# SMALL ENTIRE FUNCTIONS IN 1 C p

Alain Alain Escassut

► To cite this version:

Alain Alain Escassut. SMALL ENTIRE FUNCTIONS IN 1 C p. p-Adic Numbers, Ultrametric Analysis and Applications, 2025, Vol 17, 17 (3), pp.278-292. 10.1134/S2070046625030045. hal-05232990

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Submitted on 1 Sep 2025

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# SMALL ENTIRE FUNCTIONS IN $C_p$

ALAIN ESCASSUT

ABSTRACT. Small functions were defined in complex analysis and next in ultrametric analysis. Order of growth and type of growth were also defined in complex analysis and have a similar definition in ultrametric analysis. Here we compare these two notions in the same way, on a complete ultrametric algebraically closed field $\mathbb{K}$ of characteristic 0 such as $C_p$. The set of small functions with respect to an entire function $f$ is a ring. The set of entire functions with an order of growth strictly inferior to a number $t$ is also a ring which is included in the previous one when $f$ is regular, but not always when $f$ is not. If an entire function $h$ is small with respect to an entire function $f$, that does not imply that its order of growth is inferior to that of $f$. The cotype was defined in previous papers and is equal to the product of the order by the type for a clean function. But that is not always true for a function that is not clean. If $f$ is clean and has the same order as $h$ and if the type or the cotype of $h$ is zero, then $h$ is a small function with respect to $f$.

If $f$ is clean and the cotype of $f$ is strictly superior to the cotype of $h$, while the type of $f$ is less than the type of $h$, then $h$ is a small function with respect to $f$; similarly, if $f$ is clean and the type of $h$ is strictly superior to the type of $f$, while the cotype of $h$ is less than the cotype of $f$, then $h$ is a small function with respect to $f$. Applications are obtained to a pair of entire functions sharing 3 functions (ignoring multiplicity).

## 1. Introduction and general properties

In complex analysis, a notion of small functions with respect to a "big" holomorphic function was introduced and particularly applied to the Nevanlinna theory [2], [5]. On the other hand, a notion of order of growth was also examined and led to another kind of small functions. These studies suggest symmetric studies in ultrametric analysis: in the same way, we can define small functions with respect to a big function, and order of growth of an entire function [1], [3], [4], [7], [6].

Here we mean to compare the two notions for entire functions, on an ultrametric field such as $C_p$.

**Notations and definitions.** We denote by $\mathbb{K}$ a complete algebraically closed field of characteristic 0 such as $C_p$, by $\mathcal{A}(\mathbb{K})$ the $\mathbb{K}$-algebra of entire functions on $\mathbb{K}$ and by $\mathcal{A}^*(\mathbb{K})$ the set of $f \in \mathcal{A}(\mathbb{K})$ that are not identically zero. Given $a \in \mathbb{K}$ and $r > 0$, we denote by $d(a, r)$ the disk $\{x \in \mathbb{K} : |x - a| \le r\}$ and by $d(a, r^{-})$ the disk $\{x \in \mathbb{K} : |x - a| < r\}$. Given $r_1, r_2 \in \mathbb{R}$ such that $0 < r_1 < r_2$, we

<footnote n="0">⁰2000 Mathematics Subject Classification: 12J25; 30D35; 30G06.</footnote>

<footnote n="0">⁰Keywords: p-adic entire functions, small functions, order, type, cotype of growth</footnote>
denote by $\Gamma(0, r_1, r_2)$ the annulus $\{x \in \mathbb{K} : r_1 < |x| < r_2\}$. We denote by Log the Naperian logarithm.

Let $f \in \mathcal{A}(\mathbb{K})$. According to classical notations [3], we put $|f|(r) = \sup\{|f(x)| \mid |x| \le r\}$ [2]. Then $|\cdot|$ is an ultrametric multiplicative norm on $\mathcal{A}(\mathbb{K})$, and $|f|(r) = \sup\{|f(x)| \mid |x| \le r\} = \lim_{|x| \to r} |f(x)|$ [3].

Next, given $f \in \mathcal{A}(\mathbb{K})$, as in [6], [8], we define $\rho(f) = \limsup_{r \to +\infty} \frac{\mathrm{Log}(\mathrm{Log}(|f|(r)))}{\mathrm{Log}(r)}$,

$$ \tilde{\rho}(f) = \liminf_{r \to +\infty} \frac{\mathrm{Log}(\mathrm{Log}(|f|r)))}{\mathrm{Log}(r)}. $$

We can now state some results on the function $\rho$. Theorem (1.1) is easily proved:

**Theorem (1.1):** Let $f, g \in \mathcal{A}(\mathbb{K})$. Then

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

ii) $\rho(fg) = \max(\rho(f), \rho(g))$.

**Corollary (1.2):** Let $f \in \mathcal{A}(\mathbb{K})$. Then $\forall n \in \mathbb{N}^*, \rho(f^n) = \rho(f)$. Let $P \in \mathbb{K}[x]$. Then $\rho(P \circ f) = \rho(f)$.

**Definitions and notations:** Given $f \in \mathcal{A}(\mathbb{K})$ and $r > 0$, we put $\sigma(f, r) = \frac{\mathrm{Log}(|f|(r))}{r^{\rho(f)}}$, $\sigma(f) = \limsup_{r \to +\infty} \sigma(f, r)$ and $\tilde{\sigma}(f) = \liminf_{r \to +\infty} \sigma(f, r)$,

$\rho(f)$ is called *the order of growth*, $\tilde{\rho}(f)$ is called *the lower order of growth*, $\sigma(f)$ is called *the type of growth*, $\tilde{\sigma}(f)$ is called *the lower type of growth*.

Given $t > 0$, we denote by $\hat{\mathcal{A}}_t(\mathbb{K})$ the set of functions $h \in \mathcal{A}(\mathbb{K})$ such that $\rho(h) < t$.

**Corollary (1.3):** For any $t > 0$, $\hat{\mathcal{A}}_t(\mathbb{K})$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}(\mathbb{K})$.

**Definitions and notations:** A function $f \in \mathcal{A}(\mathbb{K})$ is said to be *regular* if $\tilde{\rho}(f) = \rho(f)$ and it is said to be *clean* if $\tilde{\sigma}(f) = \sigma(f)$.

Given $f, h \in \mathcal{A}(\mathbb{K})$, $h$ is called a small function with respect to $f$ if

$$ \lim_{r \to +\infty} \frac{\mathrm{Log}(|h|(r))}{\mathrm{Log}(|f|(r))} = 0 \quad \text{and we denote by } \mathcal{A}_f(\mathbb{K}) \text{ the set of functions } h \in \mathcal{A}(\mathbb{K}) \\ \text{that are small functions with respect to } f. $$

Given $f, h \in \mathcal{A}(\mathbb{K})$, $h$ is called a very small function with respect to $f$ if $\rho(h) < \tilde{\rho}(f)$.

**Theorem (1.4):** Let $f(x) \in \mathcal{A}(\mathbb{K})$ be such that $\tilde{\sigma}(f) > 0$. Then $f$ is regular. Moreover, if $f$ is not a constant and if $\sigma(f) < +\infty$, then $\rho(f) > 0$.

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

$$ \frac{\text{Log}(\text{Log}(|f|(r)))}{\text{Log}(r)} \geq \frac{\text{Log}(a) - \text{Log}(2)}{\text{Log}(r)} + \rho(f) \quad \forall r \geq R. $$

Thus, $\tilde{\rho}(f) \geq \rho(f)$, hence $f$ is regular. Suppose now that $\rho(f) = 0$ and that $\sigma(f) < +\infty$. Then $|f|(r)$ is bounded, hence $f$ is a constant. $\square$

**Corollary (1.5):** Let $f(x) \in \mathcal{A}(\mathbb{K})$ be clean and such that $\sigma(f) > 0$. Then $f$ is regular.

**Theorem (1.6):** Let $f(x) \in \mathcal{A}(\mathbb{K})$. Then $\mathcal{A}_f(\mathbb{K})$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}(\mathbb{K})$.

*Proof.* Suppose first $f \in \mathcal{A}(\mathbb{K})$ and let $g, h \in \mathcal{A}_f(\mathbb{K})$. Consider

$\text{Log}(|g+h|(r)) \leq \text{Log}(|g|(r) + |h|(r)) \leq \text{Log}(|g|(r)) + \text{Log}(|h|(r))$. But by hypotheses,

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

hence $g+h \in \mathcal{A}_f(\mathbb{K})$.

Next, $\text{Log}|gh|(r) \leq \text{Log}(|g|(r)|h|(r)) = \text{Log}(|g|(r)) + \text{Log}(|h|(r))$ hence

$$ \lim_{r \to +\infty} \frac{\text{Log}(|gh|(r))}{\text{Log}(|f|(r))} = 0 $$

which sows that $gh \in \mathcal{A}_f(\mathbb{K})$. Consequently, $\mathcal{A}_f(\mathbb{K})$ is a subalgebra of $\mathcal{A}(\mathbb{K})$. $\square$

**Theorem (1.7):** Let $f \in \mathcal{A}(\mathbb{K})$ be such that $\tilde{\rho}(f) > 0$. Then the $\mathbb{K}$-algebra $\widehat{\mathcal{A}_{\tilde{\rho}(f)}}(\mathbb{K})$ is included in $\mathcal{A}_f(\mathbb{K})$.

*Proof.* Let $g \in \widehat{\mathcal{A}_{\tilde{\rho}(f)}}(\mathbb{K})$. There exists $b \in ]0, \tilde{\rho}(f)[$ and $R > 0$ such that

$$ \frac{\text{Log}(\text{Log}(|g|(r)))}{\text{Log}(r)} \leq \tilde{\rho}(f) - b \quad \forall r \geq R. $$

Consequently when $r \geq R$, we have

$$ \frac{\text{Log}(\text{Log}(|g|(r)))}{\text{Log}(r)} \leq \frac{\text{Log}(\text{Log}(|f|(r)))}{\text{Log}(r)} - b $$

hence $\text{Log}(\text{Log}(|g|(r))) \leq \text{Log}(\text{Log}(|f|(r))) - b\text{Log}(r)$ which leads to

$$ \frac{\text{Log}(|g|(r))}{\text{Log}(|f|(r))} \leq r^{-b}. $$

This proves that $g \in \mathcal{A}_f(\mathbb{K})$. $\square$

**Corollary (1.8):** Let $f(x) \in \mathcal{A}(\mathbb{K})$ be regular. Then the $\mathbb{K}$-algebra $\widehat{\mathcal{A}_{\rho(f)}}(\mathbb{K})$ is included in $\mathcal{A}_f(\mathbb{K})$.

**Corollary (1.9):** Let $f(x) \in \mathcal{A}(\mathbb{K})$ be such that $\tilde{\sigma}(f) > 0$. Then $\widehat{\mathcal{A}_{\rho(f)}}(\mathbb{K})$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}_f(\mathbb{K})$.
**Corollary (1.10):** Let $f(x) \in \mathcal{A}(\mathbb{K})$ be clean and such that $\sigma(f) > 0$. Then $\widehat{\mathcal{A}_{\rho(f)}}(\mathbb{K})$ is a $\mathbb{K}$-subalgebra of $\mathcal{A}_f(\mathbb{K})$.

**Remark 1:** In other words, very small functions with respect to $f$ are small functions with respect to $f$. As shown in Remark 2 below, the converse does not hold in the general case.

**Remark 2:** Let $f \in \mathcal{A}(\mathbb{K})$ and $g \in \mathcal{A}_f(\mathbb{K})$. Can we tell that $g \in \mathcal{A}_{\rho(f)}(\mathbb{K})$? Suppose that $\rho(f) = 1$ and consider a function $h \in \mathcal{A}(\mathbb{K})$ such that $\text{Log}(|h|(r))$ is of the form $\omega(r)\text{Log}(|f|(r))$ with $\omega(r) \in [\frac{a}{\text{Log}(\text{Log}(|f|(r)))}, \frac{b}{\text{Log}(\text{Log}(|f|(,r)))}]$ and $0 < a < b$. Then

$$ \frac{\text{Log}(\text{Log}(|h|(r)))}{\text{Log}(\text{Log}(|f|(r)))} = \frac{\text{Log}(\omega(r)) + \text{Log}(\text{Log}(|f|(r)))}{\text{Log}(\text{Log}(|f|(r)))} = \frac{\text{Log}(\omega(r))}{\text{Log}(\text{Log}(|f|(r)))} + 1 $$

and that belongs to

$$ \left[ \frac{\text{Log}(a) - \text{Log}(\text{Log}(\text{Log}(|f|(r)))))}{\text{Log}(\text{Log}(|f|(r)))} + 1, \frac{\text{Log}(b) - \text{Log}(\text{Log}(\text{Log}(|f|(r)))))}{\text{Log}(\text{Log}(|f|(r)))} + 1 \right], $$

therefore

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

and hence $\rho(h) = \rho(f)$.

On the other hand, by definition,

$$ \frac{\text{Log}(|h|(r))}{\text{Log}(|f|(r))} = \omega(r) \in \left[ \frac{a}{\text{Log}(\text{Log}(|f|(r)))}, \frac{b}{\text{Log}(\text{Log}(|f|(r)))} \right], $$

$$ \text{hence } \lim_{r \to +\infty} \frac{\text{Log}(|h|(r))}{\text{Log}(|f|(r))} = 0 \text{ and hence } h \text{ belongs to } \widehat{\mathcal{A}}_f(\mathbb{K}). $$

The same reasoning works in $\mathbb{C}$.

**Remark 3:** When $f$ is not regular, a function $h \in \widehat{\mathcal{A}_{\rho(f)}}(\mathbb{K})$ does not always lie in $\mathcal{A}_f(\mathbb{K})$. Indeed, suppose that $\rho(f) - \tilde{\rho}(f) = 2b$, with $b > 0$. We can construct a function $h \in \mathcal{A}(\mathbb{K})$ such that $\rho(h) \le \rho(f) - b$ and such that there exists a sequence $(r_n)_{n \in \mathbb{N}}$ where $\lim_{n \to +\infty} r_n = +\infty$, satisfying

$$ \frac{\text{Log}(\text{Log}(|f|(r_n)))}{\text{Log}(r_n)} \le \frac{\text{Log}(\text{Log}(|h|(r_n)))}{\text{Log}(r_n)} \le \frac{\text{Log}(\text{Log}(|f|(r_n)))}{\text{Log}(r_n)} - b $$

So, we have

$$ \text{Log}(\text{Log}(|f|(r_n))) \le \text{Log}(\text{Log}(|h|(r_n))) \le \text{Log}(\text{Log}(|f|(r_n))) - b\text{Log}(r_n). $$

Thus $\text{Log}(|f|(r_n)) \le \text{Log}(|h|(r_n)$ and hence $1 \le \frac{\text{Log}(|h|(r_n))}{\text{Log}(|f|(r_n))}$ which proves that $h \notin \mathcal{A}_f(\mathbb{K})$, although $h \in \widehat{\mathcal{A}_{\rho(f)}}(\mathbb{K})$.

**Corollary (1.11):** Consider the differential equation

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

Let us also remark the following theorem whose proof is immediate:

**Theorem (1.12):** Let $f \in \mathcal{A}(\mathbb{K})$ be clean and such that $\sigma(f) > 0$ and let $h \in \mathcal{A}(\mathbb{K})$ such that $\rho(h) = \rho(f) > 0$. Then, $h \in \mathcal{A}_f(\mathbb{K})$ if and only if $\sigma(h) = 0$.

*Proof.* Let $t = \rho(f)$. Since $f$ is clean, we have

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

□

## 2. Properties of the growth

We can now examine the deep role of the order of growth. For convenience, we need to define the counting function of zeros of an entire function: let $f \in \mathcal{A}(\mathbb{K})$. We denote by $Z(r, f)$ the counting function of zeros of $f$ in $d(0, r)$ defined in the following way.

**Notations:** Let $s_0(f)$ be the order of multiplicity of 0 if it is a zero of $f$ and let $s_0(f) = 0$ else. Let $(a_n)$, $(1 \le n \le q(r))$ be the finite sequence of zeros of $f$ such that $0 < |a_n| \le r$, of respective order $s_n$.

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

In order to define the counting function of zeros of $f$ ignoring multiplicity, we put $\overline{\omega}_0(f) = 0$ if $\omega_0(f) = 0$ and $\overline{\omega}_0(f) = 1$ if $\omega_0(f) \ge 1$. Now, we denote by $\bar{Z}(f, r)$ the counting function of zeros of $f$ ignoring multiplicity:

$$ Z(r, f) = \max(\overline{\omega}_0(f), 0)\text{Log}r + \sum_{n=1}^{q(r)} (\text{Log}r - \text{Log}|a_n|) $$

Let $f \in \mathcal{A}(\mathbb{K})$. and let $d(0, r)$ be the disk of $\mathbb{K}$ {$x \in \mathbb{K} \mid |x| \le r$}. We denote by $s(r, f)$ the number of zeros of $f$ in $d(0, r)$ each counted with its multiplicity.

Assuming that $0 < \rho(f) < +\infty$, here we put $\psi(f, r) = \frac{s(r, f)}{r^{\rho(f)}}$, $\psi(f) = \limsup_{r \to +\infty} \frac{s(r, f)}{r^{\rho(f)}}$,

which is called the *cotype of $f$* and we put $\tilde{\psi}(f) = \liminf_{r \to +\infty} \frac{s(r, f)}{r^{\rho(f)}}$.

Given 3 functions $f, g, h \in \mathcal{A}(\mathbb{K})$, $f$ and $g$ are said to share $h$, ignoring multiplicity if $f(x) = h(x)$ is equivalent to $g(x) = h(x)$.

**Theorem (2.1)** Let $f \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$. Then $\sigma(f) < +\infty$ if and only if $\psi(f) < +\infty$. Suppose that these hypotheses are satisfied. Then

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

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

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

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

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

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

hence

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

because $\mathrm{Log}(r) - \mathrm{Log}(c_n) \leq \alpha \ \forall n \in [k(re^{-\alpha}), k(r)] \cap \mathbb{N}$. Consequently,

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

Therefore,

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

hence

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

Thus we have

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

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

Now by (1), we have

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

hence passing to inferior limits on both sides, we deduce

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

Thus

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

and hence, when $\alpha u = 1$, we obtain

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

We will now show the inequality

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

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

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

Thus,

$$ (4) \qquad \sigma(f, r) \ge \alpha e^{-u\alpha} \sum_{n=0}^{k(re^{-\alpha})} \frac{w_n}{(re^{-\alpha})^u} + e^{-u\alpha} \sum_{j=0}^{k(re^{-\alpha})} \frac{w_n(\text{Log}(re^{-\alpha}) - \text{Log}(c_n))}{(re^{-\alpha})^u} $$

and hence

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

Therefore, we can deduce

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

and therefore

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

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

$$ (7) \qquad \psi(f) \le \rho(f)(e\sigma(f) - \tilde{\sigma}(f)) $$
which is the left hand inequality of the general conclusion.

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

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

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

therefore

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

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

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

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

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

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

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

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

$$\rho(f)\sigma(f) \leq \psi(f) \leq \rho(f)\tilde{\sigma}(f)$$

therefore, $\sigma(f) = \tilde{\sigma}(f)$, since $\rho(f) > 0$.

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

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

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

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

When a function is not clean, we can ask whether the relation $\psi(f) = \rho(f)\sigma(f)$ holds anytime.

Now, similarly as Theorem 1.12, we can prove Theorem 2.2:
**Theorem (2.2):** Let $f \in \mathcal{A}(\mathbb{K})$ be clean and let $h \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(h) = \rho(f) < +\infty$. If $\psi(f) > \psi(h) = 0$, then $h \in \mathcal{A}_f(\mathbb{K})$. Moreover, if $h$ is clean, then $h \in \mathcal{A}_f(\mathbb{K})$ if and only if $\psi(f) > \psi(h) = 0$.

*Proof.* Since $\rho(h) > 0$, if $\psi(h) = 0$, by Theorem 2.1, we have $\sigma(h) = 0$ and hence by Theorem 1.12, $h \in \mathcal{A}_f(\mathbb{K})$. Now, suppose that $h$ is clean and suppose that $\psi(h) > 0$. Since $h$ is clean we have $\sigma(h) > 0$, hence by Theorem 1.12 $h$ does not belong to $\mathcal{A}_f(\mathbb{K})$. $\square$

We can now examine the question asked after Theorem (2.1). We will need the following basic lemmas:

**Lemma (2.3):** Let $f_1, f_2$ be two functions from $\mathbb{R}_+$ to $\mathbb{R}_+$ such that

$$ \lim_{x \to +\infty} f_1(x) = \lim_{x \to +\infty} f_2(x) = +\infty $$

and

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

Then

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

**Lemma (2.4):** Let $\alpha, \beta \in \mathbb{R}_+$ and let $g(x) = e^{-x}(\alpha x - \beta)$. Then $g'$ has a unique zero at $1 + \frac{\beta}{\alpha}$ and $g(1 + \frac{\beta}{\alpha}) = \alpha e^{-(1 + \frac{\beta}{\alpha})}$. Moreover, $g$ is increasing in $[0, 1 + \frac{\beta}{\alpha}]$ and is decreasing in $[1 + \frac{\beta}{\alpha}, +\infty[$ and tends to 0 when $x$ tends to $+\infty$.

Now let us recall the definition of a divisor in $\mathbb{K}$.

**Definition:** We call divisor in $\mathbb{K}$ a sequence $(a_n, v_n)_{n \in \mathbb{N}}$ where $(a_n)_{n \in \mathbb{N}}$ is a sequence in $\mathbb{K}$ such that $\lim_{n \to +\infty} |a_n| = +\infty$ and each $v_n$ belongs to $\mathbb{N}^*$ [6].

The following Lemma (2.5) is classical (see for example Corollary B.18.5 of [6]).

**Lemma (2.5):** Given a divisor $(a_n, v_n)_{n \in \mathbb{N}}$ in $\mathbb{K}$, there exist functions admitting each $a_n$ as a zero of order $v_n$ and no other zero and two such functions are proportional.

**Theorem (2.6):** Suppose that $\mathbb{K}$ is of characteristic 0. There exist regular non-clean functions $f \in \mathcal{A}(\mathbb{K})$ such that $\psi(f) > \rho(f)\sigma(f)$.

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

In $[0, +\infty[$, we define $g_k(\nu) = e^{-\nu}(\alpha_k\nu - \beta_k)$ up to the rank $m$ and suppose that the function $g_k$ satisfies $1 \le g_k(\nu_{2k}) \le 1 + \frac{1}{4k^2}$ and $1 \le g_k(\nu_{2k+2}) \le 1 + \frac{1}{4(k+1)^2}$ $\forall k = 1, ..., m-1$ and $g_{k-1}(\nu_{2k}) = g_k(\nu_{2k})$.
By Lemma (2.4), $g_k$ is increasing in $[\frac{\beta_k}{\alpha_k}, 1+\frac{\beta_k}{\alpha_k}]$ from 0 to a maximum equal to $\alpha_k e^{1+\frac{\beta_k}{\alpha_k}}$ and is decreasing to 0 when $\nu$ tends to $+\infty$. Hence $g_k$ takes the value 1 at a unique point $\lambda_{2k}$ in $[\frac{\beta_k}{\alpha_k}, 1+\frac{\beta_k}{\alpha_k}]$ and at a unique point $\lambda_{2k+2} \in [1+\frac{\beta_k}{\alpha_k}, +\infty[$. We then have $g_k(\lambda_{2k}) = e^{-\lambda_{2k}}(\alpha_k\lambda_{2k}-\beta_k)$ and $g_k(\lambda_{2k+2}) = e^{-\lambda_{2k+2}}(\alpha_k\lambda_{2k+2}-\beta_m)$. Hence, $\lambda_{2k} = \frac{e^{\lambda_{2k}}+\beta_k}{\alpha_k}$ and $\lambda_{2k+2} = \frac{e^{\lambda_{2k+2}}+\beta_k}{\alpha_k} > \nu_{2k+1}$ and we can take the value $r_{2k+2} \in |\mathbb{K}|$ close enough to $e^{\lambda_{2k+2}}$ such that, putting $r_{2k+2} = e^{\nu_{2k+2}}$, we then have

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

and

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

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

Then by Lemma (2.4), $g_m$ has a maximum at

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

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

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

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

$$(4) \qquad \Theta_m \ge \frac{1}{2} - \frac{1}{8(m)^2} > \frac{15}{32}.$$
We can now define by induction the sequences $(r_m)$, $(\nu_m)$, $(g_m)$, $(y_m)$, $(\Theta_m)$ and then $\lim_{n \to +\infty} r_n = +\infty$. Consequently by (3) and (4),

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

We now obtain

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

and hence, by (2),

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

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

We can now define a function $g$ in $[0, +\infty[$ as $g(\nu) = g_m(\nu)$ when $\nu \in [\text{Log}(r_{2m}), \text{Log}(r_{2m+2})[$. So, by (5) we have

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

Thus, we can check that

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

Now, by Lemma (2.5) we can consider the entire function $f$ admitting $q_m$ zeros on each circle $C(0, r_{2m})$ and no other zero. Let $(a_{j,m})_{(1 \le j \le q_m)}$ be the zeros of $f$ on the circle $C(0, r_{2m})$.

Then, when $2m \le r < r_{2m+2}$, the counting functions of zeros of $f$ (counting multiplicity) is of the form

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

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

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

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

Moreover, by (9) and Lemma (2.3), we have

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

hence $\rho(f) = 1$.

Furthermore, since $\limsup_{r \to +\infty} \frac{\mathrm{Log}(|f|(r))}{r} = \frac{2}{\sqrt{e}}$ and

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

More precisely, by construction, for every $r \in [r_{2m}, r_{2m+2}]$, we have $\psi(f, r) = \frac{2r_{2m}+\eta_m}{r}$ and hence $\psi(f, r_{2m})$ is of the form $2+y_m$ where $(y_m)_{m \in \mathbb{N}}$ is a sequence of limit 0. Therefore $\psi(f) \ge 2$, while $\rho(f) = 1$. This shows that $f$ does not satisfy the relation $\psi(f) = \rho(f)\sigma(f)$ and hence, this is not always satisfied when a function $f$ is not clean. $\square$

### 3. Sharing small entire functions

**Theorem (3.1):** Let $f \in \mathcal{A}(\mathbb{K})$ be clean and such that $0 < \rho(f) < +\infty$ and let $h \in \mathcal{A}(\mathbb{K})$ be such that $\psi(h) < \psi(f)$ and $+\infty > \sigma(h) \ge \sigma(f)$ (resp. $\psi(h) \le \psi(f)$ and $+\infty > \sigma(h) > \sigma(f)$). Then $f$ is regular, $\rho(h) < \rho(f)$ and $h \in \mathcal{A}_f(\mathbb{K})$.

*Proof.* Suppose first that $\psi(h) < \psi(f)$ and $\sigma(h) \ge \sigma(f)$. We notice that $\psi(f) > 0$ hence, since $f$ is clean, by Theorem A we have $\sigma(f) > 0$. Now, we have $\psi(f) = \rho(f)\sigma(f) > \psi(h) \ge \rho(h)\sigma(h) \ge \rho(h)\sigma(f)$. Thus, $\psi(f) > 0$. But since $\sigma(f) > 0$, then $\rho(f) > \rho(h)$. And since $f$ is clean, $\sigma(f) = \tilde{\sigma}(f)$ and hence by Theorem (1.4), $f$ is regular. Finally, since $\rho(h) < \rho(f)$, by Corollary (1.8), we obtain $h \in \mathcal{A}_f(\mathbb{K})$.

Suppose now that $\psi(h) \le \psi(f)$ and $\sigma(h) > \sigma(f)$. Let us first show that $\rho(h) > 0$. If $\rho(h) = 0$, then by Theorem (1.7), $h \in \mathcal{A}_f(\mathbb{K})$. Suppose now that $\rho(h) > 0$. Since $\sigma(h) > 0$, then $\psi(h) > 0$ and hence $\psi(f) > 0$. Therefore $\sigma(f) > 0$ because $f$ is clean. By Theorem (2.1), we have $\psi(f) = \rho(f)\sigma(f) \ge \psi(h) \ge \rho(h)\sigma(h)$. Since $\sigma(f) > 0$, by Theorem (1.4), $f$ is regular. We can obviously assume that $h$ is not a constant and, by hypothesis, $\sigma(h) < +\infty$. Then we have $\psi(f) = \rho(f)\sigma(f) \ge \psi(h) \ge \rho(h)\sigma(h) > \rho(h)\sigma(f)$ and hence $\rho(f) > \rho(h)$; therefore, by Corollary (1.10), $h \in \mathcal{A}_f(\mathbb{K})$. $\square$

In the proof of Theorems (3.3), (3.4), (3.5), (3.6), we will need the counting function of zeros of an entire function.
By [7] and Corollary 4 in [2] which is also Theorem C.9.19 in [3] we can state the following theorem which is an easy consequence of the Nevanlinna Theorem on 3 small functions for entire functions.

**Theorem (3.2):** Let $f \in \mathcal{A}(\mathbb{K})$ and let $w_1, w_2 \in \mathcal{A}_f(\mathbb{K})$ be distinct. Let $S(r) = \max(Z(r, w_1), Z(r, w_2))$. Then $Z(r, f) \le \bar{Z}(r, f - w_1) + \bar{Z}(r, f - w_2) + 7S(r) + O(1)$.

Next, we prove the following theorem:

**Theorem (3.3):** Let $f \in \mathcal{A}(\mathbb{K})$ be transcendental and let $w_j \in \mathcal{A}_f(\mathbb{K})$ ($j = 1, \dots, q$) be $q$ distinct small functions with respect to $f$. Then

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

*Proof.* By Theorem 3.2, for every pair $(i, j)$ such that $1 \le i \le j \le q$, we have

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

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

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

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

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

and hence

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

□

**Theorem (3.4):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental and share $q$ distinct small functions ignoring multiplicity. $w_j \in \mathcal{A}_f(\mathbb{K}) \cap \mathcal{A}_g(\mathbb{K})$ ($j = 1, \dots, q$) other than the constant $\infty$. Then

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

*Proof.* 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 $h \ne i$. 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 $\mathcal{A}_f(\mathbb{K})$. 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 $d(0, r^{-1})$
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 $\mathcal{A}_f(\mathbb{K})$, clearly, we have

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

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,

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

□

**Theorem 3.5:** Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental, be distinct, and share $q$ distinct small functions ignoring multiplicity: $w_j \in \mathcal{A}_f(\mathbb{K}) \cap \mathcal{A}_g(\mathbb{K})$ ($j = 1, ..., q$) Then

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

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

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 $\mathcal{A}_f(\mathbb{K})$. 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 $\mathcal{A}_f(\mathbb{K})$, clearly, we have

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

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

□

Using Theorem (3.3), we can now prove Theorem (3.6) below:

**Theorem (3.6):** Let $f, g \in \mathcal{A}(\mathbb{K})$ share 3 distinct small entire functions with respect to $f$ and $g$, ignoring multiplicity. Then $f = g$.

*Proof.* We put $V(r) = \max(Z(r, f), Z(r, g))$. Suppose that $f$ and $g$ are distinct and share $q$ small functions I.M. $w_j$, ($1 \leq j \leq q$).

By Theorem (3.3), we have

$$ qZ(r, f) \leq 2 \sum_{j=1}^{q} \bar{Z}(r, f - w_j) + o(Z(r, f)). $$

But using Theorem 3.5, we can show that

$$ qZ(r, f) \leq 2Z(r, f - g) + o(Z(r, f)) $$

and similarly

$$ qZ(r, g) \leq 2Z(r, f - g) + o(Z(r, g)). $$

Thus $qV(r) \leq 2Z(r, f - g) + o(V(r))$ and hence $qV(r) \leq 2V(r) + o(V(r))$, which is obviously absurd whenever $q \geq 3$ and proves that $f = g$. □

The following Theorem (3.7) now is just a corollary of Theorem (3.6):

**Theorem (3.7):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be such that $0 < \tilde{\rho}(f) < +\infty$, $0 < \tilde{\rho}(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, for $j = 1, 2, 3$, such that $\rho(h_j) < \min(\tilde{\rho}(f), \tilde{\rho}(g))$. Then $f = g$ and each $h_j$ is a small function with respect to $f$.

And by Theorem (1.4), we have Corollary (3.8):

**Corollary (3.8):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be regular, such that $0 < \rho(f) < +\infty$, $0 < \rho(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, for $j = 1, 2, 3$ and such that $\max(\rho(h_j) < \min(\rho(f), \rho(g)))$. Then $f = g$ and each $h_j$ is a small function with respect to $f$.
**Corollary (3.9):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be regular, such that $0 < \rho(f) < +\infty$, $0 < \rho(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, for $j = 1, 2, 3$ and such that $\max(\rho(h_1) < \min(\rho(f), \rho(g)))$ and that $\text{Log}(\max(|h_2|(r), |h_3|(r))) \le O(\text{Log}(|h_1|(r)))$. Then $f = g$ and each $h_j$ is a small function with respect to $f$.

**Theorem (3.10):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be clean, such that $0 < \rho(f) < +\infty$, $0 < \rho(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, for $j = 1, 2, 3$, such that $\rho(f) = \rho(g) = \rho(h_j)$, $j = 1, 2, 3$ and such that $\sigma(h_j) = 0$, for $j = 1, 2, 3$. Then $f = g$ and each $h_j$ is a small function with respect to $f$.

*Proof.* Indeed by Theorem 1.12, each $h_j$ is a small function with respect to $f$ and $g$. Consequently by Theorem 3.6, $f = g$. □

**Corollary (3.11):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be clean, such that $0 < \rho(f) < +\infty$, $0 < \rho(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, for $j = 1, 2, 3$, such that $\rho(f) = \rho(g) = \rho(h_j)$, $j = 1, 2, 3$ and such that $\psi(h_j) = 0$, for $j = 1, 2, 3$. Then $f = g$ and each $h_j$ is a small function with respect to $f$.

*Proof.* Indeed, since $\psi(h_j) = 0$ and $\rho(h_j) \ne 0$, by Theorem (2.1) we have $\sigma(h_j) = 0$. □

**Corollary (3.12):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be clean, such that $0 < \rho(f) < +\infty$, $0 < \rho(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, for $j = 1, 2, 3$, such that $\rho(f) = \rho(g) = \rho(h_1)$ and such that $\psi(h_1) = 0$ and that
$$ \text{Log}(\max(|h_2|(r), |h_3|(r))) \le O(\text{Log}(|h_1|(r))). $$
Then $f = g$ and each $h_j$ is a small function with respect to $f$.

*Proof.* Indeed, since $\psi(h_1) = 0$ and $\rho(h_1) \ne 0$, by Theorem (2.1) we have $\sigma(h_1) = 0$. Next, $h_2$, $h_3$ also are small functions. □

**Theorem (3.13):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be clean, such that $0 < \rho(f) < +\infty$, $0 < \rho(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, $j = 1, 2, 3$. Suppose that for each $j = 1, 2, 3$, one of the following two hypotheses is satisfied:

i) $\psi(h_j) < \min(\psi(f), \psi(g))$ and $\max(\sigma(f), \sigma(g))) \le \sigma(h_j)$,

ii) $\psi(h_j) \le \min(\psi(f), \psi(g))$ and $\max(\sigma(f), \sigma(g))) < \sigma(h_j)$.

Then $f = g$, and each $h_j$ is a small function with respect to $f$.

Moreover, if i) is satisfied, then f is regular.

*Proof.* Indeed, by Theorem (3.1) each function $h_j$ is a small function with respect to both $f$ and $g$. Consequently, by Theorem (3.6), we have $f = g$. Next, we may notice that if i) is satisfied, then $f$ is regular. Indeed $\psi(f) > 0$, hence so is $\sigma(f)$ because $f$ is clean. Consequently, $f$ is regular by Corollary 1.5. □
**Corollary (3.14):** Let $f, g \in \mathcal{A}(\mathbb{K})$ be clean, such that $0 < \rho(f) < +\infty$, $0 < \rho(g) < +\infty$, and $f, g$ share 3 distinct functions, ignoring multiplicity: $h_j \in \mathcal{A}(\mathbb{K})$, for $j = 1, 2, 3$. Suppose that one of the following two hypotheses is satisfied:
i) $\psi(h_1) < \min(\psi(f), \psi(g))$ and $\max(\sigma(f), \sigma(g)) \le \sigma(h_1)$,
ii) $\psi(h_1) \le \min(\psi(f), \psi(g))$ and $\max(\sigma(f), \sigma(g)) < \sigma(h_1)$,
and that $\mathrm{Log}(\max(|h_2|(r), |h_3|(r))) \le O(\mathrm{Log}(|h_1|(r)))$.
Then $f = g$ and each $h_j$ is a small function with respect to $f$. Moreover, if i) is satisfied, then $f$ is regular.

**Acknowledgement:** I am very grateful to the referee who pointed out several abstraction mistakes.

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UNIVERSITÉ CLERMONT AUVERGNE, UMR CNRS 6620, LMBP, F-63000 CLERMONT-FERRAND, FRANCE.

Email address: alain.escassut@uca.fr