|a| < r < R$ and $x \in d(0, R)$, $y \in d(0, r)$. By Lemma 50.11, we have

$$ |U(x + y)V(x) - U(x)V(x + y)| \le \frac{R|U'V - UV'|(R)}{e(\log R - \log r)} . $$
Note that $U(a) \neq 0$ because $U$ and $V$ have no common zero. Now set $l = \max(1, |a|)$ and take $r \ge l$. Setting $c_1 = \frac{1}{e|U(a)|}$, we have

$$|V(a + y)| \le c_1 \frac{R|P|(R)|\tilde{V}|(R)}{\text{Log}R - \text{Log}r}.$$

Then taking the supremum of $|V(a + y)|$ inside the disk $d(0, r)$, we can derive

$$(1) \qquad |V|(r) \le c_1 \frac{R|P|(R)|\tilde{V}|(R)}{\text{Log}R - \text{Log}r}.$$

Let us apply Corollary 22.30, by taking $R = r + \frac{1}{r^q}$, after noting that the number of zeros of $\tilde{V}(R)$ is bounded by $s(r, V)$. So, we have

$$(2) \qquad |\tilde{V}|(R) \le \left(1 + \frac{1}{r^{q+1}}\right)^{\beta((r+\frac{1}{r^q}),V)} |\tilde{V}|(r).$$

Now, due to the hypothesis, $s(r, V) = t(r, f) \le cr^q$ in $[1, +\infty[$, we have

$$(3) \qquad \left(1 + \frac{1}{r^{q+1}}\right)^{\beta((r+\frac{1}{r^q}),V)} \le \left(1 + \frac{1}{r^{q+1}}\right)^{[c(r+\frac{1}{r^q})^m]} \\ = \text{Exp}\left[c(r + \frac{1}{r^q})^q \text{Log}\left(1 + \frac{1}{r^{q+1}}\right)\right].$$

The function $h(r) = c(r+\frac{1}{r^m})^m \text{Log}(1+\frac{1}{r^{m+1}})$ is continuous on $]0, +\infty[$ and equivalent to $\frac{c}{r}$ when $r$ tends to $+\infty$. Consequently, it is bounded on $[l, +\infty[$. Therefore, by (2) and (3), there exists a constant $M > 0$ such that, for all $r \in [l, +\infty[$, by (3), we obtain

$$(4) \qquad |\tilde{V}|\left(r + \frac{1}{r^q}\right) \le M|\tilde{V}|(r).$$

On the other hand,

$$\text{Log}\left(r + \frac{1}{r^q}\right) - \text{Log}r = \text{Log}\left(1 + \frac{1}{r^{q+1}}\right)$$

clearly satisfies an inequality of the form

$$\text{Log}\left(1 + \frac{1}{r^{q+1}}\right) \ge \frac{c_2}{r^{q+1}}$$
in $[l, +\infty[$ with $c_2 > 0$. Moreover, we can obviously find positive constants $c_3, c_4$ such that

$$ \left(r + \frac{1}{r^q}\right) |P|\left(r + \frac{1}{r^q}\right) \le c_3 r^{c_4}. $$

Consequently, by (1) and (4), we can find positive constants $c_5, c_6$ such that $|V|(r) \le c_5 r^{c_6} |\tilde{V}|(r) \quad \forall r \in [l, +\infty[$. Thus, writing again $V = \overline{V}\tilde{V}$, we have $|\overline{V}|(r)|\tilde{V}|(r) \le c_5 r^{c_6} |\tilde{V}|(r)$, and hence

$$ |\overline{V}|(r) \le c_5 r^{c_6} \quad \forall r \in [l, +\infty[$. $$

Consequently, by Corollary 22.31, $\overline{V}$ is a polynomial of degree $\le c_6$ and hence it has finitely many zeros and so does $V$. But then, by Theorem 50.8, $f$ must be a rational function. $\square$

**Corollary 50.13.** Let $f$ be a meromorphic function on $\mathbb{K}$ such that, for some $c, q \in ]0, +\infty[$, $t(r, f)$ satisfies $t(r, f) \le cr^q$ in $[1, +\infty[$. If for some $b \in \mathbb{K}$ $f' - b$ has finitely many zeros, then $f$ is a rational function.

**Proof.** Suppose $f' - b$ has finitely many zeros. Then $f - bx$ satisfies the same hypothesis as $f$, hence it is a rational function and so is $f$. $\square$

**Corollary 50.14.** Let $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}(x)$ admitting primitives and be such that $\log(t(r, f)) = O(\log r)$. Then $f$ has no quasi-exceptional value.

**Proof.** Indeed, since $f$ admits primitives, all poles are multiple, hence, given a primitive $F$ of $f$, we have $t(r, F) \le t(r, f)$. Consequently, by hypothesis, we have $\log(t(r, F)) \le O(\log(r))$, hence thanks to Corollary 50.13, $F'$ has no quasi-exceptional value. $\square$

Thanks to Theorem 31.5, we can derive Corollaries 50.15 and 50.16:

**Corollary 50.15.** Let $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}(x)$, admitting primitives, be of the form $\frac{g}{h}$ and be such that $\rho(h) < +\infty$ and $\psi(h) < +\infty$. Then $f$ has no quasi-exceptional value.

**Corollary 50.16.** Let $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}(x)$, admitting primitives, be of the form $\frac{g}{h}$ with $g, h \in \mathcal{A}(\mathbb{K})$ and be such that $\rho(h) < +\infty$ and $\sigma(h) < +\infty$. Then $f$ has no quasi-exceptional value.
**Corollary 50.17.** Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$ be of the form $\frac{q}{h}$ with $g, h \in \mathcal{A}(\mathbb{K})$ and be such that $\rho(h) < +\infty$ and $\sigma(h) < +\infty$. Then $f'$ has no quasi-exceptional value.

**Proof.** Let $f = \frac{q}{h}$ with $\sigma(h) < +\infty$. Then, $f' = \frac{q'h-h'g}{h^2}$ and $\sigma(h^2) = 2\sigma(h)$. Thus, one can apply Corollary 50.16 to $f'$. $\square$

**Corollary 50.18.** Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$ be of the form $\frac{q}{h}$ with $g, h \in \mathcal{A}(\mathbb{K})$ and be such that $\rho(h) < +\infty$ and $\sigma(h) < +\infty$. If $f$ has a quasi-exceptional value, then it has a non-zero residue.

**Example 1.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $|a_n| \le |a_{n+1}|$ and $\lim_{n \to +\infty} |a_n| = +\infty$ and let

$$f(x) = \sum_{n=0}^{\infty} \frac{b_n}{(x - a_n)^{s_n}} \quad \text{with } |b_n| \le 1, s_n \ge 2 \quad \forall n \text{ and } s_n = 2 \quad \forall n \ge t.$$

Then the function $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{(x-a_n)^{s_n}}$ admits primitives and has no quasi-exceptional value by Theorem 50.7.

**Example 2.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $|a_n| < |a_{n+1}|$ and $\lim_{n \to +\infty} |a_n| = +\infty$ and suppose that $\log(n) = O(\log|a_n|)$. Then the function $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{(x-a_n)^{s_n}}$ with $|b_n| \le 1, s_n \ge 2 \quad \forall n$, admits primitives and has no quasi-exceptional value by Corollary 50.14.

**Example 3.** Let $h \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}[x]$ be a function having only zeros of order 1 and let $P(x) \in \mathbb{K}[x]$. Let $f(x) = \frac{P(x)}{h(x)^2}$. Then $f$ has no primitive.

Indeed, suppose that $f$ have a primitive $F = \frac{U}{V}$, where $U$ and $V$ lie in $\mathcal{A}(\mathbb{K})$ have no common zeros. Since the zeros of $h$ are of order 1, it is seen that all zeros of $V$ are of order 1 and are all the zeros of $h$. Consequently, $\tilde{V} = 1$, $\overline{V} = V$, and $F' = \frac{U'V-UV'}{V^2}$ admits no simplification. Therefore, $U'V - UV' = P$. But then, by Theorem 50.5, $U$ and $V$ are polynomials and $V^2 = h^2$, a contradiction to the hypothesis $h \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}[x]$.

**Remark 1.** In Example 3, the function $f$ certainly has residues different from 0 because if all residues were null, the function then would have primitives.
**Remark 2.** Let $f$ be a non-rational meromorphic function in $\mathbb{C}$ admitting primitives. Then the second main theorem lets easily show that $f$ has at most one Picard value [66].

Now, by Theorems 50.7 and 50.9 and Corollary 50.14, it seems that a function admitting primitives has no quasi-exceptional value. Actually, as we saw in Theorem 37.12, a meromorphic function in $\mathbb{K}$ admitting primitives is a meromorphic function whose residues are all null. Thus, the following conjecture appears likely:

**Conjecture.** *A transcendental meromorphic function in $\mathbb{K}$ whose residues are all null has no quasi-exceptional value.*
Chapter 51

Small Functions

Throughout this chapter, the field $K$ is supposed to be of characteristic 0.

Small functions with respect to a meromorphic function are well known in the general theory of complex functions. Particularly, one knows the Nevanlinna theorem on three small functions. Here we construct a similar theory.

**Definition and notation.** Throughout this chapter, we set $a \in K$ and $R \in [0, +\infty[$ and we still denote by $D$ the set $K\setminus d(0, R^{-})$. For each $f \in M(K)$ (respectively $f \in M(d(a, R^{-}))$, respectively $f \in M(D)$), we denote by $M_f(K)$ (respectively $M_f(d(a, R^{-}))$, respectively $M_f(D)$) the set of functions $h \in M(K)$ (respectively $h \in M(d(a, R^{-}))$, respectively $M(D)$) such that $T(r, h) = o(T(r, f))$ when $r$ tends to $+\infty$ (respectively $T(r, h) = o(T(r, f))$ when $r$ tends to $R$, respectively $T_R(r, h) = o(T_R(r, f))$ when $r$ tends to $+\infty$). Similarly, if $f \in A(K)$ (respectively $f \in A(d(a, R^{-}))$, $f \in A(D)$), we denote by $A_f(K)$ (respectively $A_f(d(a, R^{-}))$, respectively $A_f(D)$) the set $M_f(K) \cap A(K)$ (respectively $M_f(d(a, R^{-})) \cap A(d(a, R^{-}))$, respectively $M_f(D) \cap A(D)$).

The elements of $M_f(K)$ (respectively $M_f(d(a, R^{-}))$, respectively $M_f(D)$) are called small meromorphic functions with respect to $f$, small functions in brief. Similarly, if $f \in A(K)$ (respectively $f \in A(d(a, R^{-}))$, respectively $f \in A(D)$), the elements of $A_f(K)$ (respectively $A_f(d(a, R^{-}))$, respectively $A_f(D)$) are called small analytic functions with respect to $f$, small functions in brief.
Theorems 51.1 and 51.2 are immediate consequences of Theorems 43.3 and 46.8:

**Theorem 51.1.** Let $a \in \mathbb{K}$ and $r > 0$. $\mathcal{A}_f(\mathbb{K})$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}(\mathbb{K})$, $\mathcal{A}_f(d(a, R^-))$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}(d(a, R^-))$, $\mathcal{A}_f(D)$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}(D)$, $\mathcal{M}_f(\mathbb{K})$ is a subfield field of $\mathcal{M}(\mathbb{K})$, $\mathcal{M}_f(d(a, R^-))$ is a subfield of field of $\mathcal{M}(a, R^-)$, and $\mathcal{M}_f(D)$ is a subfield field of $\mathcal{M}(D)$. Moreover, $\mathcal{A}_b(d(a, R^-))$ is a subalgebra of $\mathcal{A}_f(d(a, R^-))$ and $\mathcal{M}_b(d(a, R^-))$ is a subfield of $\mathcal{M}_f(d(a, R^-))$.

**Theorem 51.2.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(0, R^-))$, respectively $f \in \mathcal{M}(\mathbb{K})$) and let $g \in \mathcal{M}_f(\mathbb{K})$ (respectively $g \in \mathcal{M}_f(d(0, R^-))$, respectively $g \in \mathcal{M}_f(D)$). Then, $T(r, fg) = T(r, f) + o(T(r, f))$ and $T(r, \frac{f}{g}) = T(r, f) + o(T(r, f))$ (respectively $T(r, fg) = T(r, f) + o(T(r, f))$ and $T(r, \frac{f}{g}) = T(r, f) + o(T(r, f))$, respectively $T_R(r, fg) = T_R(r, f) + o(T_R(r, f))$ and $T_R(r, \frac{f}{g}) = T_R(r, f) + o(T_R(r, f))$).

Here we can mention some precisions to Theorem 51.1 that are useful later:

**Theorem 51.3.** Let $f \in \mathcal{A}(\mathbb{K})$ (respectively let $f \in \mathcal{A}_u(d(a, r^-))$, respectively let $f \in \mathcal{A}(D)$). Let $g, h \in \mathcal{A}_f(\mathbb{K})$ (respectively let $g, h \in \mathcal{A}_f(d(a, r^-))$, respectively let $g, h \in \mathcal{A}_f(D)$) with $g$ and $h$ not identically zero. If $gh$ belongs to $\mathcal{A}_f(\mathbb{K})$ (respectively to $\mathcal{A}_f(d(a, r^-))$, respectively to $\mathcal{A}_f(D)$), then so do $g$ and $h$.

**Proof.** Concerning the claim on $f \in \mathcal{A}_u(d(a, r^-))$, we can obviously assume $a=0$. By Theorems 43.3 and 46.7, we have $T(r, g.h) = T(r, g) + T(r, h) + O(1)$. Consequently, $T(r, g.h) = o(T(r, f))$ if and only if $T(r, g) = o(T(r, f))$ and $T(r, h) = o(T(r, f))$. $\square$

**Theorem 51.4.** Let $f, g \in \mathcal{A}(\mathbb{K})$ (respectively let $f, g \in \mathcal{A}_u(d(0, r^-))$, respectively let $f, g \in \mathcal{A}(D)$) and let $q \in \mathbb{N}^*$. If $\frac{f}{g}$ is not a qth root of 1, then $f^q - g^q$ does not belong to $\mathcal{A}_f(\mathbb{K})$ (respectively to $\mathcal{A}_f(d(0, r^-))$, respectively to $\mathcal{A}_f(D)$).

**Proof.** Concerning the claim on $f \in \mathcal{A}_u(d(a, r^-))$, we can obviously assume $a=0$. Let $h = \frac{f}{g}$. Since $h$ is not a qth root of 1, neither $f-g$ nor the function $F(x) = \sum_{j=0}^{q-1} f^j g^{q-1-j}$ is identically zero.
Suppose that $f^q - g^q \in \mathcal{A}_f(\mathbb{K})$ (respectively $f^q - g^q \in \mathcal{A}_f(d(0, r^{-}))$, respectively $\mathcal{A}_f(D)$). So, by Theorem 51.3, both $f-g$ and $F$ belong to $\mathcal{A}_f(\mathbb{K})$ (respectively to $\mathcal{A}_f(d(0, r^{-}))$, respectively to $\mathcal{A}_f(D)$). Let $w = f-g$, hence $g = f+w$. Then, $F(x) = \sum_{j=0}^{q-1} f^j(f+w)^{q-1-j}$. Thus, we can check that $F(x)$ is of the form $f^{q-1}+P(f(x))$ with $P(Y)$ a polynomial in $Y$ of degree at most $q-2$, with coefficients in $\mathcal{A}_f(\mathbb{K})$ (respectively in $\mathcal{A}_f(d(0, r^{-}))$, respectively in $\mathcal{A}_f(D)$). Consequently, by Theorems 43.3 and 46.7, $T(r, F(x))$ is of the form $(q-1)T(r, f) + o(T(r, f))$ because $T(r, P(f(x))) \le (q-2)T(r, f) + o(T(r, f))$, which proves that $F$ does not belong to $\mathcal{A}_f(\mathbb{K})$ (respectively to $\mathcal{A}_f(d(0, r^{-}))$, respectively to $\mathcal{A}_f(D))$. $\square$

In the proof of Theorem 51.5 and in the sequel, we have to use the following notation:

**Notation.** Let $h \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$ (respectively $h \in E(x)\backslash E$) and let $\mathcal{Z}(h)$ be the set of zeros $c$ of $h'$ such that $h(c) \ne h(d)$ for every zero $d$ of $h'$ other than $c$. If $\mathcal{Z}(h)$ is finite, we denote by $\Upsilon(h)$ its cardinal, and if $\mathcal{Z}(h)$ is not finite, we put $\Upsilon(h) = +\infty$.

Theorem 51.5 is a wide generalization of Theorem 43.8. It consists of the following claim: given a meromorphic function $f$ and a rational function $G$ of degree $n$ whose coefficients are small functions with respect to $f$, then $T(r, G(f))$ is equivalent to $nT(r, f)$. The big difficulty consists of showing that $T(r, G(f))$ is not smaller than $nT(r, f)$. The proof, based on an elementary property of Bezout's theorem, was given in $\mathbb{C}$ by F. Gackstatter and I. Laine and was made in a field such as $\mathbb{K}$ by C.C. Yang and Peichu Hu [67].

**Theorem 51.5.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}(d(0, R^{-}))$, respectively let $f \in \mathcal{M}(D)$). Let $G(Y) \in \mathcal{M}_f(\mathbb{K})(Y)$ (respectively $G \in \mathcal{M}_f(d(0, R^{-}))(Y)$, respectively $G(Y) \in \mathcal{M}_f(D)(Y)$) and let $n = \deg(G)$. Then $T(r, G(f)) = nT(r, f) + o(T(r, f))$ (respectively $T(r, G(f)) = nT(r, f) + o(T(r, f))$, respectively $T_R(r, G(f)) = nT_R(r, f) + o(T_R(r, f))$).

**Proof.** Let $G = \frac{P}{Q}$ with $P, Q$ relatively prime in the ring $\mathcal{M}_f(\mathbb{K})$ (respectively $\mathcal{M}_f(d(a, R^{-}))$, $\mathcal{M}_f(D)$). Suppose first $G(Y) \in \mathcal{M}_f(\mathbb{K})[Y]$ (respectively $G \in \mathcal{M}_f(d(0, R^{-}))[Y]$, respectively $G(Y) \in \mathcal{M}_f(D)[Y]$), hence $G = P$. Let $P(X) = \sum_{j=0}^{n} b_j X^j$ with $c_j \in \mathcal{M}_f(\mathbb{K})$ (respectively $c_j \in \mathcal{M}_f(d(0, R^{-})), c_j \in \mathcal{M}_f(D)$). By Theorems 43.8
and 46.7, we have $T(r, P(f)) = T(r, b_n^{-1}P(f)) + o(T(r, f))$ (respectively $T(r, P(f)) = T(r, b_n^{-1}P(f)) + o(T(r, f))$, respectively $T_R(r, P(f)) = T_R(r, b_n^{-1}P(f)) + o(T_R(r, f))$). Consequently, without loss of generality, we may also assume that $P$ is monic.

Let $\hat{\mathbb{K}}$ be an algebraically closed spherically complete extension of $\mathbb{K}$. Given $a \in \mathbb{K}$, we denote by $\hat{d}(a, R^{-})$ the disk $\{x \in \hat{\mathbb{K}} | |x - a| < R\}$.

Suppose first $f \in \mathcal{M}(\mathbb{K})$ or $f \in \mathcal{M}(D)$. We can write $f = \frac{h}{l}$ with $h, l \in \mathcal{A}(\mathbb{K})$ having no common zero. Now suppose that $f \in \mathcal{M}(d(0, R^{-}))$. Since $f$ has continuation to a function $\hat{f}$ meromorphic in the disk $\hat{d}(0, R^{-})$ of the field $\hat{\mathbb{K}}$, by Lemma 43.1, we know that the Nevanlinna function $T(r, \hat{f})$ in $\hat{\mathbb{K}}$ is exactly this of $f$ in $\mathbb{K}$. Consequently, without loss of generality, we may assume that $\mathbb{K}$ is spherically complete. Thus, we can write $f$ in the form $f = \frac{h}{l}$ with $h, l \in \mathcal{A}(d(0, R^{-}))$ having no common zero. Then $P(f)$ is in the form $\frac{\sum_{j=0}^{n} b_j h^j l^{n-j}}{l^n}$ with $B, b_j \in \mathcal{A}_f(\mathbb{K})$ (respectively $B, b_j \in \mathcal{A}_f(d(0, R^{-}))$, respectively $B, b_j \in \mathcal{A}_f(D)$). Clearly, we have $Z(r, \frac{\sum_{j=0}^{n} b_j h^j l^{n-j}}{l^n}) \le Z(r, \sum_{j=0}^{n} b_j h^j l^{n-j})$ and by Theorem 43.8 (respectively by Theorem 43.8, respectively by Theorem 46.7)

$$Z\left(r, \sum_{j=0}^{n} b_j h^j l^{n-j}\right) \le \max_{0 \le j \le n} Z(r, b_j h^j l^{n-j}) \le nT(r, f) + o(T(r, f))$$

(respectively)

$$Z\left(r, \sum_{j=0}^{n} b_j h^j l^{n-j}\right) \le \max_{0 \le j \le n} Z(r, b_j h^j l^{n-j}) \le nT(r, f) + o(T(r, f)),$$

respectively

$$Z_R\left(r, \sum_{j=0}^{n} b_j h^j l^{n-j}\right) \le \max_{0 \le j \le n} Z_R(r, b_j h^j l^{n-j}) \\ \le nT_R(r, f) + o(T_R(r, f)).$$
On the other hand, $N(r, P(f)) \le nN(r, f) + o(T(r, f))$ (respectively $N(r, P(f)) \le nN(r, f) + o(T(r, f))$, respectively $N_R(r, P(f)) \le nN_R(r, f) + o(T_R(r, f))$), hence $T(r, P(f)) \le nT(r, f) + o(T(r, f))$ (respectively $T(r, P(f)) \le nT(r, f) + o(T(r, f))$, respectively $T_R(r, P(f)) \le nT_R(r, f) + o(T_R(r, f))$).

So, now it remains us to prove the reverse inequality. Indeed, suppose that inequality does not hold. For simplicity, we first suppose that $f$ lies either in $M(\mathbb{K})$ or in $M(d(0, R^{-}))$.

Then there exists $\rho \in ]0, 1[$ and a sequence of intervals $]r'_m, r''_m[$ such that $\lim_{m \to +\infty} r'_m = +\infty$ (respectively $\lim_{m \to +\infty} r'_m = R$, respectively $\lim_{m \to +\infty} r'_m = +\infty$) and that $T(r, P(f)) \le \rho nT(r, f) \quad \forall r \in \bigcup_{m=1}^{\infty} ]r'_m, r''_m[$. Particularly, we have $N(r, P(f)) \le \rho nT(r, f) \quad \forall r \in \bigcup_{m=1}^{\infty} ]r'_m, r''_m[$.

Let us write again $P(f)$ in the form $\sum_{j=0}^{n} c_j f^j$. We shall prove the following

$$ (1) \quad N(r, P(f)) \ge nN(r, f) - n \sum_{j=0}^{n} N(r, c_j) = nN(r, f) + o(T(r, f)). $$

Indeed, let $\alpha$ be a pole of $f$ and suppose it is not a pole of order $\ge -n\omega_\alpha(f)$ of $P(f)$. We can check that there must exist $j \in \{0, \dots, n-1\}$ such that $-\omega_\alpha(c_j) \ge -\omega_\alpha(f)$. Consequently, at $\alpha$, we have $-\omega_\alpha(P(f)) \ge -n\omega_\alpha(f) - \max_{j=0, \dots, n-1} (-n\omega_\alpha(c_j))$, hence

$$ N(r, f | \omega_{\alpha}(P(f)) < n\omega_{\alpha}(f)) \le n \sum_{j=0}^{n-1} N(r, c_j), $$

hence (1) follows clearly.

Consequently, there exists $q \in \mathbb{N}$ such that $nN(r, f) \le \rho nT(r, f) \quad \forall r \in \bigcup_{m=q}^{\infty} ]r'_m, r''_m[$ and therefore there exists $s \in \mathbb{N}$ and $\lambda \in ]0, 1[$ such that

$$ (2) \quad N(r, Bl^n) < n\lambda T(r, f) \quad \forall r \in \bigcup_{m=s}^{\infty} ]r'_m, r''_m[$. $$

In particular, there exists $t \in \mathbb{N}$ such that

$$ T(r, b_j h^j l^{n-j}) < \lambda T(r, b_n h^n) \quad \forall j = 0, \dots, n-1, \forall r \in \bigcup_{m=t}^{\infty} ]r'_m, r''_m[$. $$

Now, since $T(r, b_n h^n) > T(r, b_j h^j l^{n-j}) \quad \forall j = 0, \dots, n-1, \forall r \in \bigcup_{m=t}^{\infty} ]r'_m, r''_m[$, we note that $|\sum_{j=0}^{n} b_j h^j l^{n-j}|(r) = |b_n|(r) \forall r \in \bigcup_{m=t}^{\infty} ]r'_m, r''_m[$,
hence

$$T\left(r, \sum_{j=0}^{n} b_j h^j l^{n-j}\right) = T(r, b_n h^n) \quad \forall r \in \bigcup_{m=t}^{\infty} ]r'_m, r''_m[.$$

Consequently, there exists $\ell \in ]\rho, 1[$ and $u \in \mathbb{N}$ ($u \ge t$) such that

$$(3) \quad T\left(r, \sum_{j=0}^{n} b_j h^j l^{n-j}\right) > \ell n T(r, h) \quad \forall r \in \bigcup_{m=u}^{\infty} ]r'_m, r''_m[.$$

Now, by (2), we have $T(r, h) = T(r, f) \quad \forall r \in \bigcup_{m=u}^{\infty} ]r'_m, r''_m[$ and

$$T(r, P(f)) = T\left(r, \sum_{j=0}^{n} b_j h^j l^{n-j}\right) \quad \forall r \in \bigcup_{m=u}^{\infty} ]r'_m, r''_m[,$$

hence finally, by (3), we obtain $T(r, P(f)) > \ell n T(r, f) \quad \forall r \in \bigcup_{m=u}^{\infty} ]r'_m, r''_m[$, a contradiction which proves the claim when $G$ is a polynomial.

Similarly, replacing $N$ by $N_R$, $T$ by $T_R$, we can make the same reasoning when $f$ belongs to $M(D)$.

We now consider the general case $G(Y) \in \mathcal{M}_f(\mathbb{K})(Y)$ (respectively $G \in \mathcal{M}_f(d(0, R^-))(Y)$). Without loss of generality, we may assume $\deg(G) = \deg(P)$.

Since $P$ and $Q$ are relatively prime, by Bezout's theorem in a ring of polynomials on a field, we can find $A, B \in \mathcal{M}_f(\mathbb{K})[Y]$ (respectively $A, B \in \mathcal{M}_f(d(0, R^-))[Y]$) such that $AQ + PB = 1$. Since $\deg(Q) \le \deg(P)$, of course, $\deg(B) \le \deg(A)$, hence $\deg(\frac{A}{B}) = \deg(A)$. Now,

$$T\left(r, \frac{B(f)}{A(f)} + \frac{Q(f)}{P(f)}\right) = T\left(r, \frac{1}{A(f)P(f)}\right) = T(r, A(f)P(f)) + O(1).$$

Consequently, by the theorem already proven when $G$ is a polynomial, we have

$$T\left(r, \frac{B(f)}{A(f)} + \frac{Q(f)}{P(f)}\right) = (\deg(A) + \deg(P))T(r, f) + o(T(r, f)),$$

and since $\deg(P) = \deg(G)$, actually we have

$$(4) \quad T\left(r, \frac{B(f)}{A(f)} + \frac{Q(f)}{P(f)}\right) = (\deg(A) + \deg(G))T(r, f) + o(T(r, f)).$$
Now, $T(r, \frac{B(f)}{A(f)} + \frac{Q(f)}{P(f)}) \le T(r, \frac{B(f)}{A(f)}) + T(r, \frac{Q(f)}{P(f)}) + o(T(r, f))$, and by the first inequality already proven above, we obtain

$$T\left(r, \frac{B(f)}{A(f)} + \frac{Q(f)}{P(f)}\right) \\ \le \deg\left(\frac{B}{A}\right) T(r, f) + T\left(r, \frac{Q(f)}{P(f)}\right) + o(T(r, f)).$$

But since $\deg(B) \le \deg(A)$, actually we have

$$T\left(r, \frac{B(f)}{A(f)} + \frac{Q(f)}{P(f)}\right) \le \deg(A)T(r, f) + T\left(r, \frac{Q(f)}{P(f)}\right) + o(T(r, f))$$

i.e.

$$(5) \qquad T\left(r, \frac{B(f)}{A(f)} + \frac{Q(f)}{P(f)}\right) \\ \le \deg(A)T(r, f) + T(r, G(f)) + o(T(r, f)).$$

Now, by (4) and (5), we can see that $\deg(G)T(r, f) \le T(r, G(f)) + o(T(r, f))$. Similarly, when $f$ belongs to $M(D)$ and $G$ belongs to $M_f(D)(Y)$, we can make the same reasoning, as above. This completes the proof. □

**Theorem 51.6.** Let $a \in \mathbb{K}$ and $r > 0$. Let $f \in M(\mathbb{K})\backslash\mathbb{K}(x)$ (respectively $f \in M_u(d(a, R^{-}))$, respectively $f \in M^c(D)$). Then, $f$ is transcendental over $M_f(\mathbb{K})$ (respectively over $M_f(d(a, R^{-}))$, respectively over $M_f(D)$).

**Proof.** Suppose there exists a polynomial $P(Y) = \sum_{j=0}^n a_j Y^j \in M_f(\mathbb{K})[Y] \ne 0$ such that $P(f) = 0$. If $f$ belongs to $M_u(d(a, R^{-}))$, we may obviously suppose that $a = 0$. By Theorem 51.5, we have $T(r, a_n f^n) = nT(r, f) + o(T(r, f))$ whenever $f$ belongs to $M(\mathbb{K})\backslash\mathbb{K}(x)$ or to $M_f(d(0, R^{-}))$ and $T_R(r, a_n f^n) = nT_R(r, f) + o(T_R(r, f))$ whenever $f$ belongs to $M^c(\mathbb{K})$, whereas $T(r, \sum_{j=0}^{n-1} a_j f^j) = (n-1)T(r, f) + o(T(r, f))$, a contradiction. □

**Corollary 51.7.** Let $a \in \mathbb{K}$ and $r > 0$. Let $f \in M(\mathbb{K})\backslash\mathbb{K}(x)$ (respectively $f \in M_u(d(a, R^{-}))$, respectively $f \in M^c(D)$). Then, $f$ is transcendental over $\mathbb{K}(x)$. □
A function $h \in \mathcal{M}_b(d(a, R^-))$ is obviously small with respect to any function $f \in \mathcal{M}_u(d(a, R^-))$. So, we have the following corollary:

**Corollary 51.8.** Let $a \in \mathbb{K}$ and $r > 0$ and let $f \in \mathcal{M}_u(d(a, R^-))$. Then, $f$ is transcendental over $\mathcal{M}_b(d(a, R^-))$. Let $f \in \mathcal{M}^c(D)$. Then $f$ is transcendental over $\mathcal{M}_0(D)$.

By Corollary 36.23, we know that a meromorphic function $f \in \mathcal{M}(\mathbb{K})$ or $f \in \mathcal{M}(d(0, R^-))$ admits at most one quasi-exceptional value. Here we generalize that statement.

**Theorem 51.9.** Let $a \in \mathbb{K}$ and $r > 0$. Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$ (respectively $f \in \mathcal{M}_u(d(a, R^-)$), respectively $f \in \mathcal{M}^c(D)$). There exists at most one function $g \in \mathcal{M}_f(\mathbb{K})$ (respectively $g \in \mathcal{M}_f(d(a, R^-)$), respectively $g \in \mathcal{M}_f(D)$) such that $f-g$ have finitely many zeros. Moreover, if $f$ belongs to $\mathcal{A}(\mathbb{K})\backslash\mathbb{K}[x]$ (respectively to $\mathcal{A}_u(d(a, R^-)$), respectively to $\mathcal{A}^c(D)$), then there exists no function $g \in \mathcal{M}_f(\mathbb{K})\backslash\mathbb{K}(x)$ (respectively $g \in \mathcal{M}_f(d(a, R^-)$), respectively $g \in \mathcal{M}_f(D)$) such that $f-g$ have finitely many zeros.

**Proof.** Concerning claims on $\mathcal{M}_u(d(a, R^-))$, we can obviously assume $a = 0$. Suppose that there exist two distinct functions $g_1, g_2 \in \mathcal{M}_f(\mathbb{K})$ (respectively $g_1, g_2 \in \mathcal{M}_f(d(0, R^-))$, respectively $g_1, g_2 \in \mathcal{M}_f(D)$) such that $f-g_k$ has finitely many zeros. So, there exist $P_1, P_2 \in \mathbb{K}[x]$ and $h_1, h_2 \in \mathcal{A}(\mathbb{K})$ (respectively $h_1, h_2 \in \mathcal{A}(d(0, R^-)))$ such that $f-g_k = \frac{P_k}{h_k}$, $k = 1, 2$, and hence we note that

$$ (1) \qquad \begin{aligned} T(r, f) &= T\left(r, \frac{P_k}{h_k}\right) + o(T(r, f)) \\ &= T(r, h_k) + o(T(r, f)) \quad k = 1, 2. \end{aligned} $$

Consequently, putting $g = g_2 - g_1$, we have

$$ \frac{P_1}{h_1} = \frac{P_2}{h_2} + g, $$

and by Theorem 51.1, $g$ belongs to $\mathcal{M}_f(\mathbb{K})$ (respectively to $\mathcal{M}_f(d(0, R^-)))$. Therefore, $P_1h_2 - P_2h_1 = gh_1h_2$, and hence,

$$ (2) \qquad T(r, P_1h_2 - P_2h_1) = T(r, gh_1h_2). $$
Now, by Theorem 43.3, we have

$$ 
\begin{aligned}
T(r, P_1 h_2 - P_2 h_1) &\le \max(T(r, P_1 h_2), T(r, P_2 h_1)) \\
&\le \max(T(r, h_1), T(r, h_2)) + o(T(r, f)),
\end{aligned}
$$

and hence by (1), we obtain

$$ (3) \qquad T(r, P_1 h_2 - P_2 h_1) \le T(r, f) + o(T(r, f)). $$

On the other hand, by Theorem 51.2, we have

$$ T(r, gh_1 h_2) = T(r, h_1 h_2) + o(T(r, h_1 h_2)) = 2T(r, f) + o(T(r, f)), $$

a contradiction to (3).

Now, if $f$ belongs to $M^c(D)$, we can make the same reasoning with $T_R$ instead of $T$.

Suppose now that $f$ belongs to $A(\mathbb{K})\backslash\mathbb{K}[x]$ and that there exists a function $w \in M_f(\mathbb{K})$ such that $f - w$ has finitely many zeros. Set $w = \frac{l}{t}$, where $l$ and $t$ belong to $A_f(\mathbb{K})$ and have no common zeros. Thus, $f - w = \frac{tf - l}{t}$ and each zero of $tf - l$ cannot be a zero of $t$, hence is zero of $f - w$. Consequently, since $f - w$ has finitely many zeros, $tf - l$ has finitely many zeros and hence is a polynomial. But since $l$ belongs to $A_f(\mathbb{K})$, when $r$ is big enough, we have $|f|(r) > |l|(r)$ and hence $|tf|(r) > |l|(r)$, therefore $|tf - l|(r) = |tf|(r)$. And since $f$ is transcendental, by Corollary 22.23, for every fixed $q \in \mathbb{N}$, $|f|(r) > r^q$ when $r$ is big enough. Similarly, $|tf - l|(r) > r^q$ when $r$ is big enough. Consequently, by Corollary 22.23, $tf - l$ is not a polynomial, which proves that $w$ does not exist.

Suppose now that $f$ belongs to $A_u(d(0, R^-))$ and that there exists a function $w \in M_f(d(0, R^-))$ such that $f - w$ has finitely many zeros. Without loss of generality, we can assume that the field $\mathbb{K}$ is spherically complete because both $f$ and $w$ have continuation to an algebraically closed spherically complete extension of $\mathbb{K}$ where their zeros are the same as in $\mathbb{K}$. Consequently, we can write $w = \frac{l}{t}$, where $l$ and $t$ have no common zeros. Now, the zeros of $f - w$ are those of $tf - l$, hence $tf - l$ has finitely many zeros and hence is bounded in $d(0, R^-)$. But since $w$ belongs to $M_f(d(0, R^-))$, so does $l$ and hence $|tf|(r) > |l|(r)$ when $r$ tends to $R$. Consequently, $|tf - l|(r) = |tf|(r)$ is not bounded in $d(0, R^-)$, a contradiction proving again that $w$ does not exist.
Suppose finally that $f$ belongs to $\mathcal{A}^c(D)$. We can make the same reasoning as in $\mathcal{A}(\mathbb{K})$ by replacing $T$ by $T_R$. $\square$

Theorem 51.10 is known as second main theorem on three small functions in $p$-adic analysis [67]. It holds as well as in complex analysis, where it was showed first [67]. Note that this theorem was generalized to any finite set of small functions by K. Yamanoi in complex analysis [94], through methods that have no equivalent on a $p$-adic field.

**Theorem 51.10.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}_u(d(0, R^{-}))$, respectively $f \in \mathcal{M}^c(D)$) and let $w_1, w_2, w_3 \in \mathcal{M}_f(\mathbb{K})$ (respectively $w_1, w_2, w_3 \in \mathcal{M}_f(d(0, R^{-}))$, respectively $w_1, w_2, w_3 \in \mathcal{M}_f(D)$) be pairwise distinct and let $S(r) = \max(T(r, w_j), j = 1, 2, 3)$. Then, $T(r, f) \le \sum_{j=1}^3 \bar{Z}(r, f - w_j) + 13S(r)$, respectively $T(r, f) \le \sum_{j=1}^3 \bar{Z}(r, f - w_j) + 13S(r)$, respectively $T_R(r, f) \le \sum_{j=1}^3 \bar{Z}_R(r, f - w_j) + 13S(r)$.

**Proof.** We make the proof when $f$ belongs to $\mathcal{M}(\mathbb{K})$ or to $\mathcal{M}_u(d(0, R^{-}))$. Let $\phi(x) = \frac{(f(x)-w_1(x))(w_2(x)-w_3(x))}{(f(x)-w_3(x))(w_2(x)-w_1(x))}$.

By Theorem 45.7, we have

$$ (1) \quad T(r, \phi) \le \bar{Z}(r, \phi) + \bar{Z}(r, \phi - 1) + \bar{N}(r, \phi) + O(1). $$

On the other hand, we have $T(r, f) \le T(r, f - w_3) + T(r, w_3)$, hence

$$ \begin{aligned} T(r, f) &\le T\left(r, \frac{1}{f - w_3}\right) + T(r, w_3) \\ &\le T\left(r, \frac{w_3 - w_1}{f - w_3}\right) + T(r, w_1 - w_3) + T(r, w_3) + O(1), \end{aligned} $$

thereby

$$ \begin{aligned} (2) \quad T(r, f) &\le T\left(r, \frac{w_3 - w_1}{f - w_3} + 1\right) + 3S(r) + O(1) \\ &= T\left(r, \frac{f - w_1}{f - w_3}\right) + 3S(r) + O(1). \end{aligned} $$
Now, $T(r, \frac{w_2-w_1}{w_2-w_3}) \le 4S(r) + O(1)$. Consequently, by writing $\frac{f-w_1}{f-w_3} = \phi(\frac{w_2-w_1}{w_2-w_3})$, we have $T(r, \frac{f-w_1}{f-w_3}) \le T(r, \phi) + T(r, \frac{w_2-w_1}{w_2-w_3}) \le T(r, \phi) + 4S(r) + O(1)$, and finally $T(r, f) \le T(r, \phi) + 7S(r) + O(1)$. Thus, by (1), we obtain

$$ (3) \quad T(r, f) \le \bar{Z}(r, \phi) + \bar{Z}(r, \phi - 1) + \bar{N}(r, \phi) + 7S(r) + O(1). $$

Now, we can check that

$$ \begin{aligned} & \bar{Z}(r, \phi) + \bar{Z}(r, \phi - 1) + \bar{N}(r, \phi) \\ & \le \sum_{j=1}^{3} \bar{Z}(r, f - w_j) + \sum_{1 \le j < k \le 3} \bar{Z}(r, w_k - w_j) \\ & \le \sum_{j=1}^{3} \bar{Z}(r, f - w_j) + 6S(r), \end{aligned} $$

which, by (3), completes the proof when $f$ belongs to $\mathcal{M}(\mathbb{K})$ or to $\mathcal{M}_u(d(0, R^-))$. When $f$ belongs to $\mathcal{M}(D)$, we can make a similar proof just by replacing $T$ by $T_R$ and $Z$ by $Z_R$. $\square$

**Theorem 51.11.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}_u(d(0, R^-)$), respectively $f \in \mathcal{M}^c(D)$) and let $w_1, w_2 \in \mathcal{M}_f(\mathbb{K})$ (respectively $w_1, w_2 \in \mathcal{M}_f(d(0, R^-)$), respectively $w_1, w_2 \in \mathcal{M}_f(D)$) be distinct. Let $S(r) = \max(T(r, w_1), T(r, w_2))$ (respectively $S(r) = \max(T(r, w_1), T(r, w_2))$), respectively $S(r) = \max(T_R(r, w_1), T_R(r, w_2))$. Then, $T(r, f) \le \bar{Z}(r, f - w_1) + \bar{Z}(r, f - w_2) + \bar{N}(r, f) + 7S(r)$, respectively $T(r, f) \le \bar{Z}(r, f - w_1) + \bar{Z}(r, f - w_2) + \bar{N}(r, f) + 7S(r)$, respectively $T_R(r, f) \le \bar{Z}_R(r, f - w_1) + \bar{Z}_R(r, f - w_2) + \bar{N}_R(r, f) + 7S(r)$).

**Proof.** Suppose first $f \in \mathcal{M}(\mathbb{K})$ or $f \in \mathcal{M}_u(d(0, R^-))$. Let $g = \frac{1}{f}$, $h_j = \frac{1}{w_j}$, $j = 1, 2$, $h_3 = 0$. Clearly,

$$ T(r, g) = T(r, f) + O(1), \quad T(r, h_j) = T(r, w_j), \quad j = 1, 2, $$

so we can apply Theorem 51.10 to $g$, $h_1$, $h_2$, $h_3$. So, we have $T(r, g) \le \bar{Z}(r, g - h_1) + \bar{Z}(r, g - h_1) + \bar{Z}(r, g) + 7S(r)$. But we note that $\bar{Z}(r, g - h_j) = \bar{Z}(r, f - w_j)$ for $j = 1, 2$ and that $\bar{Z}(r, g) = \bar{N}(r, f)$. Moreover, we know that $T(r, g) = T(r, f) + O(1)$. Consequently, when $w_1w_2$ is not identically zero, the claim is proved.
Now, suppose that $w_2 = 0$. Take $\lambda \in \mathbb{K}^*$, let $l = f + \lambda$ and $\theta_j = w_j + \lambda$ $j = 1, 2$. Thus, we have $T(r, l) = T(r, f) + O(1)$, $T(r, \theta_j) = T(r, w_j) + O(1)$, $j = 1, 2$, and $\overline{N}(r, l) = \overline{N}(r, f)$. We note that $\max_{j=1,2} T(r, \theta_j) = S(r) + O(1)$.

By the claim already proved when $w_1w_2 \neq 0$, we can write $T(r, l) \le \overline{Z}(r, l - \theta_1) + \overline{Z}(r, l - \theta_2) + \overline{N}(l, r) + 7S(r)$, hence $T(r, f) \le \overline{Z}(r, f - w_1) + \overline{Z}(r, f) + \overline{N}(f, r) + 7S(r) + O(1)$.

Suppose now $f \in \mathcal{M}(D)$. By replacing $T$ by $T_R$, $Z$ by $Z_R$, and $N$ by $N_R$, we can check that the same reasoning applies. $\square$

**Corollary 51.12.** Let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}_u(d(0, R^-))$, respectively $f \in \mathcal{A}^c(D)$) and let $w_1, w_2 \in \mathcal{A}_f(\mathbb{K})$ (respectively $w_1, w_2 \in \mathcal{A}_f(d(0, R^-))$, respectively $w_1, w_2 \in \mathcal{A}_f(D)$) be distinct. Let $S(r) = \max(T(r, w_1), T(r, w_2))$ (respectively $S(r) = \max(T(r, w_1), T(r, w_2))$, respectively $S(r) = \max(T_R(r, w_1), T_R(r, w_2))$). Then, $T(r, f) \le \overline{Z}(r, f - w_1) + \overline{Z}(r, f - w_2) + 7S(r) + O(1)$, respectively $T(r, f) \le \overline{Z}(r, f - w_1) + \overline{Z}(r, f - w_2) + 7S(r) + O(1)$, respectively $T_R(r, f) \le \overline{Z}_R(r, f - w_1) + \overline{Z}_R(r, f - w_2) + 7S(r) + O(1)$.

Here is now an application of that theory:

**Theorem 51.13.** Let $h, w \in \mathcal{A}_b(d(a, R^-))$ and let $m, n \in \mathbb{N}^*$ be such that $\min(m, n) \ge 2$, $\max(m, n) \ge 3$. Then the functional equation

$$ (\mathcal{E}) \quad (g(x))^n = h(x)(f(x))^m + w(x) $$

has no solution in $\mathcal{A}_u(d(a, R^-))$.

**Proof.** Without loss of generality, we can obviously assume $a = 0$. Let $F(x) = g(x)^n$. Thanks to Corollary 51.12, we can write

$$ (1) \quad T(r, F) \le \overline{Z}(r, F) + \overline{Z}(r, F - w) + o(T(r, F)). $$

Now, it appears that $\overline{Z}(r, F) \le \frac{1}{n}Z(r, F)$. Moreover, since $h$ is bounded, $Z(r, h)$ is bounded, hence $\overline{Z}(r, hf^m) \le Z(r, f) + Z(r, h) = Z(r, f) + O(1)$, hence

$$ (2) \quad \overline{Z}(r, hf^m) \le \frac{1}{m}Z(r, hf^m) + O(1) = \frac{1}{m}Z(r, F) + O(1). $$
On the other hand, $Z(r, F) = Z(r, F - w) + O(1) = T(r, F) + O(1)$.
Consequently, by (1) and (2), we can derive

$$T(r, F) \le \left( \frac{1}{m} + \frac{1}{n} \right) T(r, F) + o(T(r, F)).$$

Therefore, we have $\frac{1}{m} + \frac{1}{n} \ge 1$, a contradiction to the hypothesis which implies $\frac{1}{m} + \frac{1}{n} \le \frac{5}{6}$. $\square$

**Theorem 51.14.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental (respectively $f \in \mathcal{M}_u(d(0, R^-))$, respectively $f \in \mathcal{M}^c(D)$) and let $w_j \in \mathcal{M}_f(\mathbb{K})$ ($j = 1, \dots, q$) (respectively $w_j \in \mathcal{M}_f(d(a, R^-))$, respectively $w_j \in \mathcal{M}_f(D)$) be $q$ distinct small functions. Then,

$$qT(r, f) \le 3 \sum_{j=1}^{q} \bar{Z}(r, f - w_j) + o(T(r, f)),$$

(respectively

$$qT(r, f) \le 3 \sum_{j=1}^{q} \bar{Z}(r, f - w_j) + o(T(r, f)),$$

respectively

$$qT_R(r, f) \le 3 \sum_{j=1}^{q} \bar{Z}_R(r, f - w_j) + o(T_R(r, f)).$$

Moreover, if $f$ has finitely many poles in $\mathbb{K}$ (respectively in $d(0, R^-)$, respectively in $D$), then

$$qT(r, f) \le 2 \sum_{j=1}^{q} \bar{Z}(r, f - w_j) + o(T(r, f)),$$

(respectively

$$qT(r, f) \le 2 \sum_{j=1}^{q} \bar{Z}(r, f - w_j) + o(T(r, f)),$$

respectively

$$qT_R(r, f) \le 2 \sum_{j=1}^{q} \bar{Z}_R(r, f - w_j) + o(T_R(r, f)).$$
**Proof.** Suppose first that $f$ and $w_j$ ($j = 1, \dots, q$) belong to $\mathcal{M}(\mathbb{K})$ or to $\mathcal{M}(d(0, R^{-}))$. By Theorem 51.10, for every triplet $(i, j, k)$ such that $1 \le i \le j \le k \le q$, we can write

$$T(r, f) \le \bar{Z}(r, f - w_i) + \bar{Z}(r, f - w_j) + \bar{Z}(r, f - w_k) + o(T(r, f)).$$

The number of such inequalities is $C_q^3$. Summing up, we obtain

$$(1) \quad C_q^3 T(r, f) \le \sum_{(i,j,k), 1 \le i \le j \le k \le q} \bar{Z}(r, f - w_i) + \bar{Z}(r, f - w_j) + \bar{Z}(r, f - w_k) + o(T(r, f)).$$

In this sum, for each index $i$, the number of terms $\bar{Z}(r, f - w_i)$ is clearly $C_{q-1}^2$. Consequently, by (1), we obtain

$$C_q^3 T(r, f) \le C_{q-1}^2 \sum_{i=1}^q \bar{Z}(r, f - w_i) + o(T(r, f)),$$

and hence

$$\frac{q}{3} T(r, f) \le \sum_{i=1}^{q} \bar{Z}(r, f - w_i) + o(T(r, f)).$$

Suppose now that $f$ has finitely many poles. By Theorem 51.11, for every pair $(i, j)$ such that $1 \le i \le j \le q$, we have

$$T(r, f) \le \bar{Z}(r, f - w_i) + \bar{Z}(r, f - w_j) + o(T(r, f)).$$

The number of such inequalities is then $C_q^2$. Summing up we now obtain

$$(2) \quad C_q^2 T(r, f) \le \sum_{(i,j, 1 \le i \le j \le q)} \bar{Z}(r, f - w_i) + \bar{Z}(r, f - w_j) + o(T(r, f)).$$

In this sum, for each index $i$, the number of terms $\bar{Z}(r, f - w_i)$ is clearly $C_{q-1}^1 = q - 1$. Consequently, by (1), we obtain

$$C_q^2 T(r, f) \le (q-1) \sum_{i=1}^q \bar{Z}(r, f - w_i) + o(T(r, f)),$$

and hence

$$\frac{q}{2} T(r, f) \le \sum_{i=1}^{q} \bar{Z}(r, f - w_i) + o(T(r, f)).$$
Now, if $f$ and $w_j$ ($j = 1, \dots, q$) belong to $M(D)$, we can make the same reasoning with $T_R$ instead of $T$ and $Z_R$ instead of $Z$. $\square$

**Definition.** A small function $\theta$ with respect to a function $f \in M(\mathbb{K})$ is said to be perfectly branched for $f$ if all zeros of $f - \theta$ are multiple except finitely many.

**Theorem 51.15.** Let $f \in M(\mathbb{K})$ be transcendental and have finitely many poles. Then $f$ has at most one perfectly branched rational function.

**Proof.** Suppose that $f$ has two distinct perfectly branched rational functions $P$ and $Q$. Then, $f - Q$ has two distinct perfectly branched rational functions $P - Q$ and $0$. So, without loss of generality, we can assume $Q = 0$. Now, let $g = \frac{f}{P}$. We can see that $g$ satisfies all hypotheses of Theorem 49.9 and has two perfectly branched values: $0$ and $1$. Consequently, $g$ is not transcendental and therefore neither is $f$, a contradiction. $\square$

Concerning entire functions, it is useful to examine the link between small functions and order of growth. Let us recall that when $f \in \mathcal{A}(\mathbb{K})$, then $T(r, f) = \text{Log}(|f(r)|)$.

**Theorem 51.16.** Let $f, h \in \mathcal{A}(\mathbb{K})$ be such that $\rho(h) < \tilde{\rho}(f)$. Then $h$ is a small function with respect to $f$.

**Proof.** By hypothesis, there exists $\lambda > 0$ and $R > 0$ such that

$$ \frac{\text{Log}(\text{Log}(|h|(r))}}{\text{Log}(r)} + 2\lambda < \tilde{\rho}(f) \quad \forall r > R, $$

and hence there exists $R' > R$ such that

$$ \text{Log}(\text{Log}(|h|(r))) + \lambda \text{Log}(r) < \text{Log}(\text{Log}(|f|(r))) \quad \forall r > R', $$

therefore $r^{\lambda} \text{Log}(|h|(r)) < \text{Log}(|f|(r))$, which proves that

$$ \lim_{r \to +\infty} \frac{\text{Log}(|h|(r))}{\text{Log}(|f|(r))} = 0. \quad \square $$

**Corollary 51.17.** Let $f, h \in \mathcal{A}(\mathbb{K})$, with $f$ regular and such that $\rho(h) < \rho(f)$. Then $h$ is a small function with respect to $f$.
**Theorem 51.18.** Let $f, h \in \mathcal{A}(\mathbb{K})$ be such that $\rho(h) = \rho(f)$ and $\sigma(h) = 0 < \tilde{\sigma}(f)$. Then $h$ is a small function with respect to $f$.

**Proof.** By hypothesis, we have $\lim_{r \to +\infty} \frac{\sigma(h, r)}{\tilde{\sigma}(f, r)} = 0$, hence

$$ \lim_{r \to +\infty} \frac{\mathrm{Log}(|h|(r))}{\mathrm{Log}(|f|(r))} = 0, $$

therefore $h$ is a small function with respect to $f$. □

**Theorem 51.19.** Let $f \in \mathcal{A}(\mathbb{K})$ and let $h \in \mathcal{A}(\mathbb{K})$ satisfy $|f|(r) > |h|(r) \forall r \ge R$. Then, $\rho(f+h) = \rho(f)$ and $\sigma(f+h) = \sigma(f)$. Moreover, $\psi(f+h) = \psi(f)$.

**Proof.** By construction, we have $\rho(f+h) = \rho(f)$. Then,

$$ \limsup_{r \to \infty} \frac{\mathrm{Log}(|f+h|(r))}{r^{\rho(f+h)}} = \limsup_{r \to \infty} \frac{\mathrm{Log}(|f|(r))}{r^{\rho(f)}} = \sigma(f), $$

hence $\sigma(f+h) = \sigma(f)$.

Now, by Corollary 22.8, the number of zeros of $f+h$ in each disk $d(0, r)$ equals the number of zeros of $f$ in $d(0, r)$, for every $r > R$. Consequently, since $\rho(f+h) = \rho(f)$, we have $\psi(f+h) = \psi(f)$. □

**Corollary 51.20.** Let $f \in \mathcal{A}(\mathbb{K})$ and let $h \in \mathcal{A}(\mathbb{K})$ be a small function with respect to $f$. Then, $\rho(f+h) = \rho(f)$ and $\sigma(f+h) = \sigma(f)$. Moreover, $\psi(f+h) = \psi(f)$.
# Chapter 52

## Meromorphic Functions Sharing Some Small Ones

Throughout this chapter, the field $K$ is supposed to be of characteristic 0.

**Definition.** Two functions $f, g$ defined in a set $E$ are said to share a function $h$, ignoring multiplicity if $f(x) - h(x) = 0$ is equivalent to $g(x) - h(x) = 0$.

This kind of problem was considered in [95].

**Theorem 52.1.** Let $f, g \in \mathcal{M}(K)$ be transcendental (respectively $f, g \in \mathcal{M}_u(d(a, R^{-}))$, respectively $f, g \in \mathcal{M}^c(D)$), be distinct and share $q$ distinct small functions ignoring multiplicity. $w_j \in \mathcal{M}_f(K) \cap \mathcal{M}_g(K)$ ($j = 1, \dots, q$) (respectively $w_j \in \mathcal{M}_f(d(a, R^{-})) \cap \mathcal{M}_g(d(a, R^{-}))$ ($j = 1, \dots, q$) (respectively $w_j \in \mathcal{M}_f(D) \cap \mathcal{M}_g(D)$ ($j = 1, \dots, q$)), other than the constant $\infty$. Then,

$$ \sum_{j=1}^{q} \bar{Z}(r, f - w_j) \leq \bar{Z}(r, f - g) + o(T(r, f)) + o(T(r, g)). $$

**Proof.** Suppose that $f$ and $g$ belong to $\mathcal{M}(K)$, are distinct, and share $q$ distinct small functions I.M. $w_j \in \mathcal{M}_f(K) \cap \mathcal{M}_g(K)$ ($j = 1, \dots, q$).

Let $b$ be a zero of $f - w_i$ for a certain index $i$. Then it is also a zero of $g - w_i$. Suppose that $b$ is counted several times in the sum $\sum_{j=1}^q \bar{Z}(r, f - w_j)$, which means that it is a zero of another
function $f - w_h$ for a certain index $h \neq i$. Then we have $w_i(b) = w_h(b)$ and hence $b$ is a zero of the function $w_i - w_h$ which belongs to $M_f(\mathbb{K})$. Now, put $\tilde{Z}(r, f - w_1) = \bar{Z}(r, f - w_1)$, and for each $j > 1$, let $\tilde{Z}(r, f - w_j)$ be the counting function of zeros of $f - w_j$ in the disk $d(0, r^{-})$ ignoring multiplicity and avoiding the zeros already counted as zeros of $f - w_h$ for some $h < j$. Consider now the sum $\sum_{j=1}^q \tilde{Z}(r, f - w_j)$. Since the functions $w_i - w_j$ belong to $M_f(\mathbb{K})$, clearly, we have

$$ \sum_{j=1}^{q} \bar{Z}(r, f - w_j) = \sum_{j=1}^{q} \tilde{Z}(r, f - w_j) = o(T(r, f)). $$

It is clear, from the assumption, that $f(x) - w_j(x) = 0$ implies $g(x) - w_j(x) = 0$ and hence $f(x) - g(x) = 0$. Since $f - g$ is not the identically zero function, it follows that

$$ \sum_{j=1}^{q} \bar{Z}(r, f - w_j) \leq \bar{Z}(r, f - g). $$

Consequently,

$$ \sum_{j=1}^{q} \bar{Z}(r, f - w_j) \leq \bar{Z}(r, f - g) + o(T(r, f)) + o(T(r, g)). $$

Now, if $f$ and $g$ belong to $M(d(0, R^{-}))$ or to $M(D)$, the proof is exactly the same. □

**Theorem 52.2.** Let $f, g \in M(\mathbb{K})$ be transcendental (respectively $f, g \in M_u(d(a, R^{-}))$, respectively $f, g \in M^c(D)$), be distinct, have finitely many poles, and share three distinct small functions, ignoring multiplicity. $w_j \in M_f(\mathbb{K}) \cap M_g(\mathbb{K})$ ($j = 1, 2, 3$) (respectively $w_j \in M_f(d(a, R^{-})) \cap M_g(d(a, R^{-}))$ ($j = 1, 2, 3$), respectively $w_j \in M_f(D) \cap M_g(D)$ ($j = 1, 2, 3$)). Then, $f = g$.

**Proof.** We put $V(r) = \max(T(r, f), T(r, g))$. Suppose that $f$ and $g$ are distinct and share $q$ small function I.M. $w_j$ ($1 \leq j \leq q$). By Theorem 51.14, we have

$$ qT(r, f) \leq 3 \sum_{j=1}^{q} \bar{Z}(r, f - w_j) + o(T(r, f)). $$
But thanks to Theorem 52.1, we can derive

$$qT(r, f) \leq 3T(r, f - g) + o(T(r, f)),$$

and similarly

$$qT(r, g) \leq 3T(r, f - g) + o(T(r, g)),$$

hence

$$(1) \qquad qV(r) \leq 3T(r, f - g) + o(V(r)).$$

Suppose now that $f$ and $g$ have finitely many poles. By Theorems 51.14, Relation (1) gives us

$$qV(r) \leq 2V(r) + o(V(r)),$$

which is obviously absurd whenever $q \geq 3$ and proves that $f = g$ when $f$ and $g$ belong to $\mathcal{M}(\mathbb{K})$ as well as when $f$ and $g$ belong to $\mathcal{M}_u(d(0, R^-))$ or to $\mathcal{M}^c(D)$, after replacing $T$ by $T_R$ and $Z$ by $Z_R$. $\square$

<figure><img src="image_22.png" /></figure>

By Theorem 51.16, Corollary 52.3 is easy:

**Corollary 52.3.** Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental, distinct, and share three distinct functions, ignoring multiplicity, $\omega_j \in \mathcal{A}(\mathbb{K})$, $j = 1, 2, 3$, such that $\max_{1 \leq j \leq 3} \rho(\omega_j) < \min(\tilde{\rho}(f), \tilde{\rho}(g))$. Then, $f = g$.

**Corollary 52.4.** Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental, regular, distinct, and share three distinct functions, ignoring multiplicity, $\omega_j \in \mathcal{A}(\mathbb{K})$, $j = 1, 2, 3$, such that $\max_{1 \leq j \leq 3} \rho(\omega_j) < \min(\rho(f), \rho(g))$. Then, $f = g$.

In the same way, by Theorem 51.18, Corollary 52.5 is easy:

**Corollary 52.5.** Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental, clean, and share three distinct functions, ignoring multiplicity, $\omega_j \in \mathcal{A}(\mathbb{K})$, $j = 1, 2, 3$, such that $\rho(\omega_j) = \rho(f) = \rho(g)$, $j = 1, 2, 3$, and $\sigma(\omega_j) = 0 < \min(\sigma(f), \sigma(g))$. Then, $f = g$.

After this first result in [56] letting find that two meromorphic functions sharing seven small functions are equal, Ta Thi Hoai An and Nguyen Phuong obtained a much better estimation. We denote by $E$ the disk $d(0, R^-)$.
**Lemma 52.6 (Ta Thi Hoai An-Nguyen Phuong [5]).** Let $f, g$ be distinct and belong to $\mathcal{M}(\mathbb{K})$ (respectively to $\mathcal{M}_u(E)$, respectively to $\mathcal{M}^c(D)$) sharing $q$ distinct small functions ignoring multiplic-ity: $h_1, \dots, h_q$ with $q \ge 5$. Then, for every subset $\{i_1, i_2, i_3, i_4\}$ of $\{1, \dots, q\}$, we have

$$ \sum_{j \in \{1, \dots, q\} \setminus \{i_1, \dots, i_4\}} \overline{Z}(r, f - h_j) + \overline{Z}(r, g - h_j) \\ \le o(T((r, f)) + o(T((r, g))) $$

(respectively

$$ \sum_{j \in \{1, \dots, q\} \setminus \{i_1, \dots, i_4\}} \overline{Z}(r, f - h_j) + \overline{Z}(r, g - h_j) \le o(T(r, f)) + o(T(r, g)), $$

respectively

$$ \sum_{j \in \{1, \dots, q\} \setminus \{i_1, \dots, i_4\}} \overline{Z}_R(r, f - h_j) + \overline{Z}_R(r, g - h_j) \\ \le o(T(r, f)) + o(T(r, g)). $$

**Proof.** Let us first suppose $f, g \in \mathcal{M}(\mathbb{K})$, or $f, g \in \mathcal{M}_u(E)$. Clearly, we only have to prove that

$$ \sum_{j=5}^{q} \overline{Z}(r, f - h_j) \le o(T(r, f)) + o(T(r, g)). $$

Thus, we assume that

$$ \sum_{j=5}^{q} \overline{Z}(r, f - h_j) \neq o(T(r, f)) + o(T(r, g)). $$

Given any meromorphic function $w \in \mathcal{M}(\mathbb{K})$ (respectively in $\mathcal{M}_u(E)$), we put

$$ L(w) = \frac{(w - h_1)(h_3 - h_2)}{(w - h_2)(h_3 - h_4)} $$

and $F = L(f), G = L(g)$.
Without loss of generality, we may assume that $h_1 = \infty$, $h_2 = 0$, $h_3 = 1$, and we put $h = h_4$. Now, we set

$$B = \frac{f'(h'g - hg')(f - g)}{f(f - 1)(g - h)} - \frac{g'(h'f - hf')(f - g)}{g(g - 1)(f - h)}$$

and

$$\begin{aligned} Q &= f'(h'g - hg')(f - h)(g - 1) - g'(h'f - hf')(g - h)(f - 1) \\ &= h'f f'g^2 - h'f'g - h(h - 1)f f'g' - hh'f'g^2 + hh'f'g \\ &\quad - h'f^2gg' + h'fgg' + h(h - 1)f'gg' + hh'f^2g' - hh'fg'. \end{aligned}$$

Then,

$$B = \frac{(f - g)Q}{f(f - 1)g(g - 1)}.$$

Suppose first that $B$ is identically zero. Then we have

$$(1) \qquad \frac{f'(h'g - hg')(f - g)}{f(f - 1)(g - h)} = \frac{g'(h'f - hf')(f - g)}{g(g - 1)(f - h)}.$$

If $h$ is a constant, then $f = g$ which contradicts our hypothesis. Hence, $h$ is not a constant, and then, by (1), we have

$$\frac{(f - 1)(g - h)}{g - 1)(f - h)} - 1 = \frac{f'(h'g - hg')}{g'(h'f - hf')} - 1,$$

therefore

$$\frac{(f - g)(g - h)}{(g - 1)(f - h)} = \frac{h'(f' - g')g - (f - g)g'}{g'(h'f - hf')}$$

and hence

$$(2) \qquad \frac{f' - g'}{f - g} = \frac{(1 - h)g'(h'f - hf')}{h'g(g - 1)(f - h)} + \frac{g'}{g}.$$

Let us fix $j \ge 5$, $j \le q$. By (1), there exists a common zero $\alpha \in \mathbb{K}$ (respectively $\alpha \in D$) of $f - h_j$ and $g - h_j$ which is not a zero or a pole of $h, h', h_j - 1, h_j - h$. Then, $\alpha$ must be a pole of the left-hand side of (2) but cannot be a pole of the right-hand side of (2). This is a contradiction showing that $B$ cannot be identically zero.
Now, let us fix $j \ge 5$ such that $j \le q$ and suppose that $\beta$ is a common zero of $f - h_j$ and $g - h_j$ but is not a zero or a pole of $h, h_j, h_j - 1, h_j - h$. Then it is a zero of $f - g$ and it is not a pole of

$$ \frac{Q}{f(f-1)(f-h)g(g-1)(g-h)}, $$

and hence it is a zero of $B$.

Now, since $f$ and $g$ share each $h_j$ ignoring multiplicity, we have

$$ (3) \quad \sum_{j=1}^{5} \overline{Z}(r, g - h_j) \\ = \sum_{j=1}^{5} \overline{Z}(r, f - h_j) \le Z(r, B) + o(T(r, f)) + o(T(r, g)) \\ \le T(r, B) + o(T(r, f)) + o(T(r, g)). $$

We estimate $T(r, B) - N(r, B)$. By computation, $B$ can be written as

$$ \begin{aligned} B &= \frac{f'(h'g - hg')}{(f-1)g(g-h)} - \left(\frac{f'}{f-1} - \frac{f'}{f}\right) \left(\frac{h'g - hg'}{g-h}\right) \\ &\quad + \frac{g'(h'f - hf')}{(g-1)f(f-h)} - \left(\frac{g'}{g-1} - \frac{g'}{g}\right) \left(\frac{h'f - hf'}{f-h}\right) \\ &= \left(\frac{f'}{f-1}\right) \left(\frac{g'}{g} - \frac{g' - h'}{g-h}\right) - \left(\frac{f'}{f-1} - \frac{f'}{f}\right) \left(h' - \frac{h(g' - h')}{g-h}\right) \\ &\quad + \left(\frac{g'}{g-1}\right) \left(\frac{f'}{f} - \frac{f' - h'}{f-h}\right) - \left(\frac{g'}{g-1} - \frac{g'}{g}\right) \left(h' - \frac{h(f' - h')}{f-h}\right). \end{aligned} $$

In this last relation, $B$ appears as a sum of logarithmic derivative, and hence, by Corollary 43.18, we can see that

$$ (4) \qquad T(r, B) - N(r, B) \le 0. $$

Next, we estimate $N(r, B)$. Clearly, the poles of $B$ can only occur at the zeros of $f, g, f-1, g-1, f-h, g-h$ and at the poles of $f, g, h$.

Let $A$ be the set of all zeros, 1-points, and poles of $h$. Then we first estimate the counting function of poles of $B$ when we are in $A$. Recall that $f$ and $g$ share the constants 0, 1, and $\infty$.
Let $\gamma$ be a zero of $h$. Then we can check that $B$ has no pole at $\gamma$.

Let $\gamma$ be a 1-value of $h$. If $B$ had a pole, then $f-1$ would have a zero of order $s$, $g-1$ would have a zero of order $t$, $f-g$ would have a zero, and $Q$ would have a zero of order $s+t$, hence $B$ would have a zero, not a pole.

Now, let $\gamma$ be a pole of $h$ of order $n$.

If $f(\gamma) = 1$, then $g(\gamma) = 1$ because $f$ and $g$ share $h_3$, hence $B$ has no pole at $\gamma$.

If $f$ has a pole of order $s$ at $\gamma$, then $g$ also has a pole of order $t$ because $f$ and $g$ share $h_1$. We can assume $s \ge t$, then $B$ has a pole of order at most $n+2$, hence the counting function of poles of $B$ when $\gamma$ is a pole of $f$, $g$, and $h$ and hence is a pole of $B$ of order $n+2$, is bounded by thrice the counting function of the poles of $h$, and hence is of the form $o(T(r, f)) + o(T(r, g))$.

Next, suppose that $f(\gamma) \ne 1, \infty$, hence $g(\gamma) \ne 1, \infty$. If $f$ and $g$ have no zero at $\gamma$, it is clear that $B$ has at most a pole of order $n$, hence the counting function of poles of $B$ when $f(\gamma) \ne 0, 1, \infty$ (hence $g(\gamma) \ne 0, 1, \infty$) is of the form $o(T(r, f)) + o(T(r, g))$. Finally, if $f(\gamma) = 0$, then $g(\gamma) = 0$ because $f$ and $g$ share $h_2$ and then we can check that $B$ may admit a pole at $\gamma$ of order at most $2n+2$. Consequently, the counting function of poles of $B$ when $\gamma$ lies in $A$ and is a pole of $h$ is bounded by thrice the counting function of the poles of $h$ and hence is of the form $o(T(r, f)) + o(T(r, g))$.

Thus, the counting function of poles of $B$ when $\gamma$ lies in $A$ is of the form $o(T(r, f)) + o(T(r, g))$.

Consider now the poles of $B$ when $\gamma \notin A$.

Let $\gamma \notin A$ be a common pole of $f$ and $g$ of order $s_1$ and $t_1$, respectively. Then, $\gamma$ is a pole of $Q$ of order at most $2s_1 + 2t_1 + 1$. Next, $\gamma$ is a pole of $f-g$ of order $\max(s_1, t_1)$. Hence, from the definition of $B$, $\gamma$ is a pole of the numerator of $B$ of order at most $2s_1 + 2t_1 + 1 + \max(s_1, t_1)$, and on the other hand, it is a pole of the denominator of $B$ of order $3s_1 + 3t_1$. But since

$$2s_1 + 2t_1 + 1 + \max(s_1, t_1) - 3(s_1 + t_1) = 1 + \max(s_1, t_1) - s_1 - t_1 \le 0,$$

it follows that $\gamma$ is not a pole of $B$.

Let $\gamma \notin A$ be a common zero of $f$ and $g$ of order $s_2$ and $t_2$, respectively. We can check that $\gamma$ is a zero of $Q$ of order at least $s_2 + t_2 + 1$ and that $\gamma$ is a zero of $f-g$ of order at least $\min(s_2, t_2)$.
On the other hand, we can check that the denominator of B has a zero at $\gamma$ of order at most $s_2 + t_2$. Consequently, B has no pole at $\gamma$.

Suppose now that $\gamma \notin A$ is a common zero of $f - h$ and $g - h$. Then, $\gamma$ is a zero of $f - g$. On the other hand, $\gamma$ is a pole of order 1 of $\frac{f'}{f-h}$ and $\frac{g'}{g-h}$. Consequently, by the definition of B, we can check that $\gamma$ is not a pole of B. Similarly, when $\gamma$ is a common zero of $f - 1$ and $g - 1$, $\gamma$ is not a pole of B.

Suppose now that $\gamma \notin A$ is a pole of $f$ but is not a pole of $g$ and is not a zero of $g$, $g - 1$, and $g - h$. Since $f$ and $g$ share $h_0$, that situation does not occur.

Suppose now that $\gamma \notin A$ is a zero of $f$ but is not a pole of $g$ and is not a zero of $g$. Since $f$ and $g$ share $h_0$, that situation does not occur.

Suppose now that $\gamma \notin A$ is a zero of $f - h_i$ with $i = 3, 4$ but is not a pole of $g$ and is not a zero of $g$, $g - 1$ and $g - h$. Then $\gamma$ is a pole of B of order at most 1. Similarly, if $\gamma$ is a zero of $g - h_i$ with $i = 3, 4$ but is not a pole of $f$ and is not a zero of $f$, $f - 1$ and $f - h$. Then $\gamma$ is a pole of B of order at most 1. However, since $f$ and $g$ share $h_3$ and $h_4$, such a situation does not occur.

Suppose now that $\gamma \notin A$ is a common zero of $f - h_i$ and $g - h_j$ for some $1 \le i \le 4$ and $1 \le j \le 4$ with $i \ne j$ and $i \ne 2, j \ne 2$. Then $\gamma$ is a pole of B of order at most 2. However, in such a situation, since $f$ and $g$ share $h_i$, then $\gamma$ is a zero of $h_i - h_j$. But since $h_i$ and $h_j$ are small with respect to $f$ and $g$, the counting function of points $\gamma$ that are common zeros of $f - h_i$ and $g - h_j$ for some $1 \le i \le 4$ and $1 \le j \le 4$ with $i \ne j$ and $i \ne 2, j \ne 2$ is bounded by $o(T(r, f)) + o(T(r, g))$.

Finally, suppose that $\gamma \notin A$ is a zero of $f - h_i$ and a pole of $g$ for some $i = 2, 3, 4$ or a zero of $g - h_i$ and a pole of $f$ for some $i = 2, 3, 4$. But since $f$ and $g$ share $h_2, h_3, h_4$, such a situation is impossible.

In conclusion, we have

$$N(r, B) \le o(T(r, f)) + o(T(r, g)),$$

and therefore, by (3) and (4), that completes the proof of Lemma 52.6 when $f, g \in \mathcal{M}(\mathbb{K})$ or $f, g \in \mathcal{M}_u(E)$. Now, when $f, g \in \mathcal{M}^c(D)$, the proof is similar. $\square$

**Corollary 52.7.** Let $f, g$ be distinct and belong to $\mathcal{M}(\mathbb{K})$ (respectively to $\mathcal{M}_u(E)$, respectively to $\mathcal{M}^c(D)$) sharing $q$ distinct small functions ignoring multiplicity: $h_1, \dots, h_q$ (with $q \ge 5$). Then, for
every subset {$i_1, i_2, i_3, i_4$} of {$1, \dots, q$}, we have

$$ \bar{Z}(r, f - h_j) + \bar{Z}(r, g - h_j) \le o(T((r, f)) + o(T(r, g))) \quad \forall j = 1, \dots, q $$

(respectively

$$ \bar{Z}(r, f - h_j) + \bar{Z}(r, g - h_j) \le o(T(r, f)) + o(T(r, g)) \quad \forall j = 1, \dots, q $$

respectively

$$ \bar{Z}_R(r, f - h_j) + \bar{Z}_R(r, g - h_j) \le o(T(r, f)) + o(T(r, g)) \quad \forall j = 1, \dots, q). $$

**Theorem 52.8 (Ta Thi Hoai An-Nguyen Phuong [5]).** Let $f, g \in \mathcal{M}(\mathbb{K})$ be transcendental (respectively $f, g \in \mathcal{M}_u(E)$, respectively $f, g \in \mathcal{M}^c(D)$), be distinct, and share five distinct small meromorphic functions ignoring multiplicity, $w_j \in \mathcal{M}_f(\mathbb{K}) \cap \mathcal{M}_g(\mathbb{K})$ ($j = 1, \dots, 5$) (respectively $w_j \in \mathcal{M}_f(D) \cap \mathcal{M}_g(D)$ ($j = 1, \dots, 5$), respectively $w_j \in \mathcal{M}_f(E) \cap \mathcal{M}_g(E)$ ($j = 1, \dots, 5$)). Then, $f = g$.

**Proof.** Suppose first that $f, g \in \mathcal{M}(\mathbb{K})$ or $f, g \in \mathcal{M}_u(E)$. Suppose that $f$ and $g$ are not identical. Applying Corollary 52.7 with $q = 5$, we have $\bar{Z}(r, f - h_j) = o(T(r, f)) + o(T(r, g))$ and $\bar{Z}(r, g - h_j) = o(T(r, f)) + o(T(r, g))$ for every $j \in \{1, \dots, 5\}$. Therefore,

$$ (1) \quad \sum_{j=1}^{5} \bar{Z}(r, f - h_j) + \bar{Z}(r, g - h_j) = o(Tf, r) + o(T(r, g)). $$

Now, by Theorem 51.14, we have

$$ \begin{aligned} & 5(T(r, f) + T(r, g)) \\ & \le 3 \sum_{j=1}^{5} \bar{Z}(r, f - h_j) + \bar{Z}(r, g - h_j) + o(T(r, f)) + o(T(r, g)), \end{aligned} $$

hence, by (1), $5(T(r, f) + T(r, g)) \le 2(o(T(r, f)) + o(T(r, g)))$, which is absurd and proves that $f = g$.

Suppose now that $f, g \in \mathcal{M}^c(D)$. Replacing each symbol $T(r, \cdot)$ by $T_R(r, \cdot)$ and $\bar{Z}(r, \cdot)$ by $\bar{Z}_R(r, \cdot)$, we can make the same reasoning and conclude in the same way $f = g$. □
We now show an improvement of Theorem 51.14 in the case when $q \ge 5$.

**Notation.** Given $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}_u(E)$, respectively $f \in \mathcal{M}^c(D)$), we put $M(r, f) = T(r, f) - N(r, f)$.

**Lemma 52.9 (Ta Thi Hoai An-Nguyen Phuong).** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}_u(E)$, respectively let $f \in \mathcal{M}^c(D)$) be non-constant and let $h_1, \dots, h_5$ be distinct small functions with respect to $f$. We have

$$2T(r, f) \le \sum_{i=1}^{5} \bar{Z}(r, f - h_i) + S(r, f).$$

**Proof.** We first suppose $f \in \mathcal{M}(\mathbb{K})$, or $f \in \mathcal{M}_u(E)$. Let $g = \frac{(f-h_2)(h_3-h_1)}{(f-h_1)(h_3-h_2)}$. Then it is easily seen that

$$\begin{aligned}
\bar{Z}(r, g) &= \bar{Z}(r, f - h_2) + o(T(r, f)), \\
\bar{N}(r, g) &= \bar{Z}(r, f - h_1) + o(T(r, f)), \\
\bar{Z}(r, g - 1) &= \bar{Z}(r, f - h_3) + o(T(r, f)).
\end{aligned}$$

Consequently, in order to prove Lemma 52.9, it is sufficient to prove the following inequality:

$$(1) \quad 2T(r, f) \le \bar{N}(r, f) + \bar{Z}(r, f) + \bar{Z}(r, f-1) + \bar{Z}(r, f-h_4) \\ \qquad + \bar{Z}(r, f-h_5) + o(T(r, f)),$$

where $h_1 = \infty$, $h_2 = 0$, $h_3 = 1$, and $h_4$ and $h_5$ are two small functions with respect to $f$ different from $0, 1, \infty$, and $h_4 \ne h_5$.

If one of the functions $h_4, h_5$ is a constant, then (1) is immediate by Theorem 45.7. Consequently, now we can assume that both $h_4, h_5$ are non-constant small functions.

Let

$$H = \det \begin{pmatrix} ff' & f' & f(f-1) \\ h_4h'_4 & h'_4 & h_4(h_4-1) \\ h_5h'_5 & h'_5 & h_5(h_5-1) \end{pmatrix}.$$
By a simple computation, we get

$$ (2) \qquad H = f(f-1)h_4(h_4-1)h_5(h_5-1) \\ \times \left( \left( \frac{h'_4}{h_4} - \frac{h'_5}{h_5} \right) \left( \frac{f'}{f-1} - \frac{h'_5}{h_5-1} \right) \right. \\ \left. - \left( \frac{h'_4}{h_4-1} - \frac{h'_5}{h_5-1} \right) \left( \frac{f'}{f} - \frac{h'_5}{h_5} \right) \right). $$

Suppose first that $H$ is identically zero. Since $f$ is not a constant and since $h_4$, $h_5$ are not identically 0 or 1, it follows from the definition of $H$ that

$$ (3) \qquad \left(\frac{h'_4}{h_4} - \frac{h'_5}{h_5}\right) \frac{f'}{f-1} - \left(\frac{h'_4}{h_4-1} - \frac{h'_5}{h_5-1}\right) \frac{f'}{f} \\ = \left(\frac{h'_4}{h_4} - \frac{h'_5}{h_5}\right) \frac{fh'_5}{h_5-1} - \left(\frac{h'_4}{h_4-1} - \frac{h'_5}{h_5-1}\right) \frac{h'_5}{h_5}. $$

We must now distinguish four cases.

**Case 1.** $\frac{h'_4}{h_4} = \frac{h'_5}{h_5}$. It follows from (3) that either $\frac{h'_4}{h_4-1} = \frac{h'_5}{h_5-1}$ or $\frac{f'}{f} = \frac{h'_5}{h_5}$. If $\frac{h'_4}{h_4-1} = \frac{h'_5}{h_5-1}$, then $h_4$ and $h_5$ are constants, which is excluded by hypothesis. Hence, $\frac{f'}{f} = \frac{h'_5}{h_5}$ and then $f$ is of the form $c \cdot h_5$, where $c$ is a constant, which contradicts our hypothesis: $h_5$ is a small function with respect to $f$.

**Case 2.** $\frac{h'_4}{h_4-1} = \frac{h'_5}{h_5-1}$. Similar to Case 1, we have a contradiction.

**Case 3.** $\frac{h'_4}{h_4} - \frac{h'_5}{h_5} = \frac{h_4}{h_4-1} - \frac{h'_5}{h_5-1} \neq 0$. It follows from (3) that

$$ \frac{f'}{f-1} - \frac{f'}{f} = \frac{h'_5}{h_5-1} - \frac{h'_5}{h_5}, $$

which implies

$$ \frac{f-1}{f} = \frac{C(h_5-1)}{h_5}, $$

where $C$ is a constant. Thus, we obtain

$$ \frac{1}{f} = 1 - \frac{C(h_5-1)}{h_5}, $$
and hence $T(r, \frac{1}{f}) = o(T(r, f))$, which is absurd because
$T(r, \frac{1}{f}) = T(r, f) + o(T(r, f))$.

**Case 4.**

$$ \frac{h'_4}{h_4} \neq \frac{h'_5}{h_5}, \quad \frac{h'_4}{h_4-1} \neq \frac{h'_5}{h_5-1}, \quad \text{and} \quad \frac{h'_4}{h_4} - \frac{h'_5}{h_5} \neq \frac{h'_4}{h_4-1} - \frac{h'_5}{h_5-1}. $$

Then it follows from (3) that the zeros of $f-1$ can only occur when $h_j$ takes value 0 or 1, for $j = 4, 5$ or when $\frac{h'_4}{h_4} - \frac{h'_5}{h_5}$ has a zero. In the same way, the zeros of $f$ can only occur when $h_j$ takes value 0 or 1, for $j = 4, 5$ and when $\frac{h'_4}{h_4-1} - \frac{h'_5}{h_5-1}$ has a zero. Moreover, by (3), we can see that the poles of $f$ can only occur when $h_j$ takes value 1 or when $h_j$ has a pole, for $j = 4, 5$, or when $\frac{h'_4}{h_4} - \frac{h'_5}{h_5} - \frac{h'_4}{h_4-1} + \frac{h'_5}{h_5-1}$ has a zero. Therefore, we have

$$ (4) \qquad \bar{N}(r, f) + \bar{Z}(r, f) + \bar{Z}(r, f-1) = o(T(r, f)). $$

Applying the second main Theorem 45.7 to $f$, we can derive

$$ T(r, f) \le \bar{Z}(r, f) + \bar{Z}(r, f-1) + \bar{N}(r, f) - \log(r) = o(T(r, f)), $$

which is absurd and finishes proving that $H \ne 0$.

Now, given $r > 0$, we put

$$ \delta(r) = \min(1, |h_4|(r), |h_5|(r), |h_4 - 1|(r), |h_5 - 1|(r), |h_4 - h_5|(r)). $$

Then we have

$$
\begin{align*}
& \log^+ \left( \frac{1}{\delta(r)} \right) \\
\le & \log^+ \left( \max \left( 1, \frac{1}{|h_4|(r)}, \frac{1}{|h_5|(r)}, \frac{1}{|h_4-1|(r)}, \frac{1}{|h_5-1|(r)}, \frac{1}{|h_4-h_5|(r)} \right) \right) \\
\le & \log^+ \left( \frac{1}{|h_4|(r)} \right) + \log^+ \left( \frac{1}{|h_5|(r)} \right) + \log^+ \left( \frac{1}{|h_4-1|(r)} \right), \\
& + \log^+ \left( \frac{1}{|h_5-1|(r)} \right) + \log^+ \left( \frac{1}{|h_4-h_5|(r)} \right) + \log(6).
\end{align*}
$$
Then, by Theorem 43.2, we have

$$ \log^+ \left( \frac{1}{\delta(r)} \right) \le M \left( r, \frac{1}{h_4} \right) + M \left( r, \frac{1}{h_5} \right) + M \left( r, \frac{1}{h_4 - 1} \right) \\ + M \left( r, \frac{1}{h_5 - 1} \right) + M \left( r, \frac{1}{h_4 - h_5} \right) + \log(6). $$

Consequently, we can see that

$$ (5) \qquad \log^+ \left( \frac{1}{\delta(r)} \right) = o(T(r, f)). $$

First, we consider the case when $|f - h_j|(r) > \frac{\delta(r)}{2}$ for $2 \le j \le 5$. Then, we have

$$ (6) \quad M \left( r, \frac{1}{f} \right) + M \left( r, \frac{1}{f-1} \right) + M \left( r, \frac{1}{f-h_4} \right) + M \left( r, \frac{1}{f-h_5} \right) \\ \le 5 \log^+ \left( \frac{1}{\delta(r)} \right) + O(1) = s(r, f). $$

Now, let $i$ be an index such that $2 \le i \le 5$ such that $|f - h_i|(r) \le \frac{\delta(r)}{2}$. Then for every $j \ne i$, with $2 \le j \le 5$, we have

$$ \delta(r) \le |h_i - h_j|(r) \le |f - h_i|(r) + |f - h_j|(r) \le \frac{\delta(r)}{2} + |f - h_j|(r), $$

hence

$$ \frac{\delta(r)}{2} \le |f - h_j|(r). $$

Therefore, for $i \ne j$, we have

$$ \sum_{\substack{j=2 \\ j \ne i}}^{5} M \left( r, \frac{1}{f-h_j} \right) = \sum_{\substack{j=2 \\ j \ne i}}^{5} \log^{+} \left( \frac{1}{|f-h_j|(r)} \right) \le 3 \log^{+} \left( \frac{1}{\delta(r)} \right). $$

Combining (5) and the last inequality, we have

$$ (7) \qquad \sum_{\substack{j=2, \\ j \ne i}}^{5} M \left( r, \frac{1}{f-h_j} \right) = o(T(r, f)). $$
On the other hand, for $2 \le i \le 5$, we can write

$$ 
\begin{aligned}
ff' &= (f - h_i)(f' - h'_i) + h'_i(f - h_i) + h_i(f' - h'_i) + h_i h'_i \\
f' &= (f' - h'_i) + h'_i,
\end{aligned}
$$

$$ f(f - 1) = f^2 - f = (f - h_i)^2 + (2h_i - 1)(f - h_i) + h_i^2 - h_i. $$

And now we put $g_i = (f - h_i)(f' - h'_i) + h'_i(f - h_i) + h_i(f' - h'_i)$ and $l_i = (f - h_i)^2 + (2h_i - 1)(f - h_i)$ and then, thanks to properties of determinants, we obtain

$$ (8) \qquad H = \det \begin{pmatrix} g'_i & f' - h'_i & l_i \\ h_4 h'_4 & h'_4 & h_4(h_4 - 1) \\ h_5 h'_5 & h'_5 & h_5(h_5 - 1) \end{pmatrix}. $$

Now, we have $\log^+(\delta(r)) \le \log^+(1 + |h_i|(r)) \quad \forall i = 2, \dots, 5$ because it is obvious from the definition for $i = 4, 5$ and also for $i = 2$ and $i = 3$ because $h_2 = 0$ and $h_3 = 1$. Consequently, for every $i = 2, \dots, 5$, we have $\log^+(\delta(r)) \le \log^+(|h_i|(r)) + \log(2) = M(r, h_i) + \log(2) = o(T(r, f))$. Then, by (8) and using Lemma 3.5, we obtain for every $i = 2, \dots, 5$:

$$ 
\begin{aligned}
\log^+ \left| \frac{H}{f - h_i} \right| &\le \log^+ \left( \left| \frac{f' - h'_i}{f - h_i} \right| (r) \right) + \log^+ (|f - h_i|(r)) \\
&\quad + O \left( \log^+ (|h_i|(r) + \log^+ (|h'_i|(r)) + \log^+ (|h_4|(r)) \right. \\
&\qquad \left. + \log^+ (|h'_4|(r)) + \log^+ (|h_5|(r)) + \log^+ (|h'_5|(r)) \right) \\
&\le m \left( r, \frac{f' - h'_i}{f - h_i} \right) + \log^+ (\delta(r)) + o(T(r, f)) \\
&= o(T(r, f)).
\end{aligned}
$$

Hence, for every $i = 2, \dots, 5$, we obtain

$$ 
\begin{aligned}
(9) \qquad m \left( \frac{1}{f - h_i} \right) &= \log^+ \left( \frac{1}{|f - h_i|(r)} \right) \\
&\le \log^+ \left| \frac{H}{f - h_i} \right| (r) + \log^+ \left| \frac{1}{H} \right| (r) \\
&\le m \left( r, \frac{1}{H} \right) + o(T(r, f)).
\end{aligned}
$$
Then, by (6), (7), (8), and (9), we can check that in all cases we
have

$$
(10) \quad
\begin{aligned}
& M\left(r, \frac{1}{f}\right) + M\left(r, \frac{1}{f-1}\right) + M\left(r, \frac{1}{f-h_4}\right) + M\left(r, \frac{1}{f-h_5}\right) \\
& \le M\left(r, \frac{1}{H}\right) + o(T(r, f)).
\end{aligned}
$$

Now, by Theorem 43.3, we can write

$$
\begin{aligned}
4T(r, f) ={}& M\left(r, \frac{1}{f}\right) + M\left(r, \frac{1}{f-1}\right) + M\left(r, \frac{1}{f-h_4}\right) \\
& + M\left(r, \frac{1}{f-h_5}\right) + N\left(r, \frac{1}{f}\right) + N\left(r, \frac{1}{f-1}\right) \\
& + N\left(r, \frac{1}{f-h_4}\right) + N\left(r, \frac{1}{f-h_5}\right) + o(T(r, f)),
\end{aligned}
$$

hence by (10),

$$
\begin{align*}
(11) \quad 4T(r,f) &\le M\left(r, \frac{1}{H}\right) + N\left(r, \frac{1}{f}\right) + N\left(r, \frac{1}{f-1}\right) \\
&\qquad + N\left(r, \frac{1}{f-h_4}\right) + N\left(r, \frac{1}{f-h_5}\right) + o(T(r,f)) \\
&\le T(r,H) - Z(r,H) + Z(r,f) + Z(r,f-1) \\
&\qquad + Z(r,f-h_4) + Z(r,f-h_5) + o(T(r,f)).
\end{align*}
$$

Now, given a zero $\alpha$ of order $s$ of any function $f - h_i$ ($2 \le i \le 5$),
then $\alpha$ is also a zero of $H$ of order at least $s-1$. Then, from (11),
we can see that we have

$$
(12) \quad 4T(r, f) \le \bar{Z}(r, f) + \bar{Z}(r, f-1) + \bar{Z}(r, f-h_4) + \bar{Z}(r, f-h_5) \\
\phantom{(12) \quad 4T(r, f) \le } + T(r, H) + o(T(r, f)).
$$

Next, by (2), we can check that

$$
\begin{align*}
M(r, H) &\le 2m(r, f) + o(T(r, f)) \\
N(r, H) &\le 2N(r, f) + \overline{N}(r, f) + o(T(r, f)).
\end{align*}
$$
Consequently,

$$ (13) \qquad T(r, H) \le 2(T(r, f) + \bar{N}(r, f) + o(T(r, f))). $$

Then, by (12) and (13), we obtain

$$ 2T(r, f) \le \bar{N}(r, f) + \bar{Z}(r, f) + \bar{Z}(r, f-1) + \bar{Z}(r, f-h_4) + \bar{Z}(r, f-h_5) + o(T(r, f)), $$

which finishes proving (1) and hence ends the proof of Lemma 4.1 when $f$ belongs to $\mathcal{M}(\mathbb{K})$ or $\mathcal{M}_u(E)$. Next, when $f$ belongs to $\mathcal{M}^c(D)$, we have a similar proof, writing corresponding counting functions $T_R(r, \cdot)$ instead of $T(r, \cdot)$, $Z_R(r, \cdot)$ instead of $Z(r, \cdot)$, $N_R(r, \cdot)$ instead of $N(r, \cdot)$, etc. □

**Theorem 52.10.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}_u(E)$, respectively let $f \in \mathcal{M}^c(D)$) be non-constant and let $h_1, \dots, h_q (q \ge 5)$ be $q$ distinct small functions with respect to $f$. We have

$$ \frac{2q}{5} T(r, f) \le \sum_{i=1}^{q} \bar{Z}(r, f - h_i) + o(T(r, f)) $$

(respectively

$$ \frac{2q}{5} T(r, f) \le \sum_{s=1}^{q} \bar{Z}(r, f - h_{i_s}) + o(T(r, f)), $$

respectively

$$ \frac{2q}{5} T_R(r, f) \le \sum_{s=1}^{q} \bar{Z}_R(r, f - h_{i_s}) + o(T(r, f)). $$

**Proof.** By Lemma 52.9, for every subset $\{i_1, \dots, i_5\}$ of $\{1, \dots, q\}$ such that $1 \le i_1 < \dots < i_5 \le q$, we have

$$ 2T(r, f) \le \sum_{s=1}^{q} \bar{Z}(r, f - h_{i_s}) + o(T(r, f)) $$

(respectively

$$ 2T(r, f) \le \sum_{s=1}^{q} \bar{Z}(r, f - h_{i_s}) + o(T(r, f)), $$
respectively

$$2T_R(r, f) \le \sum_{s=1}^{q} \bar{Z}_R(r, f - h_{i_s}) + o(T(r, f)).$$

The number of such inequalities is $C_q^5$. Summing up over all sub-sets $\{i_1, \dots, i_5\}$ of $\{1, \dots, q\}$, we can get

$$2C_q^5 T(r, f) \le \sum_{\substack{i_1, \dots, i_5 \subset \{1, \dots, q\} \\ 1 \le i_1 < \dots < i_5 \le q}} (\bar{Z}(r, f - h_{i_1}) + \bar{Z}(r, h_{i_2}) + \bar{Z}(r, h_{i_3}) \\ + \bar{Z}(r, h_{i_4}) + \bar{Z}(r, h_{i_5})) + o(T(r, f))$$

(respectively

$$2C_q^5 T(r, f) \le \sum_{\substack{i_1, \dots, i_5 \subset \{1, \dots, q\} \\ 1 \le i_1 < \dots < i_5 \le q}} (\bar{Z}(r, f - h_{i_1}) + \bar{Z}(r, h_{i_2}) + \bar{Z}(r, h_{i_3}) \\ + \bar{Z}(r, h_{i_4}) + \bar{Z}(r, h_{i_5})) + o(T(r, f)),$$

respectively

$$2C_q^5 T_R(r, f) \le \sum_{\substack{i_1, \dots, i_5 \subset \{1, \dots, q\} \\ 1 \le i_1 < \dots < i_5 \le q}} (\bar{Z}_R(r, f - h_{i_1}) + \bar{Z}_R(r, h_{i_2}) + \bar{Z}_R(r, h_{i_3}) \\ + \bar{Z}_R(r, h_{i_4}) + \bar{Z}_R(r, h_{i_5})) + o(T(r, f)).$$

In each one of the last inequalities, we can check that for each index $i_k$, the term $\bar{Z}(r, f - h_{i_k})$ (respectively $\bar{Z}(r, f - h_{i_k})$, respectively $\bar{Z}_R(r, f - h_{i_k})$) intervenes $C_{q-1}^4$ times. Consequently, we can derive that

$$2C_q^5 T(r, f) \le C_{q-1}^4 \sum_{i=1}^q \bar{Z}(r, f - h_i) + o(T(r, f))$$
(respectively

$$2C_q^5 T(r, f) \le C_{q-1}^4 \sum_{i=1}^q \bar{Z}(r, f - h_i) + o(T(r, f)),$$

respectively

$$2C_q^5 T_R(r, f) \le C_{q-1}^4 \sum_{i=1}^q \bar{Z}_R(r, f - h_i) + o(T_R(r, f)).$$

Consequently, it follows that

$$\frac{2q}{5} T(r, f) \le \sum_{i=1}^q \bar{Z}(r, f - h_i) + o(T(r, f))$$

(respectively

$$\frac{2q}{5} T(r, f) \le \sum_{i=1}^q \bar{Z}(r, f - h_i) + o(T(r, f)),$$

respectively

$$\frac{2q}{5} T_R(r, f) \le \sum_{i=1}^q \bar{Z}_R(r, f - h_i) + o(T_R(r, f)).$$

This completes the proof of Theorem 52.10. □
# Chapter 53

## The *p*-adic Hayman Conjecture

Throughout this chapter, the field $\mathbb{K}$ is supposed to be of characteristic 0.

In the 1950s, Walter Hayman asked the question of whether, given a meromorphic function in $\mathbb{K}$, the function $g'g^n$ might admit a quasi-exceptional value $b \neq 0$ [66]. W. Hayman showed that $g'g^n$ has no quasi-exceptional value, whenever $n \ge 3$. Henceforth, the problem was solved for $n=2$ by E. Mues in 1979 [77] and next for $n \ge 1$, in 1995 by W. Bergweiler and A. Eremenko [8] and separately by H. Chen and M. Fang [30]. See also [95]. The same problem is posed on the field $\mathbb{K}$, both in $M(\mathbb{K})$ and in a field $M(d(a, R^-))$ ($a \in \mathbb{K}$, $R > 0$). These are the studies in [52] and [80].

We still denote by $D$ the set $\mathbb{K}\backslash d(0, R^-)$ and use the sets $\mathcal{A}(D), \mathcal{A}^c(D), \mathcal{M}(D), \mathcal{M}^c(D)$.

The following lemma is immediate.

**Lemma 53.1.** Let $g \in M(\mathbb{K})$ (respectively let $g \in M(d(a, R^-))$, $a \in \mathbb{K}$, $R > 0$, respectively let $g \in M(D)$), set $f = \frac{1}{g}$, and let $n \in \mathbb{N}^*$. Then $g'g^n$ admits a quasi-exceptional value $b \in \mathbb{K}^*$ if and only if $f' + bf^{n+2}$ has finitely many zeros that are not zeros of $f$.

**Remark.** We can also consider the same problem when $n = -1$ i.e. the question of whether $f' + bf$ has infinitely many zeros. We examine this in *p*-adic analysis. When $n = 0$, in $\mathbb{C}$, the well-known counter-example furnished by the function $\tan x$ shows that $f' - f^2$ may have no zero. On the field $\mathbb{K}$, we examine the cases $n = -1$ and $n = 0$.
Henceforth, we examine that problem by considering the set of zeros of $f' + b f^m$, with $b \neq 0$. In the field $\mathbb{K}$, two theorems are specific to p-adic analysis. Both are based on the following lemma.

**Lemma 53.2.** Let $f \in \mathcal{M}(D)$ (respectively $f \in \mathcal{M}(d(0, R^{-}))$, $R > 0$), suppose that $f$ admits infinitely many zeros and suppose that there exists a sequence of intervals $[r'_n, r''_n]$ such that $\lim_{n \to +\infty} r'_n = +\infty$ (respectively $\lim_{n \to +\infty} r'_n = \lim_{n \to +\infty} r''_n = R$) and such that $|(f' + f^m)|(r) = |f^m|(r) \forall r \in \bigcup_{n \in \mathbb{N}} [r'_n, r''_n]$. Let $m \in \mathbb{N}^*$ be $\neq 2$. Then $f' + f^m$ has infinitely many zeros that are not zeros of $f$.

**Proof.** Let $J = \bigcup_{n \in \mathbb{N}} [r'_n, r''_n]$. By Corollary 22.6, we have

$$ \nu^+(f' + f^m, \log r) = \nu^+(f^m, \log r), \quad \nu^-(f' + f^m, \log r) \\ = \nu^-(f^m, \log r) \quad \forall r \in J. $$

Consequently, when $f \in \mathcal{M}(D)$, in each annulus $\Gamma(0, R, r)$ with $r \in J$, $f^m$ and $f' + f^m$ have the same difference between the number of zeros and poles. And when $f \in \mathcal{M}(d(0, R^{-}))$, in each disk $d(0, r)$ with $r \in J$, $f^m$ and $f' + f^m$ have the same difference between the number of zeros and poles.

Now, if $m \ge 3$, the poles of $f' + f^m$ and $f^m$ are the same taking multiplicity into account. And when $m = 1$, each pole of $f$ is a pole of $f' + f$ with a strictly greater order. Consequently, when $f \in \mathcal{M}(D)$ for each $r \in J$, the number of zeros of $f' + f^m$ in when $\Gamma(0, R, r)$ is superior or equal to this of $f^m$. And similarly, when $f \in \mathcal{M}(d(0, R^{-}))$, the number of zeros of $f' + f^m$ in when $d(0, r)$ is superior or equal to this of $f^m$.

Now, for each $n \in \mathbb{N}$, let $s_n$ be the number of distinct zeros of $f$ in $\Gamma(0, R, r)$ (respectively in $d(0, r''_n)$). Since $f$ has infinitely many zeros, the sequence $s_n$ is increasing and tends to $+\infty$. On the other hand, for each zero $\alpha$ of order $u$ of $f$, either $\alpha$ is not a zero of $f' + f^m$ (when $u = 1$) or it is a zero of order $u - 1$. Consequently, the number of zeros of $f' + f^m$ in $\Gamma(0, R, r''_n)$ (respectively in $d(0, r''_n)$) which are not zeros of $f$ is at least $s_n$. Thus, we have proved that $f' + f^m$ has infinitely many zeros that are not zeros of $f$. $\square$

**Remark.** Lemma 53.2 particularly applies to functions $f \in \mathcal{M}(\mathbb{K})$.
We prove together Theorems 53.3 and 53.4.

**Theorem 53.3.** Let $f \in \mathcal{M}^c(D)$ satisfy $\limsup_{r \to \infty} |f|(r) > 0$ and let $b \in \mathbb{K}^*$. Let $m \in \mathbb{N}^*$ be $\ge 3$. Then $f' + bf^m$ has infinitely many zeros that are not zeros of $f$.

**Proof.** Without loss of generality, we can assume $b = 1$. By hypotheses, there exists a sequence of intervals $[r'_n, r''_n]$ such that $\lim_{n \to +\infty} r'_n = +\infty$ and such that, putting $J = \bigcup_{n \in \mathbb{N}} [r'_n, r''_n]$, we have $\liminf_{\substack{r \to \infty \\ r \in J}} |f|(r) > 0$.

We prove that there exists $t > 0$ such that $|f' + f^m|(r) = |f^m|(r) \quad \forall r \in J \cap [t, +\infty[$. By Theorem 37.10, we have $|f'|(r) \le \frac{|f|(r)}{r}$. Thus, when $r \in J$ is big enough, we have $r(|f|(r))^{m-1} > 1$, hence $(|f|(r))^m > |f'|(r)$. Thus, there exists $t \ge s$ such that $(|f|(r))^m > |f'|(r) \quad \forall r \in J \cap [t, +\infty[$. Let $J' = J \cap [t, +\infty[$. So we have $|f' + f^m|(r) = |f^m|(r) \quad \forall r \in J'$.

We can now conclude. For each $n \in \mathbb{N}$, let $q_n$ be the number of zeros of $f$ in $\Gamma(0, R, r_n'')$. Suppose the sequence $(q_n)_{n \in \mathbb{N}}$ is bounded. Then, $f$ has finitely many zeros, hence it is of the form $\frac{P}{h}$ with $P \in \mathbb{K}[x]$ and $h \in \mathcal{A}(E)$. Consequently, we have $\lim_{r \to +\infty} |f|(r) = 0$, a contradiction to the hypothesis. Therefore, the sequence $(q_n)_{n \in \mathbb{N}}$ which is increasing by definition, tends to $+\infty$. Now, we may apply Lemma 53.2 showing that $f' + f^m$ has infinitely many zeros that are not zeros of $f$. $\square$

We have a similar version for functions inside a disk $d(a, R^{-})$.

**Theorem 53.4.** Let $f \in \mathcal{M}_u(d(a, R^{-}))$ satisfy $\limsup_{r \to R} |f|(r) = +\infty$ and let $b \in \mathbb{K}^*$. Let $m \in \mathbb{N}^*$ be $\ge 3$. Then $f' + bf^m$ has infinitely many zeros that are not zeros of $f$.

**Proof.** Without loss of generality, we can assume $b = 1$. By hypotheses, there exists a sequence of intervals $[r'_n, r''_n]$ such that $\lim_{n \to +\infty} r'_n = \lim_{n \to +\infty} r''_n = R$ and such that, putting $J = \bigcup_{n \in \mathbb{N}} [r'_n, r''_n]$, we have $\limsup_{\substack{r \to \infty \\ r \in J}} |f|(r) + \infty$. Thus, we have $|f'|(r) \le \frac{|f|(r)}{r} \le \frac{|f|(r)}{R}$. Now, when $r$ is close enough to $R$ with $r \in J$, we have $r(|f|(r))^{m-1} > 1$, hence $(|f|(r))^m > |f'|(r)$. Thus, there exists $t > 0$ such that $(|f|(r))^m > |f'|(r) \quad \forall r \in [t, +\infty[ \cap J$. We can set $J' = J \cap [t, R]$ and then we have $|f' + f^m|(r) = |f^m|(r) \quad \forall r \in J'$.

We can now conclude in Theorem 53.4 as in Theorem 53.3. For each $n \in \mathbb{N}$, let $q_n$ be the number of zeros of $f$ in $d(0, r_n'')$. Suppose the
sequence $(q_n)_{n \in \mathbb{N}}$ is bounded. Then, $f$ has finitely many zeros, hence it is of the form $\frac{P}{h}$ with $P \in \mathbb{K}[x]$ and $h \in \mathcal{A}(d(0, R^{-}))$. Consequently, we have $\limsup_{r \to +\infty} |f|(r) < \infty$, a contradiction to the hypothesis. Therefore, the sequence $(q_n)_{n \in \mathbb{N}}$ which is increasing by definition, tends to $+\infty$. Now, we may apply Lemma 53.2 showing that $f' + f^m$ has infinitely many zeros that are not zeros of $f$. $\square$

In the case $m=1$, we can have a better conclusion in $\mathcal{M}(\mathbb{K})$.

**Theorem 53.5.** Let $f \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}(x)$. For each $b \in \mathbb{K}^*$, $f' + bf$ has infinitely many zeros that are not zeros of $f$.

**Proof.** Without loss of generality, we can assume again $b=1$. By Theorem 37.10, we have $|f'|(r) < |f|(r)$ when $r$ is big enough, and hence $|f' + f|(r) = |f|(r)$ in an interval $I = [s, +\infty[$. Suppose first that $f$ has infinitely many zeros. We can then apply Lemma 53.2 and get the conclusion.

Suppose now that $f$ has finitely many zeros. Then $f$ has infinitely many poles $c_n$ of respective order $t_n$. Since $\mathbb{K}$ has characteristic zero, $f'$ admits each $c_n$ as a pole of order $t_n + 1$ and similarly, $f' + f$ also admits each $c_n$ as a pole of order $t_n + 1$. Thus, we have $N(r, f' + f) = N(r, f) + \bar{N}(r, f)$. But since $|f' + f|(r) = |f|(r)$ holds in $I$, we have $\nu(f' + f, \log r) = \nu(f, \log r) \forall r \in I$, and hence $Z(r, f' + f) - N(r, f') = Z(r, f) - N(r, f)$, therefore $Z(r, f' + f) - (N(r, f) + \bar{N}(r, f)) = Z(r, f) - N(r, f)$, and hence $Z(r, f' + f) = Z(r, f) + \bar{N}(r, f)$. Since we have supposed that $f$ has finitely many zeros and since $f$ has infinitely many poles, $f' + f$ has infinitely many zeros and all but finitely many are not zeros of $f$. $\square$

Concerning functions $f' + bf^2$, we can obtain a first conclusion when $f$ is analytic:

**Theorem 53.6.** Let $f \in \mathcal{A}^c(D)$ have infinitely many zeros in $D$ (respectively let $a \in \mathbb{K}$, let $R \in ]0, +\infty[$, and let $f \in \mathcal{A}_u(d(a, R^{-}))$). For each $b \in \mathbb{K}^*$, $f' + bf^2$ has infinitely many zeros that are not zeros of $f$.

**Proof.** Without loss of generality, we can assume $b=1$ and $a=0$. Clearly, $\lim_{r \to +\infty} |f|(r) = +\infty$ (respectively $\lim_{r \to R} |f|(r) = +\infty$), hence when $r$ is big enough, in $]0, +\infty[$ (respectively in $]0, R[$), we have $|f' + f^2|(r) = |f^2|(r)$, therefore, by Corollary 22.6, $f^2$ and $f' + f^2$ have the same number of zeros in $C(0, r)$ (taking multiplicity
into account). Let $\alpha \in C(0, r)$ be a zero of $f$ of order $q$. When $r$ is
big enough, it is a zero of order $2q$ for $f^2$ and it is a zero of order
$q-1$ for $f'+f^2$. Consequently, by Corollary 22.6, $f'+f^2$ has at least
$q+1$ zero in $C(0, r)$ that are not zeros of $f$ (taking multiplicity into
account). This is true for every such zeros of $f$ and hence $f'+f^2$ has
infinitely many zeros that are not zeros of $f$. $\square$

**Corollary 53.7.** Let $m \in \mathbb{N}^*$ be $\ge 1$, and let $f \in \mathcal{A}(\mathbb{K})\setminus\mathbb{K}(x)$. For each $b \in \mathbb{K}^*$, $f' + bf^m$ has infinitely many zeros that are not zeros of $f$.

**Corollary 53.8.** Let $m \in \mathbb{N}$ be $\ge 2$, let $a \in \mathbb{K}$, let $R \in ]0, +\infty[$, and let $f \in \mathcal{A}_u(d(a, R^{-}))$ (respectively let $f \in \mathcal{A}(D)$ have infinitely many zeros in $D$). For each $b \in \mathbb{K}^*$, $f' + bf^m$ has infinitely many zeros in $d(a, R^{-})$ (respectively in $D$) that are not zeros of $f$.

**Proposition 53.9.** Let $f \in \mathcal{M}^c(D)$ (respectively let $a \in \mathbb{K}$ and $R \in \mathbb{R}_+^*$ and let $f \in \mathcal{M}_u(d(a, R^-)$)) and let $m \in \mathbb{N}$. If $m \ge 5$, then for each $b \in \mathbb{K}^*$, $f' + bf^m$ has infinitely many zeros that are not zeros of $f$. If $m = 4$, if $f \in \mathcal{M}(\mathbb{K})\setminus\mathbb{K}(x)$, and if $f$ admits at least $s$ multiple zeros and at least $t$ multiple poles, then $f' + bf^4$ admits a number of zeros that are not zeros of $f$ (taken account of multiplicity) which is strictly superior to $\frac{s+t}{14}$.

**Proof.** For convenience, we denote $Z(r, \phi)$ instead of $Z_R(r, \phi)$ and $N(r, \phi)$ instead of $N_R(r, \phi)$, when $\phi \in \mathcal{M}(D)$.

By Corollary 22.2, the zeros of $f' + bf^m$ in $\mathbb{K}$ are the same as in
a spherically complete algebraically closed extension $\hat{\mathbb{K}}$ of $\mathbb{K}$. So, for
simplicity, we can suppose that the field $\mathbb{K}$ is spherically complete
without loss of generality. We can also suppose that $b = 1$. Then if
$f \in \mathcal{M}^c(D)$, we can obviously write $f = \frac{h}{l}$ with $h, l \in \mathcal{A}(D)$, having
no common zeros, and if $f \in \mathcal{M}(d(a, R^{-}))$, since $\mathbb{K}$ is spherically
complete, we can also write $f = \frac{h}{l}$ with $h, l \in \mathcal{A}(d(a, R^{-}))$, having
no common zeros again.

Let $g = \frac{1}{f}$ and let $n = m - 2$. So, by Lemma 53.1, the problem
is reduced to show that $g'g^n - 1$ has infinitely many zeros. Suppose
that $g'g^n - 1$ has finitely many zeros in $D$ (respectively in $d(0, R^{-}))$.
Then, $g'g^n - 1 = \frac{(l'h-h'l)l^n-h^{n+2}}{h^{n+2}}$, and since $h, l$ have no common
zeros, this is of the form $\frac{P}{h^{n+2}}$, where $P$ is a polynomial of degree $q$. Now, set $F = (l'h - h'l)l^n$. Applying Corollary 51.12 to $F$, we have

$$ (1) \quad T(r, F) = Z(r, F) + O(1) \le \bar{Z}(r, F) + \bar{Z}(r, F - P) + 7T(r, P) + O(1). $$

By (1), we can derive

$$ Z(r, l'h - h'l) + nZ(r, l) \le \bar{Z}(r, l'h - h'l) + \bar{Z}(r, l) + \bar{Z}(r, F - P) + 7T(r, P) + O(1). $$

Actually, $\bar{Z}(r, F - P) = \bar{Z}(r, h)$, hence we have $nZ(r, l) \le \bar{Z}(r, l) + \bar{Z}(r, h) + 7T(r, P) + O(1)$, and hence $(n-1)Z(r, l) \le \bar{Z}(r, h) + 7T(r, P) + O(1)$. But since $T(r, P) = q \log r + O(1)$, we have

$$ (2) \quad (n-1)Z(r, l) \le Z(r, h) + 7q \log r + O(1). $$

Consider the hypothesis $f \in \mathcal{M}^c(D)$. By Theorem 53.3, if $\liminf_{r \to +\infty} |f|(r) > 0$ i.e. if $\liminf_{r \to +\infty} Z(r, f) - N(r, f) > -\infty$, the claim is proved. Consequently, if the claim is not true, we can assume $\liminf_{r \to +\infty} Z(r, f) - N(r, f) = -\infty$ i.e.

$$ (3) \quad \liminf_{r \to +\infty} Z(r, l) - Z(r, h) = +\infty. $$

Since $f \in \mathcal{M}^c(D)$, by (3) we notice that $l \in \mathcal{A}^c(D)$, Consequently, (2) is impossible whenever $n \ge 3$, i.e. $m \ge 5$.

Consider now the hypothesis $f \in \mathcal{M}_u d(0, R^-)$). We have the same reasonings with limits to $R^-$ instead of $+\infty$ and $h$, $l \in \mathcal{A}(d(0, R^-))$ and we find again relations (2) and (3) with $R^-$ instead of $+\infty$. Then, by Theorem 53.4, if $\limsup_{r \to R^-} |f|(r) = +\infty$, i.e. if $\limsup_{r \to R^-} Z(r, f) - N(r, f) = +\infty$, the claim is proved. So, we assume that $\limsup_{r \to R^-} Z(r, f) - N(r, f) < +\infty$. But then (2) is impossible, whenever $n \ge 3$ i.e. $m \ge 5$, which ends the proof when $f \in \mathcal{M}_u(d(0, R^-))$.

Now, suppose $m = 4$ i.e $n = 2$ and more precisely, suppose $\bar{Z}(r, l) \le Z(r, l)) - \frac{s \log(r)}{2}$ and $\bar{Z}(r, h) \le Z(r, h) - \frac{t \log(r)}{2}$, hence

$$ (4) \quad (n-1)Z(r, l) \le Z(r, h) + \left( 7q - \frac{s+t}{2} \right) \log r + O(1). $$

Then Relation (3) implies $7q - \frac{s+t}{2} > 0$, and hence $f'f^n$ admits a number of zeros strictly superior to $\frac{s+t}{14}$ which ends the proof. $\square$
**Corollary 53.10.** Let $a \in \mathbb{K}$, $R > 0$, and let $f \in \mathcal{M}_u(d(a, R^-))$. For every $n \in \mathbb{N}$, $n \ge 3$, and for every $b \in \mathbb{K}^*$, $f'f^n - b$ has infinitely many zeros.

**Corollary 53.11.** Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$. Let $b \in \mathbb{K}^*$. Then if $f$ has infinitely many multiple zeros or poles, then $f' + bf^4$ has infinitely many zeros that are not zeros of $f$.

We now thoroughly examine the situation when $m = 4$, i.e. $n = 2$, as made in [52]. This requires several basic lemmas.

**Lemma 53.12.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental and such that $f'$ has finitely many multiple zeros. Then $\frac{f''f}{(f')^2}$ has no quasi-exceptional value.

**Proof.** Let $g = \frac{f}{f'}$. A pole of $g$ is a zero of $f'$, hence by hypothesis, $g$ has finitely many multiple poles. Consequently, by Theorem 50.8, $g'$ has no quasi-exceptional value. And hence neither has $1 - g'$. But $g' = \frac{(f')^2 - f''f}{(f')^2} = 1 - \frac{f''f}{(f')^2}$. Therefore, $\frac{f''f}{(f')^2}$ has no quasi-exceptional value. $\square$

**Lemma 53.13.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental and have finitely many multiple zeros. Then $f''f + 2(f')^2$ has infinitely many zeros that are not zeros of $f$.

**Proof.** Suppose first that $f'$ has infinitely many multiple zeros. Since $f$ has finitely many multiple zeros, the zeros of $f'$ are not zeros of $f$ except at most finitely many. Hence, $f'$ has infinitely many multiple zeros that are not zeros of $f$. And then, they are zeros of $f''$, hence of $f''f + 2(f')^2$, which proves the statement.

So, we are now led to assume that $f'$ has finitely many multiple zeros. By Lemma 53.12, $\frac{f''f+2(f')^2}{(f')^2}$ has infinitely many zeros. Let $c \in \mathbb{K}$ be a pole of order $q$ of $f$. Without loss of generality, we can suppose $c = 0$. The beginning of the Laurent development of $f$ at $0$ is of the form $\frac{a-q}{x^q} + \frac{\varphi(x)}{x^{q-1}}$ whereas $\varphi \in \mathcal{M}(\mathbb{K})$ has no pole at $0$. Consequently, $\frac{f''f+2(f')^2}{(f')^2}$ is of the form

$$ \frac{(a_{-q})^2(3q^2 + q) + x\phi(x)}{(a_{-q})^2(q^2) + x\theta(x)} $$
whereas $\phi, \theta \in \mathcal{M}(\mathbb{K})$ have no pole at 0. So, the function $\frac{f''f+2(f')^2}{(f')^2}$ has no zero at 0. Therefore, each zero of $\frac{f''f+2(f')^2}{(f')^2}$ is a zero of $f''f + 2(f')^2$, and hence $f''f + 2(f')^2$ has infinitely many zeros.

Now, let us show that the zeros of $f''f + 2(f')^2$ are not zeros of $f$, except maybe finitely many. Let $c$ be a zero of $f''f + 2(f')^2$ and suppose that $c$ is a zero of $f$. Then, it is a zero of $f'$, and hence it is a multiple zero of $f$. But by hypotheses, $f$ has finitely many multiple zeros, hence the zeros of $f''f + 2(f')^2$ are not zeros of $f$, except at most finitely many. That finishes proving the claim. $\square$

**Lemma 53.14.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental and let $b \in \mathbb{K}^*$ be such that $f^2f' - b$ has finitely many zeros. Then, $N(r, f) \le Z(r, f) + O(1)$.

**Proof.** Let $F = f^2f'$. Since $F-b$ is transcendental and has finitely many zeros, it is of the form $\frac{P(x)}{h(x)}$ with $h \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}[x]$. Consequently, $|F|(r)$ is a constant when $r$ is big enough, and therefore, by Theorem 43.2, we have $Z(r, F) = N(r, F) + O(1)$ when $r$ is big enough. Now, $Z(r, F) = 2Z(r, f) + Z(r, f')$ and, by Theorem 43.13, $Z(r, f') \le Z(r, f) + \bar{N}(r, f) - \log r + O(1)$. On the other hand, by Theorem 43.13 again, we have $N(r, F) = 3N(r, f) + \bar{N}(r, f)$. Consequently, $3N(r, f) + \bar{N}(r, f) \le 3Z(r, f) + \bar{N}(r, f) - \log r + O(1)$, which proves the claim. $\square$

We can now state the solution of the Hayman conjecture in a $p$-adic field. Theorem 53.15 was published in [52].

**Theorem 53.15.** Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$. Then for each $b \in \mathbb{K}^*$, $f'f^2 - b$ has infinitely many zeros.

**Proof.** Let $b \in \mathbb{K}^*$ and suppose that the claim is wrong, i.e. $f^2f' - b$ has $k$ zeros, taking multiplicity into account. By Proposition 53.9, we may assume that $f$ has finitely many multiple zeros and finitely many poles. Set $F = f^2f'$. Then, $F' = f(f''f + 2(f')^2)$. By Lemma 53.13, $f''f + 2(f')^2$ has infinitely many zeros that are not zeros of $f$. Consequently, $F'$ admits for zeros: the zeros of $f$ and the zeros of $f''f + 2(f')^2$. And by Lemma 53.13, there exists a sequence of zeros of $f''f + 2(f')^2$ that are not zeros of $f$.

Let $S = \{0, b\}$ and let $Z_0^S(r, F')$ be the counting function of zeros of $F'$ when $F(x)$ is different from 0 and $b$. Since $F - b$ has
finitely many zeros, the zeros $c$ of $F'$ which are not zeros of $f$ cannot satisfy $F(c) = b$ except at most finitely many. Consequently, there are infinitely many zeros of $F'$ counted by the counting function $Z_0^S(r, F')$, and hence for every fixed integer $q \in \mathbb{N}$, we have

$$ (1) \qquad Z_0^S(r, F') \ge q \log r + O(1). $$

Let us apply Theorem 45.7 to $F$. We have

$$ (2) \qquad T(r, F) \le \bar{Z}(r, F) + \bar{Z}(r, F-b) + \bar{N}(r, F) - Z_0^S(r, F') - \log(r) + O(1). $$

Now, we have

$$ (3) \qquad \bar{Z}(r, F) \le Z(r, f) + Z(r, f'), $$

$$ (4) \qquad \bar{N}(r, F) = \bar{N}(r, f), $$

and since the number of zeros of $F - b$ is $k$, taking multiplicity into account, we have

$$ (5) \qquad \bar{Z}(r, F-b) \le k \log r + O(1). $$

Consequently, by (2), (3), (4), and (5), we obtain

$$ (6) \qquad T(r, F) \le Z(r, f) + Z(r, f') + \bar{N}(r, f) - Z_0^S(r, F') + (k-1) \log r + O(1). $$

On the other hand, by construction, $T(r, F) \ge Z(r, F) = 2Z(r, f) + Z(r, f')$, hence by (6), we obtain

$$ (7) \qquad Z(r, f) \le \bar{N}(r, f) - Z_0^S(r, F') + (k-1) \log r + O(1). $$

Now, by Lemma 53.14, we have $N(r, f) \le Z(r, f) + O(1)$, hence by (7), we obtain $0 \le (k-1) \log r - Z_0^S(r, F') + O(1)$, and hence, by (1), fixing $q > k-1$ we can derive $0 \le (k-1) \log r - q \log r + O(1)$, a contradiction. That finishes the proof of Theorem 53.15. $\square$
By Lemma 53.1 and Theorems 53.9 and 53.15, we can now state
the general result on the *p*-adic Hayman conjecture:

**Corollary 53.16 (52).** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental and let $b \in \mathbb{K}^*$. Then for every $n \ge 2$, $f'f^n - b$ has infinitely many zeros. For every $m \ge 4$, $f'+bf^4$ has infinitely many zeros that are not zeros of $f$.

Concerning the case $m = 3$ i.e. $n = 1$ which remains unsolved,
thanks to Theorem 50.11, Corollary 53.16 has an immediate appli-
cation to the conjecture with additional hypotheses.

**Corollary 53.17.** Let $f \in \mathcal{M}(\mathbb{K})$. Suppose that there exists $c, q \in [0, +\infty[$ such that $N(r, f) \le cr^q \quad \forall r \in [1, +\infty[$. If $f'f^n - b$ has finitely many zeros for some $b \in \mathbb{K}^*$, with $n \in \mathbb{N}$, then $f \in \mathbb{K}(x)$.

**Proof.** Suppose $f$ is transcendental. By hypothesis, $f^{n+1}$ satisfies
$Z(r, \frac{1}{f^{n+1}}) = \xi(r, f^{n+1}) \le c(n+1)r^q \quad \forall r \in [1, +\infty[$, hence by Corollary 53.16 and Theorem 50.11, $f'f^n$ has no quasi-exceptional value different from 0. $\square$

Corollary 53.17 may be written in another way:

**Corollary 53.18.** Let $f \in \mathcal{M}(\mathbb{K})\K(x)$. Suppose that there exists $c, q \in [0, +\infty[$ such that $Z(r, f) \le cr^q \quad \forall r \in [1, +\infty[$. Then for all $m \in \mathbb{N}$, $m \ge 3$, and for all $b \in \mathbb{K}$, $f' - bf^m$ admits infinitely many zeros that are not zeros of $f$.

**Proof.** We set $g = \frac{1}{f}$. Then by Corollary 53.17, $g'g^{m-2}$ has no quasi-exceptional value. Consequently, given $b \in \mathbb{K}^*$, $g'g^{m-2} + b$ has infinitely many zeros, and hence, $f' - bf^m$ has infinitely many zeros that are not zeros of $f$. Next, if $b=0$, by Theorem 50.11, $f'$ has infinitely many zeros. $\square$

Consider now the case $m = 3$, i.e. $n = 1$.

**Theorem 53.19.** Let $f \in \mathcal{M}(\mathbb{K})$. Suppose that there exists $c, q \in [0, +\infty[$ such that $s(r, f) \le cr^q \quad \forall r \in [1, +\infty[$. Then, for all $b \in \mathbb{K}$, $\frac{f'}{f^2} - b$ has infinitely many zeros.

**Proof.** Set $g = \frac{1}{f}$ again. Since the poles of $g$ are the zeros of $f$, we have $t(r, g) \le cr^q$. Consequently, by Corollary 50.12, $g'$ has no quasi-exceptional value. $\square$
**Remark.** Using Theorem 53.19 to study the zeros of $f' - bf^2$ that are not zeros of $f$ is not so immediate, as we see in the following because of residues of $f$ at poles of order 1. Of course, if $\frac{1}{f}$ is an affine function, $f' + f^2$ has no zeros, except if it is identically zero. And if it is not identically zero, the residue at the pole is not 1 in the general case.

**Lemma 53.20.** Let $f = \frac{h}{l} \in \mathcal{M}(\mathbb{K})$ with $h, l \in \mathcal{A}(\mathbb{K})$ having no common zero, let $b \in \mathbb{K}^*$, and let $a \in \mathbb{K}$ be a zero of $h'l - hl' + bh^2$ that is not a zero of $f' + bf^2$. Then $a$ is a pole of order 1 of $f$ and $\text{res}(f, a) = \frac{1}{b}$.

**Proof.** Clearly, if $l(a) \neq 0$, $a$ is a zero of $f' + bf^2$. Hence, a zero $a$ of $h'l - hl' + bh^2$ that is not a zero of $f' + bf^2$ is a pole of $f$. Now, when $l(a) = 0$, we have $h(a) \neq 0$, hence $l'(a) = bh(a) \neq 0$ and therefore $a$ is a pole of order 1 of $f$ such that $\frac{h(a)}{l'(a)} = \frac{1}{b}$. But since $a$ is a pole of order 1, we have $\text{res}(f, a) = \frac{h(a)}{l'(a)}$, which ends the proof. $\square$

Theorem 53.21 is not a result specific to p-adic analysis but it is useful in Theorem 53.23.

**Theorem 53.21.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $a \in \mathbb{K}$, and let $f \in \mathcal{M}(d(a, R^{-}))$), let $b \in \mathbb{K}^*$ and let $\alpha \in \mathbb{K}$ (respectively let $\alpha \in d(a, R^{-}))$ be a point that is not a zero of $f$ and such that the residue of $f$ at $\alpha$ is different from $\frac{1}{b}$. Then $\alpha$ is a zero of $f' + bf^2$ if and only if it is a zero of $\frac{f'}{f^2} + b$. Moreover, if it is a zero of both functions, it has the same multiplicity with both.

**Proof.** Suppose first $\alpha$ is a zero of $f' + bf^2$. If $\alpha$ is not a pole of $f$, of course it is a zero of $\frac{f'}{f^2} + b$ with same multiplicity. Suppose now that $\alpha$ is a pole of $f$: since it is not a pole of $f' + bf^2$, it must be a pole of order 1 of $f$. Without loss of generality, we may assume that $\alpha = 0$ (respectively $a = \alpha = 0$). Consider the Laurent series of $f$ at 0: $f(x) = \frac{a-1}{x} + a_0 + a_1x + x^2\phi(x)$ with $\phi \in \mathcal{M}(\mathbb{K})$ (respectively $\phi \in \mathcal{M}(d(0, R^{-}))$) and $\phi(0) \neq \infty$. Then $f' + bf^2$ is of the form

$$f'(x) + bf(x)^2 = \frac{a_{-1}(-1 + ba_{-1})}{x^2} + \frac{2ba_0a_1}{x} + a_1 + b(a_0^2 + 2a_1a_{-1}) + x\eta(x)$$
with $\eta \in \mathcal{M}(\mathbb{K})$ (respectively $\eta \in \mathcal{M}(d(0, R^{-}))$ and $\eta(0) \neq \infty$, and hence, we have $a_{-1}(-1 + ba_{-1}) = 0$, $a_0a_{-1} = 0$, $a_0^2 + 2a_1a_{-1} = 0$. Since by hypothesis $\text{res}(f, \alpha) \neq -\frac{1}{b}$, we have $(1 + ba_{-1}) \neq 0$, hence $a_{-1} = 0$, a contradiction. Consequently, every zero of $f' + bf^2$ that is not a zero of $f$ is a zero of $\frac{f'}{f^2} + b$ with the same multiplicity.

Conversely, suppose now that $\alpha$ is a zero of $\frac{f'}{f^2} + b$. If $\alpha$ is not a pole of $f$, it is a zero of $f' + bf^2$, with the same multiplicity, because by hypothesis it is not a zero of $f$. Now suppose that $\alpha$ is a zero of $\frac{f'}{f^2} + b$ and is a pole of $f$. Clearly, it is a pole of order 1, and again, we may assume that $\alpha = 0$.

Consider again the Laurent series of $f$ at $0$: $f(x) = \frac{a_{-1}}{x} + a_0 + a_1x + x^2\phi(x)$ with $\phi \in \mathcal{M}(\mathbb{K})$ and $\phi(0) \neq \infty$. Then,

$$ \frac{f'}{f^2} = \frac{\frac{-a_{-1}}{x^2} + a_1 + x\theta(x)}{\frac{(a_{-1})^2}{x^2} + \frac{2a_0a_1}{x} + a_0^2 + 2a_1a_{-1} + x\eta(x)} $$

where both $\theta$, $\eta \in \mathcal{M}(\mathbb{K})$ have no pole at 0. Clearly, $\frac{f'}{f^2}$ is analytic at 0 and its value is $\frac{-1}{a_{-1}}$. But since 0 is a zero of $\frac{f'}{f^2} + b$, we have $a_{-1} = \frac{1}{b}$, what is excluded by hypothesis. Thus, we have proved that every zero of $\frac{f'}{f^2} + b$ is a zero of $f' + bf^2$ (that is not a zero of $f$) with the same multiplicity, and this ends the proof of Theorem 53.21. $\square$

**Theorem 53.22.** Let $b \in \mathbb{K}^*$ and let $f \in \mathcal{M}(\mathbb{K})$ have finitely many zeros and finitely many residues at its simple poles equal to $\frac{1}{b}$ and be such that $f' + bf^2$ has finitely many zeros. Then $f$ belongs to $\mathbb{K}(x)$.

**Proof.** Let $f = \frac{P}{l}$ with $P \in \mathbb{K}[x]$, $l \in \mathcal{A}(\mathbb{K})$ having no common zero with $P$. Then, $f' + bf^2 = \frac{P'l-l'P+bP^2}{l^2}$. By hypothesis, this function has finitely many zeros. Moreover, if $a$ is a zero of $P'l - l'P + bP^2$ but is not a zero of $f' + bf^2$, then by Lemma 53.20 it is a pole of order 1 of $f$ such that $\text{res}(f, a) = \frac{1}{b}$. Consequently, $P'l - l'P + bP^2$ has finitely many zeros, and hence, we can write $\frac{P'l-l'P+bP^2}{l^2} = \frac{Q}{l^2}$ with $Q \in \mathbb{K}[x]$, hence $P'l - l'P = -bP^2 + Q$. But then, by Theorem 50.5, $l$ is a polynomial, which ends the proof. $\square$
**Remark.** If $f(x) = \frac{1}{x}$, the function $f' + b f^2$ has no zero whenever $b \neq 1$.

**Theorem 53.23.** Let $f \in \mathcal{M}(\mathbb{K})$ be transcendental and have finitely many zeros of order $\ge 2$ and let $b \in \mathbb{K}$. Then $\frac{f'}{f^2} + b$ has infinitely many zeros. Moreover, if $b \ne 0$, every zero $\alpha$ of $\frac{f'}{f^2} + b$ that is not a zero of $f' + b f^2$ is a pole of $f$ of order 1 such that the residue of $f$ at $\alpha$ is equal to $\frac{1}{b}$.

**Proof.** Let $g = \frac{f'}{f^2} + b$. Since all zeros of $f$ are of order 1 except maybe finitely many, $g$ has finitely many poles of order $\ge 3$, hence a primitive $G$ of $g$ has finitely many poles of order $\ge 2$. Consequently, by Theorem 50.8, $g$ has infinitely many zeros.

Now, suppose $b \ne 0$. Let $\alpha$ be a zero of $g$. If $\alpha$ is not a pole of $f$, it is a zero of $f' + b f^2$ and we can see that it is not a zero of $f$.

Finally, suppose that $\alpha$ is a pole of $f$. Then it must be a pole of order 1, and then, by Lemma 53.20, the residue of $f$ at $\alpha$ is $\frac{1}{b}$. $\square$

**Corollary 53.24.** Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$ have finitely many zeros of order $\ge 2$ and finitely many poles of order 1 and let $b \in \mathbb{K}^*$. Then $f' + b f^2$ has infinitely many zeros that are not zeros of $f$.
# Chapter 54

## Composition of Meromorphic Functions

Throughout this chapter, the field $\mathbb{K}$ is supposed to be of characteristic 0.

In order to look for meromorphic functions of uniqueness in the following chapter, here we first examine the composition of meromorphic functions which, in the general case, is not a meromorphic function.

Throughout this chapter, $\mathbb{E}$ is an algebraically closed field of characteristic 0 without any assumption on the existence of an absolute value. The field $\mathbb{K}$ is supposed to have characteristic zero.

In each main claim, instead of assuming that the equality $h(f(x)) = h(g(x))$ holds in the whole set of definition, thanks to properties of analytic sets, we check that it is sufficient to have the equality on a bounded sequence having no cluster point at the poles of $f$ and $g$. For that, we use properties of analytic elements on infra-connected subsets of $\mathbb{K}$.

In the field $\mathbb{K}$ (as in $\mathbb{C}$), the composition of two meromorphic functions $h \circ f$ is not a meromorphic function, in the general case: a pole of $f$ is currently narrowed by poles of $h \circ f$ coming from the poles of $h$. This is why we first have to study general and basic properties of such functions in the ultrametric context.

**Remark and definition.** Consider a Laurent series $\sum_{n=-\infty}^{+\infty} a_n (x-\alpha)^n$ converging for $0 < |x-\alpha| < r$.
If the Laurent series is a power series, it then defines an analytic function $f(x)$ in the disk $d(\alpha, r^{-})$. The point $\alpha$ is called a regular point for $f$. Else, $\alpha$ is called a singular point for $f$.

If the Laurent series is of the form $\sum_{n=-q}^{\infty} a_n (x-\alpha)^n$ with $a_{-q} \neq 0$, then it defines a meromorphic function in $d(\alpha, r^{-})$ and $\alpha$ is a pole of order $q$ of that function.

If $\alpha$ is a singular point but is not a pole, it is called a point of high singularity for $f$. This is the case if and only if infinitely many terms $a_n$ with $n < 0$ are not zero.

**Notation.** Let $h \in \mathcal{M}(\mathbb{K})$ (respectively $h \in \mathcal{M}(d(a, R^{-})))$. We denote by $\mathcal{P}(h)$ the set of poles of $h$ and by $\mathcal{C}(h)$ the set of zeros of $h$. Let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}(d(a, r^{-})))$. We set $\mathcal{T}(f, h) = \{x \in \mathbb{K} \mid f(x) \in \mathcal{P}(h)\}$ (respectively $\mathcal{T}(f, h) = \{x \in d(a, R^{-}) \mid f(x) \in \mathcal{P}(h)\}$) and we denote by $\mathcal{S}(f, h)$ the set $\mathcal{P}(f) \cup \mathcal{T}(f, h)$.

**Proposition 54.1** is classical in $\mathbb{C}$ as in the field $\mathbb{K}$:

**Proposition 54.1.** Let $h \in \mathcal{M}(\mathbb{K})$ and let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}(d(a, R^{-})))$. Then $\mathbb{K}\setminus(\mathcal{S}(f, h))$ is an open subset of $\mathbb{K}$ dense in $\mathbb{K}$ (respectively $d(a, R^{-})\setminus(\mathcal{S}(f, h))$ is an open subset of $d(a, R^{-})$ dense in $d(a, R^{-}))$. For each $\alpha \in \mathbb{K}\setminus\mathcal{P}(f)$ (respectively $\alpha \in d(a, R^{-})\setminus\mathcal{P}(f)$), $h \circ f(x)$ is equal to a Laurent series in $x - \alpha$ in any set of the form $d(\alpha, r)\setminus\{\alpha\}$ included in $\mathbb{K}\setminus\mathcal{P}(\alpha)$ (respectively included in $d(a, R^{-})\setminus(\mathcal{P}(f))$. If $\alpha \notin \mathcal{T}(f, h)$, the Laurent series in $x - \alpha$ of $h \circ f$ has no terms of negative index. If $\alpha \in \mathcal{T}(f, h)$, then the Laurent series of $h \circ f$ in $x - \alpha$ has finitely many terms of negative indices (i.e. $h \circ f$ is meromorphic in a disk of center $\alpha$ and has a pole at $\alpha$).

**Definition.** Let $h \in \mathcal{M}(\mathbb{K})$ and let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}(d(a, R^{-})))$. Let $\alpha \in \mathbb{K}\setminus\mathcal{P}(f)$ (respectively $\alpha \in d(a, R^{-})\setminus\mathcal{P}(f)$). We call Laurent series of $h \circ f$ at $\alpha$ the Laurent series in $x - \alpha$ equal to $h \circ f(x)$ in a neighborhood of $\alpha$, excepting $\alpha$ when the function $h \circ f$ is not defined at $\alpha$.

**Remark.** Under the notation of the last definition, if the Laurent series of $h \circ f$ at $\alpha$ is a power series, the series then defines an analytic function in a disk $d(\alpha, r^{-})$. The point $\alpha$ is then a regular point for $h \circ f$. Else, $\alpha$ is a singular point for $h \circ f$.
If the Laurent series at $\alpha \in \mathcal{T}(f, h)$ is of the form $\sum_{n=-q}^{\infty} a_n(x - \alpha)^n$ with $a_{-q} \neq 0$, then the Laurent series defines a meromorphic function in a disk $d(\alpha, r^{-})$ and $\alpha$ is a pole of order $q$ of that function.

Next, by Proposition 54.1, a singular point for $h \circ f$ which is not a pole of $h \circ f$ belongs to $\mathcal{P}(f)$ and is a point of high singularity for $h \circ f$.

Proposition 54.2 is immediate:

**Proposition 54.2.** A point of high singularity for $h \circ f$ is a pole of $f$. Conversely, a pole of $f$ is just a pole for $h \circ f$ when $h$ is a (non-constant) polynomial. And when $h$ is a rational function tending to a finite limit at infinity, then a pole of $f$ is a regular point for $h \circ f$.

**Notation.** Given a subset $A$ of $\mathbb{K}$ and positive numbers $t, r \in ]0, t[$, we set $\mathcal{D}(t, r, A) = d(0, t) \setminus \bigcup_{\alpha \in A} d(\alpha, r^{-})$.

**Lemma 54.3.** Let $h \in \mathcal{M}(\mathbb{K})$, let $f \in \mathcal{M}(\mathbb{K})$ (respectively let $f \in \mathcal{M}(d(a, R^{-}))$). Then for every $t > 0$ (respectively $t \in ]0, R[$) and $r \in ]0, t[$, both $\mathcal{P}(f) \cap d(0, t)$ and $\mathcal{T}(f, h) \cap \mathcal{D}(t, r, \mathcal{P}(f))$ are finite, any cluster point of $\mathcal{S}(f, h)$ is a pole of $f$, the number of holes of $\mathcal{D}(t, r, \mathcal{S}(f, h))$ is finite, and $h \circ f$ belongs to $H(\mathcal{D}(t, r, \mathcal{S}(f, h)))$.

**Proof.** We assume $a = 0$. Let $t > 0$ (respectively $t \in ]0, R[$) be fixed and let $r \in ]0, t[$. Let $\{\alpha_1, \dots, \alpha_q\}$ be the finite set of all poles of $f$ and $g$ in $d(0, t)$.

Now consider $\mathcal{T}(f, h) \cap \mathcal{D}(t, r, \mathcal{P}(f))$. The set $E = \mathcal{D}(t, r, \mathcal{P}(f))$ is a closed bounded infraconnected set with finitely many holes, hence by Corollary 21.3, $f$ is bounded and takes every value finitely many times. Consequently, $h$ has finitely many poles in $f(E)$ and $\mathcal{T}(f, h) \cap \mathcal{D}(t, r, \mathcal{P}(f))$ is finite. Therefore, the number of holes of $\mathcal{D}(t, r, \mathcal{S}(f, h))$ is finite.

We note that any cluster point of $\mathcal{S}(f, h)$ in $d(0, t)$ is necessarily one of the $\alpha_j$. Indeed, let $\alpha \in \mathcal{S}(f, h) \cap d(0, t)$ with $\alpha \neq \alpha_j \forall j = 1, \dots, q$. Then $f(\alpha)$ is a pole of $h$, hence there exists a disk $d(f(\alpha), s)$ such that $h$ has no singularity in $d(f(\alpha), s)$ except $f(\alpha)$ i.e. $h$ is of the form $\frac{g}{(x-f(\alpha))^u}$ with $g \in H(d(f(\alpha), s))$. And of course there exists a disk $d(\alpha, \ell)$ such that $f(d(\alpha, \ell)) \subset d(f(\alpha), s)$, which shows that $h \circ f$ has no singularity other than $\alpha$ in $d(\alpha, \ell)$, hence $\alpha$ is not a cluster point of $\mathcal{S}(f, h)$. $\square$
**Lemma 54.4.** Let $h \in \mathcal{A}(\mathbb{K})$ and let $f \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ (respectively $f \in \mathcal{A}_u(d(0, R^{-})))$. There exists $s > 0$ (respectively $s \in ]0, R[$) such that $|h \circ f|(r) = |h|(|f|(r)) \quad \forall r \ge s$ (respectively $\forall r \in [s, R[$).

**Proof.** Let $b = f(0)$. By Theorem 22.9, $f(d(0, r))$ is a disk $d(b, t)$. Suppose $f$ belongs to $\mathcal{A}(\mathbb{K})\backslash\mathbb{K}$, hence it admits a zero $\alpha \in \mathbb{K}$. Let $s = |\alpha|$ and let us take $r \ge s$. Then $f(d(0, r))$ is a disk $d(b, t)$ equal to $d(0, t)$, hence $t = |f|(r)$. Consequently, $|h \circ f|(r) = \sup\{|h(f(x))| \mid x \in d(0, r)\} = \sup\{|h(u)| \mid u \in d(0, |f|(r))\}$.

Similarly, suppose $f$ belongs to $\mathcal{A}_u(d(a, R^{-}))\backslash\mathbb{K}$. Since $f$ is unbounded, it admits a zero $\alpha \in d(a, R^{-})$. By putting again $s = |\alpha|$ we can go on as in the previous case. $\square$

**Corollary 54.5.** Let $h \in \mathcal{A}(\mathbb{K})$ and let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^{-})))$. Then there exists $s > 0$ (respectively $s \in ]0, R[$) such that $Z(r, h \circ f) - Z(|f|(r), h)$ is a constant for all $r \ge s$.

**Lemma 54.6.** Let $h \in \mathcal{M}(\mathbb{K})$ and let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^{-})))$. Then, $T(r, h \circ f) = T(|f|(r), h) + O(1)$.

**Proof.** By Lemma 36.3, there exist $\phi, \theta \in \mathcal{A}(\mathbb{K})$ with no common zeros such that $h = \frac{\phi}{\theta}$. Then $Z(r, \phi \circ f) = Z(r, h \circ f)$ and $Z(r, \theta \circ f) = N(r, h \circ f)$. On the other hand, by Corollary 54.5, we have $Z(r, \phi \circ f) = Z(|f|(r), \phi) + O(1)$, $Z(r, \theta \circ f) = Z(|f|(r), \theta) + O(1)$. Consequently, $T(r, h \circ f) = \max(Z(|f|(r), \phi), Z(|f|(r), \theta)) + O(1)$. But now, we have $\max(Z(|f|(r), \phi), Z(|f|(r), \theta)) = T(|f|(r), h) + O(1)$, which ends the proof. $\square$

**Lemma 54.7.** Let $\phi \in \mathcal{A}(\mathbb{K})$ satisfy $h(0) \ne 0$, $s > 0$ (respectively $s \in ]0, R[$) and let $\eta, \upsilon$ be increasing continuous functions from $]0, +\infty[$ (respectively from $[0, R[$) to $]0, +\infty[$ satisfying $\lim_{r \to +\infty} \eta(r) = \lim_{r \to +\infty} \upsilon(r) = +\infty$ (respectively $\lim_{r \to R} \eta(r) = \lim_{r \to R} \upsilon(r) = +\infty$) and such that $Z(\eta(r), \phi) - Z(\upsilon(r), \phi)$ is a constant in $[s, +\infty[$ (respectively in $[s, R[$). Then $\log(\eta(r)) - \log(\upsilon(r))$ is bounded in $[s, +\infty[$ (respectively in $[s, R[$). Moreover, if $h$ is not a polynomial (respectively if $\phi$ belongs to $\mathcal{A}_u(d(a, R^{-})))$, then $\log(\eta(r)) - \log(\upsilon(r))$ tends to 0 when $r$ tends to $+\infty$ (respectively to $R$).

**Proof.** Since $Z(\eta(r), \phi) = \log(|\phi|(\eta(r))) - \log(|\phi(0)|)$ and $Z(\upsilon(r), \phi) = \log(|\phi|(\upsilon(r))) - \log(|\phi(0)|)$, we can see that $\log(|\phi|(\eta(r))) - \log(|\phi|(\upsilon(r)))$ is a constant $C$ in $[s, +\infty[$ (respectively
in $[s, R]$. Then by Corollary 22.30, we know that $|\log(|\phi(\eta(r))| - \log(|\phi(\nu(r))|)|_\infty$

$$ \geq \min(\nu^+(\phi, \log(\nu(r))), \nu^+(\phi, \log(\eta(r))))|\log(\eta(r)) - \log(\nu(r))|_\infty. $$

Suppose first $\phi \in \mathcal{A}(\mathbb{K})$. If $\phi$ is not a polynomial, we have

$$ \lim_{r \to +\infty} \nu^+(\phi, \log(\eta(r))) = \lim_{r \to +\infty} \nu^+(\phi, \log(\nu(r))) = +\infty, $$

hence $\lim_{r \to +\infty} \log(\eta(r)) - \log(\nu(r)) = 0$. And if $\phi$ is a polynomial of degree $q$, then when $r$ is big enough, we have $\nu^+(\phi, \log(\eta(r))) = \nu^+(\phi, \log(\nu(r))) = q$, hence $\log(\eta(r)) - \log(\nu(r))$ is constant.

Now, if $\phi$ belongs to $\mathcal{A}_u(d(a, R^-))$, then

$$ \lim_{r \to R} \nu^+(\phi, \log(\eta(r))) = \lim_{r \to R} \nu^+(\phi, \log(\nu(r))) = +\infty, $$

hence $\lim_{r \to R}(\log(\eta(r)) - \log(\nu(r))) = 0$ again. In conclusion, $\log(\eta(r)) - \log(\nu(r)))$ is bounded in all cases. $\square$

**Lemma 54.8.** Let $h \in \mathcal{M}(\mathbb{K})$ and let $f, g \in \mathcal{A}(\mathbb{K})$ (respectively $f, g \in \mathcal{A}(d(a, R^-))$) satisfy $h \circ f = h \circ g$. Then $T(r, f) - T(r, g)$ is bounded in any interval $[b, +\infty[$ (respectively $[b, R[$).

**Proof.** Let $\phi, \theta \in \mathcal{A}(\mathbb{K})$ with no common zeros such that $h = \frac{\phi}{g}$. Then $Z(r, \phi \circ f) = Z(r, h \circ f)$ and $Z(r, \phi \circ g) = Z(r, h \circ g)$. On the other hand, by Lemma 54.5, there exists $s', s'' > 0$ (respectively $s', s'' \in ]0, R[$) such that $Z(r, \phi \circ f)$ is of the form $Z(|f|(r), \phi) + C'$ with $C' \in \mathbb{R}$ whenever $r \ge s'$ (respectively whenever $r \in [s', R[$) and similarly $Z(r, \phi \circ g)$ is of the form $Z(|g|(r), \phi) + C''$ with $C'' \in \mathbb{R}$, whenever $r \ge s''$ (respectively whenever $r \in [s'', R[$). Consequently, putting $s = \max(s', s'')$, we have $Z(|f|(r), \phi) + C' = Z(|g|(r), \phi) + C'' \forall r \ge s$, hence $Z(|f|(r), \phi) - Z(|g|(r), \phi)$ is a constant $C$ whenever $r \ge s$. Now, since the functions $|f|(r), |g|(r)$ are continuous strictly increasing functions of $r$, tending to $+\infty$ when $r$ tends to $+\infty$ (respectively when $r$ tends to $R$), then by Lemma 54.7, $\log(|f|(r)) - \log(|g|(r))$ is bounded in $[b, +\infty[$ (respectively in $[b, R[$) hence so is $T(r, f) - T(r, g)$. $\square$

**Lemma 54.9.** Let $h \in \mathcal{A}(\mathbb{K}) \setminus \mathbb{K}$ and let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}(d(a, R^-)$). Let $\alpha$ be a pole of $f$ in $\mathbb{K}$ (respectively in $d(a, r^-)$). Then $\alpha$ is a singular point for $h \circ f$.
Let $h \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}(x)$ and let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}(d(a, R^{-}))$). Let $\alpha$ be a pole of $f$ in $\mathbb{K}$ (respectively in $d(a, r^{-})$). Then $\alpha$ is a point of high singularity for $h \circ f$.

**Proof.** The first statement is immediate. Now suppose that $h$ is a meromorphic function and is not a rational function. Let us show the second statement. Let $q$ be the order of the pole $\alpha$. Without loss of generality, we can assume that $\alpha = 0$. Putting $u = \frac{1}{x}$ and $g(u) = f(x)$, we can see that $g(u)$ is equal to a Laurent series converging in a set $\mathbb{K}\backslash d(0, S)$. Moreover, by taking $S$ big enough, we have $\nu(g, \mu) = q \ \forall \mu \ge \log(S)$, while $g$ has no zero and no pole inside $\mathbb{K}\backslash d(0, S)$. Consequently, $g(u)$ is of the form $\sum_{n=-\infty}^{q} b_n u^n$ and $|g(u)| = |b_q||u_q|^q$, $g(C(0, r)) = C(0, |b_q|r^q)$.

Suppose that $h$ has a sequence of zeros (respectively poles) $(a_n)_{n \in \mathbb{N}}$ such that $|a_n| > |b_q|S^q$, $\lim_{n \to +\infty} |a_n| = +\infty$. There exists a sequence $(c_n)_{n \in \mathbb{N}}$ such that $|b_q||c_n|^q = |a_n|$ and $g(c_n) = a_n$. Now, setting $u_n = \frac{1}{c_n}$, the sequence $(u_n)$ is a sequence of zeros (respectively poles) of $h \circ f$ that converges to 0, which shows that 0 is neither a regular point nor a pole for $h \circ f$. Consequently, if $h$ is not a rational function, $h \circ f$ admits 0 as a point of high singularity. $\square$

We can now state Proposition 54.10.

**Proposition 54.10.** Let $h \in \mathcal{M}(\mathbb{K})$ and let $f, g \in \mathcal{M}(\mathbb{K})$ (respectively let $f, g \in \mathcal{M}(d(a, R^{-}))$). Let $(a_m)_{m \in \mathbb{N}}$ be a bounded sequence of $\mathbb{K}\backslash(S(f, h) \cup S(g, h))$ (respectively a sequence in $d(a, R^{-})\backslash(S(f, h) \cup S(g, h))$ such that $\sup_{m \in \mathbb{N}} |a_m - a| < R$) admitting no cluster point in $\mathcal{P}(f) \cup \mathcal{P}(g)$, satisfying further $h \circ f(a_m) = h \circ g(a_m) \ \forall m \in \mathbb{N}$. Then, $h \circ f(x) = h \circ g(x) \ \forall x \in \mathbb{K}\backslash(S(f, h) \cup S(g, h))$ (respectively $\forall x \in d(a, R^{-})\backslash(S(f, h) \cup S(g, h)))$.

**Proof.** We assume $a=0$. Suppose that $h \circ f$ and $h \circ g$ are two distinct functions. Let $s = \sup |a_m|$. Let $t > s$ (respectively $t \in ]s, R[$) be fixed.

We now prove that infinitely many terms of the sequence $(a_m)$ belong to $d(0, t)$. Suppose it is not true.

Let us first show that the sequence $(a_m)_{m \in \mathbb{N}}$ does not admit any point of $S(f, h) \cup S(g, h)$ as a cluster point. By hypothesis, it may not admit a cluster point in $\mathcal{P}(f) \cup \mathcal{P}(g)$. We show that it does not admit any $b \in \mathcal{T}(f, h)$ as a cluster point, either. Indeed, suppose that
a subsequence of the sequence $(a_m)_{m \in \mathbb{N}}$ converges to $b \in \mathcal{T}(f, h)$. So, $f(b)$ is a pole of order $q'$ of $h$, and hence, $h(f(x))$ is equal to a Laurent series of the form $\sum_{n=q'}^{\infty} \lambda'_n(x-b)^n$ in a disk of center $b$. In the same way, $h(g(x))$ is equal to a Laurent series of the form $\sum_{n=q''}^{\infty} \lambda''_n(x-b)^n$ in a disk of center $b$, therefore $h(g(x)) - h(f(x))$ is a Laurent series of the form $\sum_{n=q}^{\infty} \lambda_n(x-b)^n$ in a disk $d(b, u^-)$ and therefore belongs to $\mathcal{M}(d(b, u^-))$. But since $b$ is a cluster point of the sequence $(a_m)$, the meromorphic function $h(g(x)) - h(f(x))$ admits $b$ as a cluster point of zeros, a contradiction. Consequently, the sequence $(a_m)$ does not admit $b$ as a cluster point of $S(f, h)$. Similarly, it does not admit $b$ as a cluster point of $S(g, h)$. Thus, the sequence $(a_m)$ has no cluster point in $S(f, h) \cup S(g, h)$.

Now, since the sequence $(a_m)$ is bounded, by Corollary 2.19, we can extract from that sequence a subsequence thinner than a circular filter $\mathcal{F}$. Since $\mathcal{F}$ does not converge to a point of $S(f, h) \cup S(g, h)$, either the filter converges to a point of $d(0, t) \setminus S(f, h) \cup S(g, h)$ or it is a large circular filter secant with $d(0, t) \setminus S(f, h) \cup S(g, h)$.

If $\mathcal{F}$ converges to a point $\alpha \in d(0, t) \setminus S(f, h) \cup S(g, h)$, the function $h \circ f - h \circ g$ has a sequence of zeros converging at $\alpha$, hence it is identically zero in $d(0, \ell)$. Consider the set $E(r, t) := \mathcal{D}(t, r, S(f, h)) \cup S(g, h)$ for some $r \in ]0, \ell[$. By Lemma 54.3, $E(r, t)$ has finitely many holes, hence, by Corollary 21.4, it is an analytic subset of $\mathbb{K}$. But by Lemma 36.1, $h \circ f - h \circ g$ belongs to $H(E(r, t))$. Consequently, $h \circ f - h \circ g$ is identically zero in $E(r, t)$. Since we can choose $r$ arbitrary small and $t$ arbitrary big (with $t < R$), we can conclude that $h \circ f - h \circ g$ is identically zero.

Suppose now that $\mathcal{F}$ is a large circular filter secant with $d(0, t) \setminus S(f, h) \cup S(g, h)$. Then, when $r$ is small enough, $\mathcal{F}$ is secant with $\mathcal{D}(t, r, S(f, h)) \cup S(g, h)$ which has finitely many holes and hence is an analytic set, hence $h \circ f - h \circ g$ is identically zero in $\mathcal{D}(t, r, S(f, h)) \cup S(g, h)$ again so we are led to the same conclusion. $\square$

**Proposition 54.11.** Let $h \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$ and let $f, g \in \mathcal{M}(\mathbb{K})$ (respectively $f, g \in \mathcal{M}(d(a, R^-))$) satisfy $h \circ f(x) = h \circ g(x) \ \forall x \in \mathbb{K} \setminus (S(f, h) \cup S(g, h))$ (respectively $h \circ f(x) = h \circ g(x) \ \forall x \in d(a, R^-) \setminus (S(f, h) \cup S(g, h))$). Moreover, if $f$ or $g$ does not belong to $\mathcal{A}(\mathbb{K})$ (respectively if $f$ or $g$ does not belong to $\mathcal{A}(d(a, R^-))$), we assume that $h \notin \mathbb{K}(x) \setminus \mathbb{K}[x]$. Then $f, g$ satisfy $\mathcal{P}(f) = \mathcal{P}(g)$, $S(f, h) = S(g, h)$.
**Proof.** First we show that $\mathcal{P}(f) = \mathcal{P}(g)$. Suppose $\alpha$ is a pole of $f$. For this, we obviously suppose that $f \notin \mathcal{A}(\mathbb{K})$ (respectively $f \notin \mathcal{A}(d(a, R^-))$). Consequently, by hypothesis, $h \notin \mathbb{K}(x)\mathbb{K}[x]$. If $h \in \mathcal{A}(\mathbb{K})$ (respectively if $h \in \mathcal{A}(d(a, R^-))$), then by Lemma 54.9, $\alpha$ is a singular point for $h \circ f$, hence for $h \circ g$, therefore $\alpha$ is a singular point for $g$ and hence it is a pole for $g$. Now suppose that $h \notin \mathcal{A}(\mathbb{K})$, hence $h \in \mathcal{M}(\mathbb{K})\mathbb{K}(x)$ (respectively $h \notin \mathcal{A}(d(a, R^-))$), hence $h \in \mathcal{M}(d(a, R^-))\mathbb{K}(x)$. Since $h$ does not lie in $\mathbb{K}(x)$, by Lemma 54.9, $\alpha$ is a point of high singularity for $h \circ f$, hence for $h \circ g$. But if $\alpha$ is not a pole for $g$, it is a regular point for $g$, hence it is either a regular point or a pole for $h \circ g$, a contradiction. Consequently, $\alpha$ is a pole for $g$, and therefore, since $f$ and $g$ play the same role, we have $\mathcal{P}(f) = \mathcal{P}(g)$.

Now, suppose $\alpha \in \mathcal{T}(f, h)$, hence $h \circ f$ has a pole at $\alpha$ and so does $h \circ g$, thereby $\alpha \in \mathcal{T}(g, h)$. Consequently, $\mathcal{T}(f, h) \subset \mathcal{T}(g, h)$ hence $\mathcal{T}(f, h) = \mathcal{T}(g, h)$ and therefore $\mathcal{S}(f, h) = \mathcal{S}(g, h)$, which completes the proof. $\square$

**Remark.** In order to avoid the restriction: if $f, g$ do not belong to $\mathcal{A}(\mathbb{K})$ (respectively if $f, g$ do not belong to $\mathcal{A}(d(a, r^-))$), we assume that $h \notin \mathbb{K}(x)\mathbb{K}[x]$, we would like to show $\mathcal{P}(f) = \mathcal{P}(g)$ when $h$ is a rational function. But it is hopeless as shows the following situation. Suppose $h \in \mathbb{K}(x)$ is not a function of uniqueness for meromorphic functions and let $f, g \in \mathcal{M}(\mathbb{K})$ satisfy $h \circ f = h \circ g$ and $f(c) \neq g(c)$ for some $c \in \mathbb{K}$ (respectively $c \in d(a, r^-)$). If $c$ is a pole of $f$, this just shows $\mathcal{P}(f) \neq \mathcal{P}(g)$. Suppose $c$ is not a pole for $f$ and $g$. Let $b = f(c)$, let $\phi(x) = \frac{1}{f(x)-b}$, let $\theta(x) = \frac{1}{g(x)-b}$, and let $G(u) = h(b + \frac{1}{u})$. Then $G$ belongs to $\mathbb{K}(x)$ and we can check that $G \circ \theta = G \circ \phi$ and that $c$ lies in $\mathcal{P}(\phi)$ but not in $\mathcal{P}(\theta)$.
# Chapter 55

## Functions of Uniqueness

Throughout this chapter, $\mathbb{K}$ is of characteristic 0.

We want to study sufficient conditions on a meromorphic function $h$ assuring that if the composition of meromorphic functions of the form $h \circ f$ and $h \circ g$ are equal, then $f$ and $g$ are equal. This kind of problem follows many other problems of uniqueness studied, particularly on unique range sets with (or without) multiplicities and polynomials of uniqueness for analytic or meromorphic functions in the complex field and in an ultrametric field [3], [47], [93], [98]. Polynomials of uniqueness were introduced and studied in $\mathbb{C}$ and in a $p$-adic field by X.H Hua and C.C. Yang [98], H. Fujimoto [60], and P. Li and C.C. Yang [97]. Here most results come from [47].

Throughout this chapter, $\mathbb{E}$ is an algebraically closed field of characteristic 0 without any assumption on the existence of an absolute value. The field $\mathbb{K}$ is supposed to have characteristic zero.

In Chapter 41, we have showed that certain polynomials whose derivative has two distinct zeros are polynomials of uniqueness for meromorphic functions or for analytic functions in $\mathbb{K}$ or inside a disk. Here we look for a more systematic process in order to obtain functions of uniqueness.

Previous results obtained in $\mathbb{C}$ by H. Fujimoto in $\mathbb{C}$ and in $\mathbb{K}$ by T.T.H. An and H.H. Khoai and next by T.T.H. An, J. Wang and P.-M. Wong concerned polynomials. Here we consider a meromorphic function $h$ instead of a polynomial $P$ and we only assume that a few zeros $c_1, \dots, c_k$ of $h'$ satisfy $h(c_j) \neq h(d)$ for every other zero $d$ of $h'$. We then examine the situation in four cases in $\mathbb{K}$: $f, g$ entire or