# SHARING COMPLEX SMALL FUNCTIONS

Alain Escassut

► To cite this version:

Alain Escassut. SHARING COMPLEX SMALL FUNCTIONS. Asian-European Journal of Mathematics, In press. hal-05331609

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https://uca.hal.science/hal-05331609v1

Submitted on 25 Oct 2025

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Distributed under a Creative Commons Attribution 4.0 International License
SHARING COMPLEX SMALL FUNCTIONS

ALAIN ESCASSUT

ABSTRACT. Small functions were defined in complex analysis together with the notions of order of growth and type of growth.

Here we consider certain entire functions $f, g$ such that the logarithm of the maximum modulus is almost equivalent to the Nevanlinna characteristic and we construct an example where this equivalence is an equality. We show that if $f$ and $g$ satisfy certain property of equivalence and share 3 small entire functions with respect to $f$ and $g$, ignoring multiplicity, then $f = g$.

*In memory of Professor Jean Fresnel*

**Definition:** Let $\mathcal{A}(\mathbb{C})$ be the $\mathbb{C}$-algebra of complex entire functions. Given $f \in \mathcal{A}(\mathbb{C})$ we denote by $M(f, r)$ the maximum modulus of $f$ on the circle $|z| = r$ and we put $T(r, f) = \frac{1}{2\pi} \int_{0}^{2\pi} \text{Log}^{+}(|f(re^{i\theta})|)d\theta$. We put $\rho(f) = \limsup_{r \to +\infty} \frac{\text{Log}(T(r, f))}{\text{Log}(r)}$ and $\rho(f)$ is called the growth order of $f$. Then $\rho(f)$ is also equal to $\limsup_{r \to +\infty} \frac{\text{Log}(\text{Log}(M(f, r))}{\text{Log}(r)}$ [5].

A function $h \in \mathcal{A}(\mathbb{C})$ is called a small function with respect to $f$ if $\lim_{r \to +\infty} \frac{T(r, h)}{T(r, f)} = 0$.

Two functions $f, g \in \mathcal{A}(\mathbb{C})$ are said to share a small function $h \in \mathcal{A}(\mathbb{C})$ Ignoring Multiplicity (I.M.) (resp. a constant $b$) if $f(z) = h(z)$ (resp. $f(z) = b$) is equivalent to $g(z) = h(z)$ (resp. $g(z) = b$).

The question whether two analytic functions sharing a few small functions are equal is an old problem that received several solutions [4], [6], [8], [9], [10] and particularly in ultrametric analysis [3], [8]. Here we consider this problem applied to entire functions satisfying a specific property of equivalence between $T(r, f)$ and $\text{Log}(M(f, r))$ because such a property, satisfied by entire functions on an ultrametric field, lets find interesting results. These properties are linked to the Nevanlinna theory [5], [7]. Actually our problem is closely linked to upper bounding $\limsup_{r \to +\infty} \frac{M(f, r)}{T(r, f)}$. Here we will consider Weierstrass products without exponential factors, as in [1].

<footnote n="0">⁰2000 Mathematic Subject Classification: 30D35; 30B30, 30D10.</footnote>

<footnote n="0">⁰Keywords: complex entire functions, Nevanlinna theory</footnote>
**Definition:** Let $f \in \mathcal{A}(\mathbb{C})$ and let $\Lambda \in ]0, +\infty[$. Then $f$ will be said to be $\Lambda$-majored if $\limsup_{r \to +\infty} \frac{\mathrm{Log}(M(f, r))}{T(r, f)} < 1 + \Lambda$ and it will be said to be perfectly-majored if $\lim_{r \to +\infty} \frac{\mathrm{Log}(M(f, r))}{T(r, f)} = 1$. Next, an entire function $h \in \mathcal{A}(\mathbb{C})$ will be said to be majored if it is $\Gamma$-majored for some $\Gamma \in ]0, +\infty[$.

Suppose that $f(0) \neq 0$ and let $(a_j, m_j)_{j \in N^*}$ be the sequence of the zeros of $f$, of respective multiplicity $m_j$, with $|a_j| \le |a_{j+1}|$. For every $r > 0$, we put $Z(r, f) = \sum_{j=1}^{\infty} \mathrm{Log}^+(\frac{r^{m_j}}{|a_j|^{m_j}})$, $\bar{Z}(r, f) = \sum_{j=1}^{\infty} \mathrm{Log}^+(\frac{r}{|a_j|})$ and $Y(r, f) = \sum_{j=1}^{\infty} m_j \min(\frac{|a_j|}{r}, \frac{r}{|a_j|})$.

Then $f$ will be said to be $\Lambda$-vanishing if $\limsup_{r \to +\infty} \frac{Y(r, f)}{Z(r, f)} < \Lambda$ and $f$ will be said to be perfectly-vanishing if $\lim_{r \to +\infty} \frac{Y(r, f)}{Z(r, f)} = 0$.

The following theorem is classical [5]:

**Theorem A:** Let $f \in \mathcal{A}(\mathbb{C})$. Then, $Z(r, f) \le T(r, f) + O(1)$ and $T(r, f) \le \mathrm{Log}(M(f, r)) \le 3T(2r, f)$. Moreover, if $g \in \mathcal{A}(\mathbb{C})$, then, $T(r, f + g) \le T(r, f) + T(r, g)$, $\rho(f + g) \le \max(\rho(f), \rho(g))$ and $\rho(f,g) \le \max(\rho(f), \rho(g))$.

As shows Theorem A, an interesting question consists of giving to $\mathrm{Log}(M(f, r))$ an upper bound better than $3T(2r, f)$. This will be one of our topics.

Theorems 1 and 2 are easily proved:

**Theorem 1:** Let $f, g \in \mathcal{A}(\mathbb{C})$ be $\Lambda$-majored. Then, for every $n \in \mathbb{N}$, $f^n$ is $\Lambda$-majored and there exists $\epsilon > 0$ such that

$$T(r, f + g) \le (1 + \Lambda - \epsilon) \max(T(r, f), T(r, g)) + \mathrm{Log}(2).$$

Moreover, if $f$ and $g$ are perfectly-majored, then $f^n$ is perfectly majored and

$$T(r, f + g) \le \max(T(r, f), T(r, g)) + \mathrm{Log}(2).$$

*Proof.* Indeed, we have $T(r, f^n) = nT(r, f)$, $\mathrm{Log}(M(f^n, r)) = n\mathrm{Log}(M(f, r)) \forall n \in \mathbb{N}$, hence $f^n$ is $\Lambda$-majored. Next,

$$T(r, f + g) \le \mathrm{Log}(M(f + g, r)) \le (\mathrm{Log}(M(f, r) + M(g, r)))$$

$$\le \mathrm{Log}(2 \max(M(f, r), M(g, r))) = \mathrm{Log}(2) + \max(\mathrm{Log}(M(f, r)), \mathrm{Log}(M(g, r)))$$

when $r$ is big enough. Consequently, there exists $\epsilon > 0$ such that

$$T(r, f + g) \le (1 + \Lambda - \epsilon) \max(T(r, f), T(r, g)) + \mathrm{Log}(2).$$

The last conclusions follow.

□
**Theorem 2:** Let $f, h \in \mathcal{A}(\mathbb{C})$.

i) If $f$ is $\Lambda$-majored and if $\lim_{r \to +\infty} \frac{\log(M(h, r))}{\log(M(f, r))} = 0$, then $h$ is a small function with respect to $f$.

ii) If $h$ is a $\Lambda$-majored small function with respect to $f$, then $\lim_{r \to +\infty} \frac{\log(M(h, r))}{\log(M(f, r))} = 0$.

*Proof.* i) Indeed, by Theorem A, we have $\lim_{r \to +\infty} \frac{T(r, h)}{\log(M(f, r))} = 0$. But since $f$ is $\Lambda$-majored, we have $\log(M(f, r)) \le (1+\Lambda)T(r, f)$ when $r$ is big enough, hence $\lim_{r \to +\infty} \frac{T(r, h)}{T(r, f)} = 0$.

ii) Suppose $h$ is a $\Lambda$-majored small function with respect to $f$. Then

$$0 = \lim_{r \to +\infty} \frac{T(r, h)}{T(r, f)} \ge \lim_{r \to +\infty} \frac{\log(M(h, r))}{(1 + \Lambda)T(r, f)} \ge \lim_{r \to +\infty} \frac{\log(M(h, r))}{(1 + \Lambda)\log(M(f, r))}$$

when $r$ is big enough, which ends the proof.

<figure><img src="image_2.png" /></figure>

**Example:** Let $f(z) = e^z$. One can check that $M(f, r) = e^r$ and

$$2\pi T(r, f) = \int_{0}^{2\pi} \log^{+}(|e^{r(\cos(\theta) + i \sin(\theta))}|) d\theta = \int_{0}^{2\pi} \log^{+}(e^{r \cos(\theta)}) d\theta \\ = \int_{0}^{\frac{\pi}{2}} \log(e^{r \cos(\theta)}) d\theta + \int_{\frac{3\pi}{2}}^{2\pi} \log(e^{r \cos(\theta)}) d\theta = 2r.$$

Consequently, $f$ is $\Lambda$-majored for every $\Lambda > \pi - 1$.

Here we can notice that $T(r, f) = \frac{r}{\pi}$ and similarly, putting $g(z) = e^{-z}$, $T(r, g) = T(r, f)$. Consequently, by the classical inequality we have $T(r, f+g) \le \frac{2r}{\pi}$ whereas Theorem 1 only gives us $T(r, f+g) \le \pi T(r, f) = r$. Thus, in this case the information given by Theorem 1 is worse than the classical inequality. However, when $\Lambda < 1$, for both $f$ and $g$, we can obtain a better inequality thanks to Theorem 1.

In the same way, if $f_n(z) = e^{\frac{z}{n}}$, or $f_n(z) = e^{z^n}$, $f_n$ is $\Lambda$-majored for every $\Lambda \ge \pi - 1$.

Theorem 3 is immediate:

**Theorem 3:** Let $f \in \mathcal{A}(\mathbb{C})$ be perfectly-majored. Then $\sigma(f) = \limsup_{r \to +\infty} \frac{\log(M(f, r))}{r^{\rho(f)}}$.

**Remark:** The equality of Theorem 3 does not hold in the general case (when $f$ is not perfectly-majored) and this is mentioned in [4] and [5], contrary to a wrong claim in [2] where the equality is written for all functions $f \in \mathcal{A}(\mathbb{C})$.
**Theorem 4:** Let $f \in \mathcal{A}(\mathbb{C})$ be $\Lambda$-majored and let $h \in \mathcal{A}(\mathbb{C})$ be a majored small function with respect to $f$. Then $f+h$ is $\Lambda$-majored.

*Proof.* Since $h$ is a small function with respect to $f$, we have $T(r, h) = o(T(r, f))$, hence $T(r, f+h) = T(r, f) + o(T(r, f))$. Consequently,

$$ \limsup_{r \to +\infty} \frac{\log(M(f+h, r))}{T(r, f+h)} = \limsup_{r \to +\infty} \frac{\log(M(f+h, r))}{T(r, f)} $$

$$ \leq \limsup_{r \to +\infty} \frac{\log(M(f, r))}{T(r, f)} + \limsup_{r \to +\infty} \frac{\log(M(h, r))}{T(r, f)} $$

$$ \leq \limsup_{r \to +\infty} \frac{\log(M(f, r))}{T(r, f)} + \limsup_{r \to +\infty} \left( \frac{\log(M(h, r))}{T(r, h)} \right) \left( \limsup_{r \to +\infty} \left( \frac{T(r, h)}{T(r, f)} \right) \right). $$

But since $h$ is majored, we have

$$ \limsup_{r \to +\infty} \left( \frac{\log(M(h, r))}{T(r, h)} \right) \left( \limsup_{r \to +\infty} \left( \frac{T(r, h)}{T(r, f)} \right) \right) = 0 $$

and hence

$$ \limsup_{r \to +\infty} \frac{\log(M(f+h, r))}{T(r, f+h)} = \limsup_{r \to +\infty} \frac{\log(M(f, r))}{T(r, f)} < 1 + \Lambda. \quad \square $$

Theorem B is easily checked:

**Theorem B:** For every $n \in \mathbb{N}$, we put $f_n(x) = \prod_{j=1}^{n} \left(1 - \frac{x}{a_j}\right)$ with a sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{C}$ such that $\lim_{n \to +\infty} |a_n| = +\infty$ and $\sum_{j=1}^{\infty} \frac{1}{|a_j|} < +\infty$. Then the sequence $f_n$ converges in all disks $d(0, r)$.

Let us also notice this lemma:

**Lemma C:** Given $b > 1$, one has $\log(b+1) \leq \log(b) + \frac{1}{b}$.

**Theorem 5:** Let $f \in \mathcal{A}(\mathbb{C})$ be of the form $\prod_{j=1}^{\infty} \left(1 - \frac{x}{a_j}\right)$ with a sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{C}$ such that $|a_n| \leq |a_{n+1}|$, $\lim_{n \to +\infty} |a_n| = +\infty$ and $\sum_{j=1}^{\infty} \frac{1}{|a_j|} < +\infty$. Then

$$ \log(M(f, r)) \leq Z(r, f) + Y(r, f) \quad \forall r > 0. $$
**Proof.** Given $r > 0$, let $q(r)$ be the integer such that $|a_j| \le r$ whenever $j \le q(r)$ and $|a_j| > r$ whenever $j > q(r)$. Then

$$ \sum_{j=1}^{q(r)} \log\left(\left|\frac{x}{a_j} - 1\right|\right) \le \sum_{j=1}^{q(r)} \log\left(\frac{r}{|a_j|} + 1\right) \le \sum_{j=1}^{q(r)} \log\left(\frac{r}{|a_j|}\right) + \frac{|a_j|}{r} $$

and, by Lemma C,

$$ \sum_{j>q(r)} \log\left(\left|\frac{x}{a_j} - 1\right|\right) \le \sum_{j>q(r)} \log\left(\frac{r}{|a_j|} + 1\right) \le \sum_{j>q(r)} \frac{r}{|a_j|} $$

and hence we can check that

$$ \log(|f(x)|) \le Z(r, f) + Y(r, f) \quad \forall x \in \mathbb{C}, |x| = r $$

i.e.

$$ \log(M(f, r)) \le Z(r, f) + Y(r, f). $$

□

**Theorem 6:** Let $f \in \mathcal{A}(\mathbb{C})$ be of the form $\prod_{j=1}^{\infty} \left(1 - \frac{x}{a_j}\right)$ with a sequence $(a_n)_{n \in \mathbb{N}}$

in $\mathbb{C}$ such that $|a_n| \le |a_{n+1}|$, $\lim_{n \to +\infty} |a_n| = +\infty$ and $\sum_{j=1}^{\infty} \frac{1}{|a_j|} < +\infty$ and let it be $\Lambda$-vanishing. Then, $f$ is $\Lambda$-majored. If $f$ is perfectly-vanishing it is perfectly-majored.

**Proof.** By Theorem A, we have $T(r, f) \le \log(M(f, r))$. Now, by Theorem 5, $\log(M(f, r)) \le Z(r, f) + Y(r, f)$. Consequently, since $f$ is $\Lambda$-vanishing there exists $\epsilon > 0$ such that

$$ \log(M(f, r)) \le (1 + \Lambda - \epsilon)Z(r, f) $$

and hence by Theorem A,

$$ \log(M(f, r)) \le (1 + \Lambda - \epsilon)T(r, f) + O(1), $$

whenever $r$ is big enough hence

$$ \limsup_{r \to +\infty} \frac{\log(M(f, r))}{T(r, f)} \le 1 + \Lambda - \epsilon. $$

The last conclusion follows.

□

**Theorem 7:** Let $k$ be an integer $\ge 2$ and for every $n \in \mathbb{N}$, let $a_n \in \mathbb{C}$ be such that $|a_n| = n^k$. Let $f(z) = \prod_{n=1}^{\infty} (1 - \frac{z}{a_n})$. Then for every $\epsilon > 0$, $f$ is $\frac{2}{k^2} + \epsilon$-vanishing and $\rho(f) \le \frac{1}{k}$.
Proof. We first estimate $Z(r, f)$. Let $r \in [n^k, (n+1)^k[$. We have

$$ 
\begin{aligned}
Z(r, f) &= n\text{Log}(r) - k \sum_{j=1}^{n} \text{Log}j \ge n\text{Log}(r) - \int_{x=1}^{k(n+1)} \text{Log}(x)dx \\
&= n\text{Log}(r) - k((n+1)\text{Log}(n+1) - (n+1)) \ge k(n\text{Log}(n) - (n+1)\text{Log}(n+1) + n + 1) \\
(1) \quad &\ge k(n+1) + k(n\text{Log}(n) - (n+1)\text{Log}(n+1)) \ge k(n+1 - \text{Log}(n+1)).
\end{aligned}
$$

This is of the form $kn(1 + \phi(n))$ with $\lim_{n \to +\infty} \phi(n) = 0$.

We will now estimate $Y(r, f)$. We have

$$ 
\begin{aligned}
Y(r, f) &\le \sum_{j=1}^{n} \frac{j^k}{r} + \sum_{j=n+1}^{\infty} \frac{r}{j^k} \le \frac{1}{n^k} \int_{1}^{n+1} x^k dx + (n+1)^k \int_{n}^{\infty} \frac{1}{x^k} dx \\
&= \left(\frac{1}{n^k}\right) \left(\frac{(n+1)^{k+1}}{k+1}\right) + (n+1)^k \left(\frac{1}{(k-1)n^{k-1}}\right) \\
(2) \qquad &\qquad = \left(\frac{n+1}{n}\right)^{k-1} \left(\frac{(n+1)^2}{n(k+1)} + \frac{n+1}{k-1}\right).
\end{aligned}
$$

By (2) we can see that this is of the form $\frac{n}{k}(2 + \omega(n))$ with $\lim_{n \to +\infty} \omega(n) = 0$. Consequently, by (1) we have

$$ \limsup_{r \to +\infty} \frac{Y(r, f)}{Z(r, f)} \le \frac{2}{k^2} $$

which shows that $f$ is $\frac{2}{k^2} + \epsilon$-vanishing for every $\epsilon > 0$.

Now look at $\frac{\text{Log}(\text{Log}(M(f, r)))}{\text{Log}(r)}$ when $r \in [n^k, (n+1)^k[$. Let us put $\Lambda = \frac{2}{k^2} + \epsilon$. By (1), we have $Z(r, f) \le k(n + 1 - \text{Log}(n + 1))$, hence

$$ \text{Log}(M(f, r)) \le (1 + \Lambda)(k(n + 1 - \text{Log}(n + 1))), $$

therefore

$$ \rho(f) \le \frac{\text{Log}((k(n + 1 + \text{Log}(n + 1)))(1 + \Lambda))}{k\text{Log}(n)} $$

Consequently, $\rho(f) \le \frac{1}{k}$.

□

**Theorem 8:** For every $n \in \mathbb{N}$, let $a_n \in \mathbb{C}$ be such that $|a_n| = n^n$ and let $g(z) = \prod_{n=1}^{\infty} (1 - \frac{z}{a_n})$. Then $g$ is a perfectly-vanishing entire functions.
Proof. For every $r > 0$, we put $Z(r, g) = \sum_{j=1}^{\infty} \text{Log}^{+}(\frac{r}{|a_j|})$ and $Y(r, g) = \sum_{j=1}^{\infty} \min(\frac{|a_j|}{r}, \frac{r}{|a_j|})$.

We will first give a low bound to $Z(r, g)$ whenever $r \in [n^n, (n+1)^{n+1}]$. Indeed,

$$Z(r, g) = \sum_{j=1}^{n} \text{Log}(r) - \text{Log}(|a_j|) \geq n\text{Log}(n^n) - \sum_{j=1}^{n} \text{Log}(j^j) = n^2\text{Log}(n) - \sum_{j=1}^{n} j\text{Log}(j) \\ \geq n^2\text{Log}(n) - \int_{1}^{n} x\text{Log}(x)dx.$$

After computing, it is easily seen that $Z(r, g)$ admits a low bound of the form $\frac{n^2\text{Log}(n)(1 + \varphi(n))}{2}$, where $\lim_{n \to +\infty} \varphi(n) = 0$.

Consider now $Y(r, g)$ and put $Y_1(r, g) = \sum_{j=1}^{n} \frac{j^j}{r}$ and $Y_2(r, g) = \sum_{j=n+1}^{\infty} \frac{r}{j^j}$ so that

$Y(r, g) = Y_1(r, g) + Y_2(r, g)$ whenever $r \in [n^n, (n+1)^{n+1}]$. Clearly, $Y_1(r, g) \leq n$. And

$$Y_2(r, g) \leq ((n+1)^{n+1}) \left( \sum_{m=0}^{\infty} \frac{1}{(n+1+m)^{n+1+m}} \right) \\ \leq \sum_{m=0}^{\infty} \left( \frac{n+1}{n+1+m} \right)^{n+1} \left( \frac{1}{(n+1+m)^m} \right) \leq \sum_{m=0}^{\infty} \left( \frac{1}{(n+1+m)^m} \right) \\ \leq 1 + \frac{1}{n+1+1} + \sum_{m=2}^{\infty} \frac{1}{(n+1+m)^2}.$$

Thus, $Y_2(r, g)$ is bounded by a constant $C$ that does not depend on $n$ and hence we have

$$\frac{Y(r, g)}{Z(r, g)} \leq \frac{n + C}{n^2 \text{Log}(n)(1 + \varphi(n))}$$

which finishes proving that $g$ is perfectly vanishing. $\square$

**Remark:** Theorems 7 and 8 suggest that $\Lambda$-majored functions have a weak distribution of zeros and a weak order of growth.

The following lemma E is easily checked and was shown in an ultrametric field [3].

**Lemma E:** Let $f, g \in \mathcal{A}(\mathbb{C})$ be distinct and let $h_1, ..., h_q \in \mathcal{A}_g(\mathbb{C}) \cap \mathcal{A}_f(\mathbb{C})$ be such that, for every $j = 1, ..., q$ each zero of $g - h_j$ is a zero of $f - h_j$ and each zero of $f - h_j$ is a zero of $g - h_j$. Then, $\sum_{j=1}^{q} \bar{Z}(r, g - h_j) \leq \bar{Z}(r, f - g) + o(T(r, g)) + o(T(r, f))$ and $\sum_{j=1}^{q} \bar{Z}(r, f - h_j) \leq \bar{Z}(r, f - g) + o(T(r, g)) + o(T(r, f))$.
Proof. On one hand, it is clear that $\sum_{j=1}^q \bar{Z}(r, g - h_j) \le Z(r, f - g)$. Let us now consider the situation when a zero of some $g - h_i$ is also a zero of some $g - h_j$, with $i \ne j$. Since the $h_j$ are small functions with respect to both $f$ and $g$, it is then clear that $\sum_{j=1}^q \bar{Z}(r, g - h_j) \le \bar{Z}(r, f - g) + o(T(r, f)) + o(T(r, g))$. $\square$

By Yamanoi's theorem [11], we have this application which applies to all entire functions.

**Theorem. F:** Let $f \in \mathcal{A}(\mathbb{C})$ and let $h_1, \dots, h_n \in \mathcal{A}(\mathbb{C})$ be $n$ small functions with respect to $f$ and $g$. Then, given $\omega > 0$, we have

$$ (n - 1 - \omega)T(r, f) \le \sum_{j=1}^{n} \bar{Z}(r, f - h_j) + o(T(r, f)). $$

In the general case, let us recall this statement that is easily obtained:

**Theorem 9:** Let $f, g \in \mathcal{A}(\mathbb{C})$ and let $h_j$, $(1 \le j \le 4)$ be 4 small functions in $\mathcal{A}(\mathbb{C})$ with respect to $f$ and $g$ such that for each $j = 1, 2, 3, 4$, $f(z) = h_j(z)$ is equivalent to $g(z) = h_j(z)$. Then $f = g$.

Proof. Suppose $f$ and $g$ are distinct. Without loss of generality, we can assume that $f(0)g(0)(f - g)(0) \ne 0$. By Theorem A, we have

$$ T(r, f - g) \le 2(\max(T(r, f), T(r, g))), $$

when $r$ is big enough.

Now, by Theorem F, given $\omega > 0$, we have

$$ \begin{aligned} (3 - \omega)T(r, f) &\le \sum_{j=1}^{4} \bar{Z}(r, f - h_j) + o(T(r, f)) \le \bar{Z}(r, f - g) \le T(r, f - g) + o(T(r, f)) \\ &\le (2 \max(T(r, f), T(r, g))) + o(T(r, f)) \end{aligned} $$

when $r$ is big enough. Similarly,

$$ \begin{aligned} (3 - \omega)T(r, g) &\le \sum_{j=1}^{3} \bar{Z}(r, f - h_j) + o(T(r, g)) \le \bar{Z}(r, f - g) + o(T(r, g)) \le T(r, f - g) + o(T(r, g)) \\ &\le (2 \max(T(r, f), T(r, g))) + o(T(r, g)) \end{aligned} $$

when $r$ is big enough.

Consequently,

$$ (3 - \omega) \max(T(r, f), T(r, g)) \le 2 \max(T(r, f), T(r, g)) + o(T(r, f)) + o(T(r, g)) $$

when $r$ is big enough, and hence we have a contradiction ending the proof. $\square$
**Corollary 9.a:** Let $f, g \in \mathcal{A}(\mathbb{C})$ and let $a_j \in \mathbb{C}$ ($1 \le j \le 4$) be such that for each $j = 1, 2, 3, 4$, $f(z) = a_j$ is equivalent to $g(z) = a_j$. Then $f = g$.

Now, thanks to $\Lambda$-majored functions, here we can obtain Theorem 10 that looks like a theorem obtained in an ultrametric field [3].

**Theorem 10:** Let $f, g \in \mathcal{A}(\mathbb{C})$ be $\Lambda$-majored with $\Lambda = 1$ and let $h_j$, ($1 \le j \le 3$) be 3 small functions in $\mathcal{A}(\mathbb{C})$ with respect to $f$ and $g$ such that for each $j = 1, 2, 3$, $f(z) = h_j(z)$ is equivalent to $g(z) = h_j(z)$. Then $f = g$.

*Proof.* Suppose $f$ and $g$ are distinct. Without loss of generality, we can assume that $f(0)g(0)(f - g)(0) \ne 0$. By Theorem 1, there exists $\epsilon > 0$ such that

$$ (1) \qquad T(r, f - g) \le (2 - \epsilon)(\max(T(r, f), T(r, g)) + \text{Log}(2), $$

when $r$ is big enough.

Now, by Theorem F, given $\omega > 0$, we have

$$ \begin{aligned} (2-\omega)T(r, f) &\le \sum_{j=1}^{3} \overline{Z}(r, f-h_j)+o(T(r, f)) \le \overline{Z}(r, f-g) \le T(r, f-g)+o(T(r, f)) \\ &\le (2 - \epsilon) \max(T(r, f), T(r, g)) + \text{Log}(2) + o(T(r, f)) \end{aligned} $$

when $r$ is big enough. Similarly,

$$ \begin{aligned} (2-\omega)T(r, g) &\le \sum_{j=1}^{3} \overline{Z}(r, f-h_j)+o(T(r, g)) \le \overline{Z}(r, f-g)+o(T(r, g)) \le T(r, f-g)+o(T(r, g)) \\ &\le (2 - \epsilon) \max(T(r, f), T(r, g)) + \text{Log}(2) + o(T(r, g)) \end{aligned} $$

when $r$ is big enough.

Consequently, given $\omega > 0$, we can get

$$ (2-\omega) \max(T(r, f), T(r, g)) \le (2-\epsilon) \max(T(r, f), T(r, g))+\text{Log}(2)+o(T(r, f))+o(T(r, g)) $$

when $r$ is big enough. Now, we can choose $\omega$ such that $\epsilon > \omega$ and hence we have:

$$ (2-\omega) \max(T(r, f), T(r, g)) > (2-\epsilon) \max(T(r, f), T(r, g))+\text{Log}(2)+o(T(r, f))+o(T(r, g)) $$

when $r$ is big enough, which shows a contradiction ending the proof.

□

**Corollary 10.a:** Let $f, g \in \mathcal{A}(\mathbb{C})$ be $\Lambda$-majored with $\Lambda = 1$ and let $a_j \in \mathbb{C}$ ($1 \le j \le 3$) be such that for each $j = 1, 2, 3$, $f(z) = a_j$ is equivalent to $g(z) = a_j$. Then $f = g$.
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UNIVERSITÉ CLERMONT AUVERGNE, UMR CNRS 6620, LMBP, F-63000 CLERMONT-FERRAND, FRANCE.

Email address: alain.escassut@uca.fr