# Finiteness of meromorphic mappings from
Kähler manifold into projective space

Pham Duc Thoan  Department of Mathematics, National University of Civil
Engineering
55 Giai Phong street, Hai Ba Trung, Hanoi, Vietnam<EMAIL_ADDRESS>,
Nguyen Dang Tuyen  Department of Mathematics, National University of Civil
Engineering
55 Giai Phong street, Hai Ba Trung, Hanoi, Vietnam<EMAIL_ADDRESS>and
Noulorvang Vangty  Department of Mathmatics, National University of Education
136-Xuan Thuy str., Hanoi, Vietnam<EMAIL_ADDRESS>
###### Abstract.

The purpose of this paper is to prove the finiteness theorems for meromorphic
mappings of a complete connected Kähler manifold into projective space sharing
few hyperplanes in subgeneral position without counting multiplicity, where
all zeros with multiplicities more than a certain number are omitted. Our
results are extensions and generalizations of some recent ones.

††footnotetext: 2010 Mathematics Subject Classification: Primary 32H30, 32A22;
Secondary 30D35.
Key words and phrases: finiteness theorems, meromorphic mapping, complete
Kähler manifold.

##  1\. Introduction

Let $f$ be a non-constant meromorphic mapping of $\mathbb{C}^{m}$ into
$\mathbb{P}^{n}(\mathbb{C})$ and let $H$ be a hyperplane in
$\mathbb{P}^{n}(\mathbb{C})$. Denote by $\nu_{(f,H_{j})}(z)$ the intersecting
multiplicity of the mapping $f$ with the hyperplane $H_{j}$ at the point
$f(z)$.

For a divisor $\nu$ on $\mathbb{C}^{m}$ and for a positive integer $k$ or
$k=+\infty$, we set

$\nu_{\leqslant k}(z)=\begin{cases}0&{\text{ if }}\nu(z)>k,\\\ \nu(z)&{\text{
if }}\nu(z)\leqslant k.\end{cases}$

Similarly, we define $\nu_{>k}(z).$ If $\varphi$ is a meromorphic function,
the zero divisor of $\varphi$ is denoted by $\nu_{\varphi}.$

Let $H_{1},H_{2},\ldots,H_{q}$ be hyperplanes of $\mathbb{P}^{n}(\mathbb{C})$
(in subgeneral position or in general position) and let $k_{1},\ldots,k_{q}$
be positive integers or $+\infty$. Assume that $f$ is a meromorphic mapping
satisfying

$\dim\\{z:\nu_{(f,H_{i}),\leqslant k_{i}}(z)\cdot\nu_{(f,H_{j}),\leqslant
k_{j}}(z)\\}\leqslant m-2\ \ (1\leqslant i<j\leqslant q).$

Let $d$ be an integer number. We denote by
$\mathcal{F}(f,\\{H_{j},k_{j}\\}_{j=1}^{q},d)$ the set of all meromorphic
mappings $g:\mathbb{C}^{m}\to\mathbb{P}^{n}(\mathbb{C})$ satisfying the
following two conditions:

  * (a)

$\min(\nu_{(f,H_{j}),\leqslant k_{j}},d)=\min(\nu_{(g,H_{j}),\leqslant
k_{j}},d)$ ($1\leqslant j\leqslant q$).

  * (b)

$f(z)=g(z)$ on $\bigcup_{j=1}^{q}\\{z:\nu_{(f,H_{j}),\leqslant
k_{j}}(z)>0\\}$.

If $k_{1}=\cdots=k_{q}=+\infty$, we will simply use notation
$\mathcal{F}(f,\\{H_{j}\\}_{j=1}^{q},d)$ instead of
$\mathcal{F}(f,\\{H_{j},\infty\\}_{j=1}^{q},d).$

In 1926, Nevanlinna [8] showed that two distinct nonconstant meromorphic
functions $f$ and $g$ on the complex plane cannot have the same inverse images
for five distinct values, and that $g$ is a linear fractional transformation
of $f$ if they have the same inverse images counted with multiplicities for
four distinct values. After that, many authors have extended and improved
Nevanlinna’s results to the case of meromorphic mappings into complex
projective spaces such as Fujimoto [3, 5, 6], Smiley [15], Ru-Sogome [14],
Chen-Yan [1], Dethloff-Tan [2], Quang [16, 17, 18, 19], Nhung-Quynh [9]….
These theorems are called uniqueness theorems or finiteness theorems. The
first finiteness theorem for the case of meromorphic mappings from
$\mathbb{C}^{m}$ into complex projective space $\mathbb{P}^{n}(\mathbb{C})$
sharing $2n+2$ hyperplanes is given by Quang [17] in 2012 and its correction
[20] in 2015. Recently, he [18] extended his results and obtained the
following finiteness theorem, in which he did not need to count all zeros with
multiplicities more than certain values. Theorem A (see [18, Theorem 1.1]) Let
$f$ be a linearly nondegenerate meromorphic mapping of $\mathbb{C}^{m}$ into
$\mathbb{P}^{n}(\mathbb{C})$. Let $H_{1},\ldots,H_{2n+2}$ be $2n+2$
hyperplanes of $\mathbb{P}^{n}(\mathbb{C})$ in general position and let
$k_{1},\ldots,k_{2n+2}$ be positive integers or $+\infty$. Assume that

$\sum_{i=1}^{2n+2}\frac{1}{k_{i}+1}<\min\left\\{\frac{n+1}{3n^{2}+n},\frac{5n-9}{24n+12},\frac{n^{2}-1}{10n^{2}+8n}\right\\}.$

Then $\sharp\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{2n+2},1)\leqslant 2.$

Note that the condition
$\displaystyle\sum_{i=1}^{2n+2}\frac{1}{k_{i}+1}<\min\left\\{\frac{n+1}{3n^{2}+n},\frac{5n-9}{24n+12},\frac{n^{2}-1}{10n^{2}+8n}\right\\}$
in Theorem A becomes
$\displaystyle\sum_{i=1}^{2n+2}\frac{1}{k_{i}+1}<\frac{n+1}{3n^{2}+n}$ when
$n\geq 5.$

We now consider the general case, where $f:M\to\mathbb{P}^{n}(\mathbb{C})$ is
a meromorphic mapping of an $m$-dimensional complete connected Kähler manifold
$M$, whose universal covering is biholomorphic to a ball
$B(R_{0})=\\{z\in{\mathbf{C}}^{m}\ :\ ||z||<R_{0}\\}$
$(0<R_{0}\leqslant\infty)$, into $\mathbb{P}^{n}(\mathbb{C})$.

Let $H_{1},\ldots,H_{q}$ be hyperplanes of $\mathbb{P}^{n}(\mathbb{C})$ and
let $k_{1},\ldots,k_{q}$ be integers or $+\infty$. Then, the family
$\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},d)$ are defined similarly as above,
where $d$ is an integer number.

For $\rho\geqslant 0,$ we say that $f$ satisfies the condition $(C_{\rho})$ if
there exists a nonzero bounded continuous real-valued function $h$ on $M$ such
that

$\rho\Omega_{f}+dd^{c}\log h^{2}\geq\text{Ric}\omega,$

where $\Omega_{f}$ is the full-back of the Fubini-Study form $\Omega$ on
$\mathbb{P}^{n}(\mathbb{C})$,
$\omega=\dfrac{\sqrt{-1}}{2}\sum_{i,j}h_{i\bar{j}}dz_{i}\wedge
d\overline{z}_{j}$ is Kähler form on $M$,
$\text{Ric}\omega=dd^{c}\log(det(h_{i\overline{j}}))$,
$d=\partial+\overline{\partial}$ and
$d^{c}=\dfrac{\sqrt{-1}}{4\pi}(\overline{\partial}-\partial)$.

Very recently, Quang [19] obtained a finiteness theorem for meromorphic
mappings from such Kähler manifold $M$ into $\mathbb{P}^{n}(\mathbb{C})$
sharing hyperplanes regardless of multiplicities by giving new definitions of
”functions of small intergration” and ”functions of bounded intergration” as
well as proposing a new method to deal with the difficulties when he met on
the Kähler manifold. We would like to emphasize that Quang’s result is also
the first finiteness theorem for meromorphic mappings on the Kähler manifold,
although the uniqueness theorems were discovered early by Fujimoto [5] and
later by many authors such as Ru-Sogome [14] or Nhung-Quynh [9] and others.
Here is his result. Theorem B (see [19, Theorem 1.1]). Let $M$ be an
$m$-dimensional connected Kähler manifold whose universal covering is
biholomorphic to $\mathbb{C}^{m}$ or the unit ball $B(1)$ of $\mathbb{C}^{m}$,
and let $f$ be a linearly nondegenerate meromorphic mapping of $M$ into
$\mathbb{P}^{n}(\mathbb{C})\ (n\geqslant 2)$. Let $H_{1},\ldots,H_{q}$ be $q$
hyperplanes of $\mathbb{P}^{n}(\mathbb{C})$ in general position. Assume that
$f$ satisfies the condition $(C_{\rho})$. If

$\displaystyle
q>n+1+\frac{3nq}{6n+1}+\rho\frac{(n^{2}+4q-3n)(6n+1)}{6n^{2}+2}$

then $\sharp\mathcal{F}(f,\\{H_{i}\\}_{i=1}^{q},1)\leqslant 2.$

Unfortunately, in this result, all zeros with multiplicities must need to be
counted and hence Theorem B can not be an extension or a generalization of
Theorem A.

Our purpose in this article is to prove a similar result to Theorems A and B
for the case of a meromorphic mapping from a complete connected Kähler
manifold into projective space, in which all zeros with multiplicities more
than a certain number are omitted. However, the key used in the proof of
Theorem A is technique “rearranging counting functions” to compare counting
functions with characteristic functions, which is not valid on the Kähler
manifold. In addition, the proof of Theorem B cannot work on the case of
$k_{i}<\infty$. To overcome these difficulties, we use the technique in [22]
and the methods in [19], as well as considering new auxiliary functions to
obtain a new finiteness theorem which will generalize and extend the theorems
cited above. Namely, we will prove the following theorem.

######  Theorem 1.1.

Let $M$ be an $m$-dimensional connected Kähler manifold whose universal
covering is biholomorphic to $\mathbb{C}^{m}$ or the unit ball $B(1)$ of
$\mathbb{C}^{m}$, and let $f$ be a linearly nondegenerate meromorphic mapping
of $M$ into $\mathbb{P}^{n}(\mathbb{C})\ (n\geqslant 2)$. Let
$H_{1},\ldots,H_{q}$ be $q$ hyperplanes of $\mathbb{P}^{n}(\mathbb{C})$ in
$N$-subgeneral position and let $k_{1},\ldots,k_{q}$ be integers or $+\infty$.
Assume that $f$ satisfies the condition $(C_{\rho})$. Let $k$ be the largest
integer number not exceeding $\dfrac{q-2N-2}{2}$ and let $l$ be the smallest
integer number not less than $\dfrac{2N-2}{k+2}+2$ if $k>0$ or let $l=2N+1$ if
$k=0.$ Then $\sharp\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)\leqslant 2$ if

$\displaystyle q$
$\displaystyle>2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{4(q-n)n}{n-1}\big{)}$
$\displaystyle+\max\left\\{\frac{3nq}{2\big{(}3n+1+\frac{n-1}{l}\big{)}},\frac{4q+3nq-14}{4q+3n-14},\frac{3nq^{2}}{6nq+(n-2)(q-2)+4q-6n-8}\right\\}.$

Remark 1. It is easy to see that

$\dfrac{3nq}{2\big{(}3n+1+\frac{n-1}{l}\big{)}}<\dfrac{3nq}{6n+2}<\dfrac{3nq}{6n+1},$

and

$\dfrac{3nq^{2}}{6nq+(n-2)(q-2)+4q-6n-8}<\dfrac{3nq^{2}}{6nq+q}=\dfrac{3nq}{6n+1},\forall
n\geq 2.$

We now show that

$\frac{4q+3nq-14}{4q+3n-14}<\dfrac{3nq}{6n+1},\forall n\geq 3.$

Indeed, it suffices to prove that $12nq^{2}-9n^{2}q-69nq-4q+84n+14>0$ for all
$n\geq 3.$ Since $q\geq 2n+2$, we have $12nq^{2}-9n^{2}q-69nq-4q\geq
q(15n^{2}-45n-4)>0$ for all $n\geq 4.$ For $n=3,$ we have
$12nq^{2}-9n^{2}q-69nq-4q+84n+14=36q^{2}-292q+266>0$ since $q\geq 8.$

Hence, when $k_{1}=\cdots=k_{q}=+\infty$ and $N=n$, Theorem 1.1 is an
extension of Theorem B.

When $q=2n+2$, $M=\mathbb{C}^{n}$ and $H_{1},\ldots,H_{q}$ are in general
position, by $\rho=0$, $N=n$, $k=0$ and $l=2n+1,$ we obtain the following
corollary from Theorem 1.1.

######  Corollary 1.2.

Let $f$ be a linearly nondegenerate meromorphic mapping of $\mathbb{C}^{m}$
into $\mathbb{P}^{n}(\mathbb{C})$. Let $H_{1},\ldots,H_{2n+2}$ be $2n+2$
hyperplanes of $\mathbb{P}^{n}(\mathbb{C})$ in general position and let
$k_{1},\ldots,k_{n+2}$ be positive integers or $+\infty$. Then
$\sharp\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{2n+2},1)\leqslant 2$ provided

$\sum_{i=1}^{2n+2}\frac{1}{k_{i}+1}<\min\left\\{\frac{1}{2n},\frac{n^{3}+2n+3}{n(7n^{2}+5n+3)}\right\\}.$

In particular, if $n\geq 4$ then
$\sharp\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{2n+2},1)\leqslant 2$ provided

$\sum_{i=1}^{2n+2}\frac{1}{k_{i}+1}<\frac{1}{2n}.$

Remark 2. Consider the quantities
$A=\min\left\\{\frac{n+1}{3n^{2}+n},\frac{5n-9}{24n+12},\frac{n^{2}-1}{10n^{2}+8n}\right\\}$
in Theorem A and
$B=\min\left\\{\frac{1}{2n},\frac{n^{3}+2n+3}{n(7n^{2}+5n+3)}\right\\}$ in
Corollary 1.2. We have the following estimates.

$\bullet$ For $n\geq 5$, $A=\frac{n+1}{3n^{2}+n}<\frac{1}{2n}=B.$

$\bullet$ For $n=4$, $A=\frac{n^{2}-1}{10n^{2}+8n}<\frac{1}{2n}=B.$

$\bullet$ For $n=3$,
$A=\frac{n^{2}-1}{10n^{2}+8n}<\frac{n^{3}+2n+3}{n(7n^{2}+5n+3)}=B.$

$\bullet$ For $n=2$,
$A=\frac{5n-9}{24n+12}<\frac{n^{3}+2n+3}{n(7n^{2}+5n+3)}=B.$

In all the cases, always $A<B$. Therefore, Corollary 1.2 is a nice improvement
of Theorem A.

In order to prove our results, we first give an new estimate of the counting
function of the Cartan’s auxiliary function (see Lemma 2.8). We second improve
the algebraically dependent theorem of three meromorphic mappings (see Lemma
3.3). After that we use arguments similar to those used by Quang [19] to
finish the proofs.

##  2\. Basic notions and auxiliary results from Nevanlinna theory

We will recall some basic notions in Nevanlinna theory due to [13, 21].

2.1. Counting function. We set
$||z||=\big{(}|z_{1}|^{2}+\dots+|z_{n}|^{2}\big{)}^{1/2}$ for
$z=(z_{1},\dots,z_{n})\in\mathbb{C}^{m}$ and define

$\displaystyle B(r):=\\{z\in\mathbb{C}^{m}:||z||<r\\},\quad
S(r):=\\{z\in\mathbb{C}^{m}:||z||=r\\}\ (0<r\leqslant\infty),$

where $B(\infty)=\mathbb{C}^{m}$ and $S(\infty)=\emptyset$.

Define

$v_{m-1}(z):=\big{(}dd^{c}||z||^{2}\big{)}^{m-1}\quad\quad\text{and}$
$\sigma_{m}(z):=d^{c}\text{log}||z||^{2}\land\big{(}dd^{c}\text{log}||z||^{2}\big{)}^{m-1}\text{on}\quad\mathbb{C}^{m}\setminus\\{0\\}.$

A divisor $E$ on a ball $B(R_{0})$ is given by a formal sum
$E=\sum\mu_{\nu}X_{\nu}$, where $\\{X_{\nu}\\}$ is a locally family of
distinct irreducible analytic hypersurfaces in $B(R_{0})$ and
$\mu_{\nu}\in\mathbb{Z}$. We define the support of the divisor $E$ by setting
$\mathrm{Supp}\,(E)=\cup_{\mu_{\nu}\neq 0}X_{\nu}$. Sometimes, we identify the
divisor $E$ with a function $E(z)$ from $B(R_{0})$ into $\mathbb{Z}$ defined
by $E(z):=\sum_{X_{\nu}\ni z}\mu_{\nu}$.

Let $M,k$ be positive integers or $+\infty$. We define the truncated divisor
$E^{[M]}$ by

$\displaystyle E^{[M]}:=\sum_{\nu}\min\\{\mu_{\nu},M\\}X_{\nu},$

and the truncated counting function to level $M$ of $E$ by

$\displaystyle
N^{[M]}(r,r_{0};E):=\int\limits_{r_{0}}^{r}\frac{n^{[M]}(t,E)}{t^{2m-1}}dt\quad(r_{0}<r<R_{0}),$

where

$\displaystyle n^{[M]}(t,E):=\begin{cases}\int\limits_{\mathrm{Supp}\,(E)\cap
B(t)}E^{[M]}v_{m-1}&\text{ if }m\geqslant 2,\\\ \sum_{|z|\leqslant
t}E^{[M]}(z)&\text{ if }m=1.\end{cases}$

We omit the character [M] if $M=+\infty$.

Let $\varphi$ be a non-zero meromorphic function on $B(R_{0})$. We denote by
$\nu^{0}_{\varphi}$ (resp. $\nu^{\infty}_{\varphi}$) the divisor of zeros
(resp. divisor of poles ) of $\varphi$. The divisor of $\varphi$ is defined by

$\nu_{\varphi}=\nu^{0}_{\varphi}-\nu^{\infty}_{\varphi}.$

For a positive integer $M$ or $M=\infty$, we define the truncated divisors of
$\nu_{\varphi}$ by

$\nu^{[M]}_{\varphi}(z)=\min\
\\{M,\nu_{\varphi}(z)\\},\quad\nu^{[M]}_{\varphi,\leqslant
k}(z):=\begin{cases}\nu^{[M]}_{\varphi}(z)&\text{ if
}\nu^{[M]}_{\varphi}(z)\leqslant k,\\\ 0&\text{ if
}\nu^{[M]}_{\varphi}(z)>k.\end{cases}$

For convenience, we will write $N_{\varphi}(r,r_{0})$ and
$N^{[M]}_{\varphi,\leqslant k}(r,r_{0})$ for $N(r,r_{0};\nu^{0}_{\varphi})$
and $N^{[M]}(r,r_{0};\nu^{0}_{\varphi,\leqslant k})$ respectively.

2.2. Characteristic function. Let
$f:B(R_{0})\longrightarrow\mathbb{P}^{n}(\mathbb{C})$ be a meromorphic
mapping. Fix a homogeneous coordinates system $(w_{0}:\cdots:w_{n})$ on
$\mathbb{P}^{n}(\mathbb{C})$. We take a reduced representation
$f=(f_{0}:\cdots:f_{n})$, which means $f_{i}\ (0\leqslant i\leqslant n)$ are
holomorphic functions and $f(z)=\big{(}f_{0}(z):\dots:f_{n}(z)\big{)}$ outside
the analytic subset $\\{f_{0}=\dots=f_{n}=0\\}$ of codimension at least two.
Set $\|f\|=\big{(}|f_{0}|^{2}+\dots+|f_{n}|^{2}\big{)}^{1/2}$. Let $H$ be a
hyperplane in $\mathbb{P}^{n}(\mathbb{C})$ defined by
$H=\\{(\omega_{0},\ldots,\omega_{n}):a_{0}\omega_{0}+\cdots+a_{n}\omega_{n}=0\\}$.
We set $H(f)=a_{0}f_{0}+\cdots+a_{n}f_{n}$ and
$\|H\|=\big{(}|a_{0}|^{2}+\dots+|a_{n}|^{2}\big{)}^{1/2}.$

The characteristic function of $f$ (with respect to Fubini Study form
$\Omega$) is defined by

$\displaystyle
T_{f}(r,r_{0}):=\int_{t=r_{0}}^{r}\dfrac{dt}{t^{2m-1}}\int_{B(t)}f^{*}\Omega\wedge
v_{m-1},\quad\quad 0<r_{0}<r<R_{0}.$

By Jensen’s formula we have

$\displaystyle
T_{f}(r,r_{0})=\int_{S(r)}\log||f||\sigma_{m}-\int_{S(r_{0})}\log||f||\sigma_{m},\quad\quad
0<r_{0}<r<R_{0}.$

Through this paper, we assume that the numbers $r_{0}$ and $R_{0}$ are fixed
with $0<r_{0}<R_{0}$. By notation “$||\ P$”, we mean that the asseartion $P$
hold for all $r\in[r_{0},R_{0}]$ outside a set $E$ such that
$\int_{E}dr<\infty$ in case $R_{0}=\infty$ and
$\int_{E}\dfrac{1}{R_{0}-r}dr<\infty$ in case $R_{0}<\infty$.

2.3. Functions of small intergration. We recall some definitions due to Quang
[19].

Let $f^{1},\ldots,f^{k}$ be $k$ meromorphic mappings from the complete Kähler
manifold $B(1)$ into $\mathbb{P}^{m}(\mathbb{C})$, which satisfies the
condition $(C_{\rho})$ for a non-negative number $\rho$. For each $1\leqslant
u\leqslant k$, we fix a reduced representation
$f^{u}=(f_{0}^{u}:\cdots:f_{n}^{u})$ of $f^{u}$.

A non-negative plurisubharmonic function $g$ on $B(1)$ is said to be of small
intergration with respective to $f^{1},\ldots,f^{k}$ at level $l_{0}$ if there
exists an element $\alpha=(\alpha_{1},\ldots,\alpha_{m})\in\mathbb{N}^{m}$
with $|\alpha|\leqslant l_{0}$, a positive number $K$, such that for every
$0\leqslant tl_{0}<p<1$ then

$\int_{S(r)}|z^{\alpha}g|^{t}\sigma_{m}\leqslant
K\left(\frac{R^{2m-1}{R-r}}{\sum}_{u=1}^{m}T_{f^{u}}(r,r_{0})\right)^{p}$

for all $r$ with $0<r_{0}<r<R<1,$ where $z^{\alpha}=z_{1}^{\alpha_{1}}\cdots
z_{m}^{\alpha_{m}}.$

We denote by $S(l_{0};f^{1},\ldots,f^{k})$ the set of all non-negative
plurisubharmonic functions on $B(1)$ which are of small intergration with
respective to $f^{1},\ldots,f^{k}$ at level $l_{0}.$ We see that, if $g\in
S(l_{0};f^{1},\ldots,f^{k})$ then $g\in S(l;f^{1},\ldots,f^{k})$ for all
$l>l_{0}.$ Moreover, if $g$ is a constant function then $g\in
S(0;f^{1},\ldots,f^{k})$.

By [19, Proposition 3.2], if $g_{i}\in S(l_{i};f^{1},\ldots,f^{k})$, then
$g_{1}\cdots g_{s}\in S(\sum_{i=1}^{s}l_{i};f^{1},\ldots,f^{k})$.

A meromorphic function $h$ on $B(1)$ is said to be of bounded intergration
with bi-degree $(p,l_{0})$ for the family $\\{f^{1},\ldots,f^{k}\\}$ if there
exists $g\in S(l_{0};f^{1},\ldots,f^{k})$ satisfying

$|h|\leqslant||f^{1}||^{p}\cdots||f^{u}||^{p}\cdot g,$

outside a proper analytic subset of $B(1).$

We denote by $B(p,l_{0};f^{1},\ldots,f^{k})$ the set of all meromorphic
functions on $B(1)$ which are of bounded intergration of bi-degree $p,l_{0}$
for $\\{l_{0};f^{1},\ldots,f^{k}\\}$. We have the following assertions:

$\bullet$ For a meromorphic mapping $h$, $|h|\in S(l_{0};f^{1},\ldots,f^{k})$
iff $h\in B(0,l_{0};f^{1},\ldots,f^{k})$.

$\bullet$ $B(p,l_{0};f^{1},\ldots,f^{k})\subset B(p,l;f^{1},\ldots,f^{k})$ for
all $0\leqslant l_{0}<l.$

$\bullet$ If $h_{i}\in B(p_{i},l_{i};f^{1},\ldots,f^{k})$ then $h_{1}\cdots
h_{s}\in B(\sum_{i=1}^{s}p_{i},\sum_{i=1}^{s}l_{i};f^{1},\ldots,f^{k})$.

2.4. Some Lemmas and Propositions.

######  Lemma 2.1.

[6, Lemma 3.4] If $\Phi^{\alpha}(F,G,H)=0$ and
$\Phi^{\alpha}\left(\frac{1}{F},\frac{1}{G},\frac{1}{H}\right)=0$ for all
$\alpha$ with $|\alpha|\leqslant 1$, then one of the following assertions
holds:

(i) $F=G,G=H$ or $H=F$.

(ii) $\frac{F}{G},\frac{G}{H}$ and $\frac{H}{F}$ are all constants.

######  Proposition 2.2 (see [11, 12]).

_Let $H_{1},\ldots,H_{q}$ $(q>2N-n+1)$ be hyperplanes in
$\mathbb{P}^{n}(\mathbb{C})$ located in $N$-subgeneral position. Then there
exists a function $\omega:\\{1,\ldots,q\\}\to(0,1]$ called a Nochka weight and
a real number $\tilde{\omega}\geqslant 1$ called a Nochka constant satisfying
the following conditions:
(i) If $j\in\\{1,\ldots,q\\}$, then $0<\omega_{j}\tilde{\omega}\leqslant 1.$
(ii) $q-2N+n-1=\tilde{\omega}(\sum^{q}_{j=1}\omega_{j}-n-1).$
(iii) For $R\subset\\{1,\ldots,q\\}$ with $|R|=N+1$, then $\sum_{i\in
R}\omega_{i}\leqslant n+1.$
(iv) $\frac{N}{n}\leqslant\tilde{\omega}\leqslant\frac{2N-n+1}{n+1}.$
(v) Given real numbers $\lambda_{1},\ldots,\lambda_{q}$ with
$\lambda_{j}\geqslant 1$ for $1\leqslant j\leqslant q$ and given any
$R\subset\\{1,\ldots,q\\}$ and $|R|=N+1,$ there exists a subset $R^{1}\subset
R$ such that $|R^{1}|=\text{rank}\\{H_{i}\\}_{i\in R^{1}}=n+1$ and_

$\prod_{j\in R}\lambda_{j}^{\omega_{j}}\leqslant\prod_{i\in
R^{1}}\lambda_{i}.$

######  Proposition 2.3 (see [21], Lemma 3.2).

Let $\\{H_{i}\\}_{i=1}^{q}\ (q\geqslant n+1)$ be a set of hyperplanes of
$\mathbb{P}^{n}(\mathbb{C})$ satisfying $\cap_{i=1}^{q}H_{i}=\emptyset$ and
let $f:B(R_{0})\longrightarrow\mathbb{P}^{n}(\mathbb{C})$ be a meromorphic
mapping. Then there exist positive constants $\alpha$ and $\beta$ such that

$\alpha\|f\|\leqslant\max\limits_{i\in\\{1,\ldots,q\\}}|H_{i}(f)|\leqslant\beta\|f\|.$

######  Proposition 2.4 (see [4], Proposition 4.5).

Let $F_{1},\ldots,F_{n+1}$ be meromorphic functions on
$B(R_{0})\subset\mathbb{C}^{m}$ such that they are linearly independent over
$\mathbb{C}$. Then there exists an admissible set
$\\{\alpha_{i}=(\alpha_{i1},\ldots,\alpha_{im})\\}_{i=1}^{n+1}$ with
$\alpha_{ij}\geq 0$ being integers,
$|\alpha_{i}|=\sum_{j=1}^{m}|\alpha_{ij}|\leqslant i$ for $1\leqslant
i\leqslant n+1$ such that the generalized Wronskian
$W_{\alpha_{1},\ldots,\alpha_{n+1}}(F_{1},\ldots,F_{n+1})\not\equiv 0$ where
$W_{\alpha_{1},\ldots,\alpha_{n+1}}(F_{1},\ldots,F_{n+1})=det\left(\mathcal{D}^{\alpha_{i}}F_{j}\right)_{1\leqslant
i,j\leqslant n+1}.$

Let $L_{1},\ldots,L_{n+1}$ be linear forms of $n+1$ variables and assume that
they are linearly independent. Let
$F=(F_{1}:\cdots:F_{n+1}):B(R_{0})\to\mathbb{P}^{n}(\mathbb{C})$ be a
meromorphic mapping and $(\alpha_{1},\ldots,\alpha_{n+1})$ be an admissible
set of $F$. Then we have following proposition.

######  Proposition 2.5 (see [13], Proposition 3.3).

In the above situation, set $l_{0}=|\alpha_{1}|+\cdots+|\alpha_{n+1}|$ and
take $t,p$ with $0<tl_{0}<p<1.$ Then, for $0<r_{0}<R_{0}$ there exists a
positive constant $K$ such that for $r_{0}<r<R<R_{0},$

$\int\limits_{S(r)}\left|z^{\alpha_{1}+\cdots+\alpha_{n+1}}\dfrac{W_{\alpha_{1},\ldots,\alpha_{n+1}}(F_{1},\ldots,F_{n+1})}{L_{1}(F)\cdots
L_{n+1}(F)}\right|^{t}\sigma_{m}\leqslant
K\left(\dfrac{R^{2m-1}}{R-r}T_{F}(R,r_{0})\right)^{p},$

where $z^{\alpha}=z_{1}^{\alpha_{1}}\cdots z_{m}^{\alpha_{m}}$ for
$z=(z_{1},\ldots,z_{m})$ and $\alpha=(\alpha_{1},\ldots,\alpha_{m})$.

For convenience of presentation, for meromorphic mappings
$f^{u}:B(R)\to\mathbb{P}^{n}(\mathbb{C})$ and hyperplanes
$\\{H_{i}\\}_{i=1}^{q}$ of $\mathbb{P}^{n}(\mathbb{C})$, we denote by
$\mathcal{S}$ the closure of

$\cup_{1\leqslant u\leqslant 3}I(f^{u})\cup\cup_{1\leqslant i<j\leqslant
q}\\{z:\nu_{(f,H_{i}),\leqslant k_{i}}(z)\cdot\nu_{(f,H_{j}),\leqslant
k_{j}}(z)>0\\}.$

We see that $\mathcal{S}$ is an analysis subset of codimension two of $B(R)$.

######  Lemma 2.6.

[22, Lemma 2.6] Let $f^{1},f^{2},f^{3}$ be three mappings in
$\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$. Suppose that there exist
$s,t,l\in\\{1,\ldots,q\\}$ such that

$P:=Det\left(\begin{array}[]{ccc}(f^{1},H_{s})&(f^{1},H_{t})&(f^{1},H_{l})\\\
(f^{2},H_{s})&(f^{2},H_{t})&(f^{2},H_{l})\\\
(f^{3},H_{s})&(f^{3},H_{t})&(f^{3},H_{l})\end{array}\right)\not\equiv 0.$

Then we have

$\displaystyle\nu_{P}(z)\geq\sum_{i=s,t,l}(\min_{1\leqslant u\leqslant
3}\\{\nu_{(f^{u},H_{i}),\leqslant
k_{i}}(z)\\}-\nu^{[1]}_{(f^{1},H_{i}),\leqslant
k_{i}}(z))+2\sum_{i=1}^{q}\nu^{[1]}_{(f^{1},H_{i}),\leqslant k_{i}}(z),\forall
z\not\in\mathcal{S}.$

######  Lemma 2.7.

[22, Lemma 2.7] Let $f$ be a linearly nondegenerate meromorphic mapping from
$B(R_{0})$ into $\mathbb{P}^{n}(\mathbb{C})$ and let
$H_{1},H_{2},\ldots,H_{q}$ be $q$ hyperplanes of $\mathbb{P}^{n}(\mathbb{C})$
in $N$-subgeneral position. Set $l_{0}=|\alpha_{0}|+\cdots+|\alpha_{n}|$ and
take $t,p$ with $0<tl_{0}<p<1.$ Let $\omega(j)$ be Nochka weights with respect
to $H_{j}$, $1\leqslant j\leqslant q$ and let $k_{j}\ (j=1,\ldots,q)$ be
positive integers not less than $n$. For each $j$, we put
$\hat{\omega}(j):=\omega{(j)}\big{(}1-\frac{n}{k_{j}+1}).$ Then, for
$0<r_{0}<R_{0}$ there exists a positive constant $K$ such that for
$r_{0}<r<R<R_{0},$

$\int\limits_{S(r)}\left|z^{\alpha_{0}+\cdots+\alpha_{n}}\frac{W_{\alpha_{0}\ldots\alpha_{n}}(f)}{(f,H_{1})^{\hat{\omega}(1)}\cdots(f,H_{q})^{\hat{\omega}(q)}}\right|^{t}\bigl{(}\|f\|^{\sum_{j=1}^{q}\hat{\omega}(j)-n-1}\bigr{)}^{t}\sigma_{m}\leqslant
K\bigl{(}\frac{R^{2m-1}}{R-r}T_{f}(R,r_{0})\bigl{)}^{p},$

In fact, Lemma 2.7 is another version of Lemma 8 in [10], in which
$\omega{(j)}$ is replaced by $\hat{\omega}(j)$.

######  Lemma 2.8.

Let $M$, $f$ and $H_{1},H_{2},\ldots,H_{q}$ be as in Theorem 1.1. Let $P$ be a
holomorphic function on $M$ and $\beta$ be a positive real number such that
$P^{\beta}\in B(\alpha,l_{0};f^{1},f^{2},f^{3})$ and

$\displaystyle\sum_{u=1}^{3}\sum_{i=1}^{q}\nu^{[n]}_{H_{i}(f^{u}),\leqslant
k_{i}}\leqslant\beta\nu_{P},$

where $f^{1},f^{2},f^{3}\in\mathcal{F}(f,\\{H_{j},k_{j}\\}_{j=1}^{q},1)$. Then

$q\leqslant
2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{2}{3}l_{0}\big{)}+{\alpha}.$

###### Proof.

Let $F_{u}=(f^{u}_{0}:\cdots:f^{u}_{n})$ be a reduced representation of
$f^{u}\ (1\leqslant u\leqslant 3)$. By routine arguments in the Nevanlinna
theory and using Proposition 2.2 (i), we have

$\displaystyle\sum\limits_{i=1}^{q}\omega_{i}\nu_{H_{i}(f^{u})}(z)$
$\displaystyle-\nu_{W_{\alpha_{u,0}\cdots\alpha_{u,n}}(F_{u})}(z)$
$\displaystyle\leqslant\sum\limits_{i=1}^{q}\omega_{i}\min\\{n,\nu_{H_{i}(f^{u})}(z)\\}$
$\displaystyle=\sum\limits_{i=1}^{q}\omega_{i}\min\\{n,\nu_{H_{i}(f^{u}),\leqslant
k_{i}}(z)\\}+\sum\limits_{i=1}^{q}\omega_{i}\min\\{n,\nu_{H_{i}(f^{u}),>k_{i}}(z)\\}$
$\displaystyle\leqslant\sum\limits_{i=1}^{q}\frac{1}{\tilde{\omega}}\nu^{[n]}_{H_{i}(f^{u}),\leqslant
k_{i}}(z)+\sum\limits_{i=1}^{q}\omega_{i}\dfrac{n}{k_{i}+1}\nu_{H_{i}(f^{u})}(z).$

Hence, it is easy to see from the assumption that

(2.9)
$\displaystyle\sum_{i=1}^{q}{\hat{\omega}_{i}}(\nu_{H_{i}(f^{1})}+\nu_{H_{i}(f^{2})}+\nu_{H_{i}(f^{3})})-(\nu_{W_{\alpha_{1}}(F_{1})}+\nu_{W_{\alpha_{2}}(F_{2})}+\nu_{W_{\alpha_{3}}(F_{3})})\leqslant\frac{\beta}{\tilde{\omega}}\nu_{P},$

where $\hat{\omega}_{i}:=\omega_{i}\big{(}1-\dfrac{n}{k_{i}+1}\big{)}$ for all
$1\leqslant i\leqslant q$.

Since the universal covering of $M$ is biholomorphic to
$B(R_{0}),0<R_{0}\leqslant\infty$, by using the universal covering if
necessary, we may assume that $M=B(R_{0})\subset{\mathbf{C}}^{m}$. We consider
the following cases.

$\bullet$ First case: $R_{0}=\infty$ or $\lim\sup_{r\to
R_{0}}\dfrac{T_{f^{1}}(r,r_{0})+T_{f^{2}}(r,r_{0})+T_{f^{3}}(r,r_{0})}{\log(1/(R_{0}-r))}=\infty$.

Integrating both sides of inequality (2.9), we get

(2.10) $\displaystyle\beta N_{P}(r)$
$\displaystyle\geqslant{\tilde{\omega}}\sum_{u=1}^{3}(\sum_{i=1}^{q}{\omega_{i}}N_{H_{i}(f^{u})}(r,r_{0})-N_{W_{\alpha}(F_{u})}(r,r_{0}))-\sum_{u=1}^{3}\sum_{i=1}^{q}\frac{\tilde{\omega}\omega_{i}n}{k_{i}+1}(T_{f^{u}}(r,r_{0})+O(1).$

Applying Lemma 2.7 to $\omega_{i}\ (1\leqslant i\leqslant q),$ we have

$\int\limits_{S(r)}\left|z^{\alpha_{0}+\cdots+\alpha_{n}}\frac{W_{\alpha_{0}\ldots\alpha_{n}}(F_{u})}{H_{1}^{{\omega}_{1}}(f^{u})(z)\cdots
H_{q}^{{\omega}_{q}}(f^{u})(z)}\right|^{t_{u}}\left(\|f^{u}\|^{\sum_{i=1}^{q}{\omega}_{i}-n-1}\right)^{t_{u}}\sigma_{m}\leqslant
K\bigl{(}\frac{R^{2m-1}}{R-r}T_{f^{u}}(R,r_{0})\bigl{)}^{p_{u}}.$

By the concativity of the logarithmic function, we obtain

$\displaystyle\int\limits_{S(r)}\log|z^{\alpha_{0}+\cdots+\alpha_{n}}|\sigma_{m}$
$\displaystyle+(\sum_{i=1}^{q}{\omega}_{i}-n-1)\int\limits_{S(r)}\log||f^{u}||\sigma_{m}+\int\limits_{S(r)}\log|W_{\alpha_{0}\ldots\alpha_{n}}(F_{u})|\sigma_{m}$
$\displaystyle-\sum_{i=1}^{q}\omega_{i}\int\limits_{S(r)}\log|H_{i}(f^{u})|\sigma_{m}\leqslant\frac{p_{u}K}{t_{u}}\big{(}\log^{+}\frac{1}{R_{0}-r}+\log^{+}T_{f^{u}}(r,r_{0})\big{)}.$

By the definition of the characteristic function and the counting function, we
get the following estimate

$\displaystyle||\ (\sum_{i=1}^{q}{\omega}_{i}-n-1)T_{f^{u}}(r,r_{0})$
$\displaystyle\leqslant\sum_{i=1}^{q}\omega_{i}N_{H_{i}(f^{u})}(r,r_{0})-N_{W_{\alpha_{1}\ldots\alpha_{n}}F_{u})}(r)$
$\displaystyle+K_{1}\big{(}\log^{+}\frac{1}{R_{0}-r}+\log^{+}T_{f^{u}}(r,r_{0})\big{)}.$

Using Proposition 2.2 (ii), we get

$\displaystyle||\ (q-2N+n-1)T_{f^{u}}(r,r_{0})$
$\displaystyle\leqslant{\tilde{\omega}}\left(\sum_{i=1}^{q}{\omega_{i}}N_{H_{i}(f^{u})}(r,r_{0})-N_{W_{\alpha_{0}\ldots\alpha_{n}}(F_{u})}(r,r_{0})\right)$
$\displaystyle+{\tilde{\omega}}{K_{1}}\big{(}\log^{+}\frac{1}{R_{0}-r}+\log^{+}T_{f^{u}}(r,r_{0})\big{.}$

Combining these inequalities with (2.10) and noticing that
$\tilde{\omega}\omega_{i}\leqslant 1$, we get

(2.11) $\displaystyle||\ \beta N_{P}(r)$
$\displaystyle\geqslant(q-2N+n-1)T(r,r_{0})-\sum_{i=1}^{q}\frac{n}{k_{i}+1}T(r,r_{0})+O(1),$

where $T(r,r_{0}):=T_{f}(r,r_{0})+T_{g}(r,r_{0}).$

Since the assumption $P^{\beta}\in B(\alpha,l_{0};f^{1},f^{2},f^{3})$, there
exists $g\in S(l_{0};f^{1},f^{2},f^{3})$ satisfying

$|P|^{\beta}\leqslant||f^{1}||^{\alpha}\cdot||f^{2}||^{\alpha}\cdot||f^{3}||^{\alpha}\cdot
g,$

outside a proper analytic subset of $B(1).$ Hence, by Jensen’s formula and the
definition of the characteristic function, we have the following estimate

(2.12) $\displaystyle||\ \beta N_{P}(r)=$
$\displaystyle\int_{S(r)}\log|P|^{\beta}\sigma_{n}+O(1)$
$\displaystyle\leqslant$
$\displaystyle\int_{S(r)}({\alpha}\sum_{u=1}^{3}\log||f^{u}||+\log||g||)\sigma_{n}+O(1)$
$\displaystyle=$ $\displaystyle{\alpha}T_{f}(r,r_{0})+o(T(r,r_{0})).$

Together (2.11) with (2.12), we obtain

$\displaystyle(q-2N+n-1)T(r,r_{0})-\sum_{i=1}^{q}\frac{n}{k_{i}+1}T(r,r_{0})\leqslant{\alpha}T(r,r_{0})+o(T(r,r_{0}))$

for every $r$ outside a Borel finite measure set. Letting
$r\rightarrow\infty$, we deduce that

$q-2N+n-1-\sum_{i=1}^{q}\frac{n}{k_{i}+1}\leqslant\rho\big{(}n(2N-n+1)+\frac{2}{3}l_{0}\big{)}+{\alpha}$

with $\rho=0.$

$\bullet$ Second Case: $R_{0}<\infty$ and $\lim\sup_{r\to
R_{0}}\dfrac{T_{f^{1}}(r,r_{0})+T_{f^{2}}(r,r_{0})+T_{f^{3}}(r,r_{0})}{\log(1/(R_{0}-r))}<\infty$.
It suffices to prove the lemma in the case where $B(R_{0})=B(1)$.

Suppose that

$q>2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{2}{3}l_{0}\big{)}+{\alpha}.$

Then, we have

$q>2N-n+1+\sum_{i=1}^{q}{\tilde{\omega}}{\omega_{i}}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{2}{3}l_{0}\big{)}+\alpha.$

It follows from Proposition 2.2 ii), iv) that

$\displaystyle\sum_{i=1}^{q}{{\omega_{i}}}\big{(}1-\frac{n}{k_{i}+1}\big{)}-(n+1)-\dfrac{\alpha}{\tilde{\omega}}$
$\displaystyle>\rho\big{(}\frac{n(2N-n+1)}{\tilde{\omega}}+\frac{2}{3}\frac{l_{0}}{\tilde{\omega}}\big{)}$
$\displaystyle\geqslant\rho\big{(}n(n+1)+\frac{2}{3}\frac{l_{0}}{\tilde{\omega}}\big{)}.$

Put

$t=\dfrac{\frac{2\rho}{3}}{\displaystyle\sum_{i=1}^{q}{\hat{\omega}_{i}}-(n+1)-\dfrac{\alpha}{\tilde{\omega}}}.$

It implies that

(2.13)
$\displaystyle\big{(}\frac{3n(n+1)}{2}+\frac{l_{0}}{\tilde{\omega}}\big{)}t<1.$

Put
$\psi_{u}=z^{\alpha_{u,0}+\cdots+\alpha_{u,n}}\dfrac{W_{\alpha_{u,0}\cdots\alpha_{u,n}}(F_{u})}{H_{1}^{\hat{\omega}_{1}}(f^{u})\cdots
H_{q}^{\hat{\omega}_{q}}(f^{u})}\ \ (1\leqslant u\leqslant 3)$. It follows
from (2.9) that
$\psi_{1}^{t}\psi_{2}^{t}\psi_{3}^{t}P^{\frac{t\beta}{\tilde{\omega}}}$ is
holomorphic. Hence
$a=\log|\psi_{1}^{t}\psi_{2}^{t}\psi_{3}^{t}P^{\frac{t\beta}{\tilde{\omega}}}|$
is plurisubharmonic on $B(1)$.

We now write the given Kähler metric form as

${\omega}=\frac{\sqrt{-1}}{2\pi}\sum\limits_{i,j}h_{i\bar{j}}dz_{i}\wedge
d\bar{z}_{j}.$

From the assumption that $f^{1}$, $f^{2}$ and $f^{3}$ satisfy condition
$(C_{\rho})$, there are continuous plurisubharmonic functions $a^{\prime}_{u}$
on $B(1)$ such that

$e^{a^{\prime}_{u}}\text{det}(h_{i\bar{j}})^{\frac{1}{2}}\leqslant\|f^{u}\|^{\rho},u=1,2,3.$

Put $a_{u}=\frac{2}{3}a^{\prime}_{u}$, $u=1,2,3$ and we get

$e^{a_{u}}\text{det}(h_{i\bar{j}})^{\frac{1}{3}}\leqslant\|f^{u}\|^{\frac{2\rho}{3}}.$

Therefore, by the definition of $t$, we get

$\displaystyle e^{a+a_{1}+a_{2}+a_{3}}\text{det}(h_{i\bar{j}})$
$\displaystyle\leqslant
e^{a}\|f^{1}\|^{\frac{2\rho}{3}}\|f^{2}\|^{\frac{2\rho}{3}}\|f^{3}\|^{\frac{2\rho}{3}}$
$\displaystyle=|\psi_{1}|^{t}|\psi_{2}|^{t}|\psi_{3}|^{t}|P|^{\frac{t\beta}{\tilde{\omega}}}\|f^{1}\|^{\frac{2\rho}{3}}\|f^{2}\|^{\frac{2\rho}{3}}\|f^{3}\|^{\frac{2\rho}{3}}$
$\displaystyle\leqslant|\psi_{1}|^{t}|\psi_{2}|^{t}|\psi_{3}|^{t}\big{(}\|f^{1}\|\|f^{2}\|\|f^{3}|\big{)}^{\frac{t\alpha}{\tilde{\omega}}}\|f^{1}\|^{\frac{2\rho}{3}}\|f^{2}\|^{\frac{2\rho}{3}}\|f^{3}\|^{\frac{2\rho}{3}}\cdot|g|^{\frac{t}{\tilde{\omega}}}$
$\displaystyle=|\psi_{1}|^{t}|\psi_{2}|^{t}|\psi_{3}|^{t}\big{(}\|f^{1}\|\|f^{2}\|\|f^{3}\|\big{)}^{t(\frac{\alpha}{\tilde{\omega}}+\frac{2\rho}{3t})}\cdot|g|^{\frac{t}{\tilde{\omega}}}$
$\displaystyle=|\psi_{1}|^{t}|\psi_{2}|^{t}|\psi_{3}|^{t}\big{(}\|f^{1}\|\|f^{2}\|\|f^{3}\|\big{)}^{t(\sum_{i=1}^{q}\hat{\omega}_{i}-n-1)}\cdot|g|^{\frac{t}{\tilde{\omega}}}.$

Note that the volume form on $B(1)$ is given by

$dV:=c_{m}\text{det}(h_{i\bar{j}})v_{m};$

therefore,

$\int\limits_{B(1)}e^{a+a_{1}+a_{2}+a_{3}}dV\leqslant
C\int\limits_{B(1)}\prod_{u=1}^{3}\big{(}|\psi_{u}|\|f^{u}\|^{\sum_{i=1}^{q}\hat{\omega}_{i}-n-1}\big{)}^{t}\cdot|g|^{\frac{t}{\tilde{\omega}}}v_{m},$

with some positive constant $C.$

Setting $x=\dfrac{l_{0}/\tilde{\omega}}{3n(n+1)/2+l_{0}/\tilde{\omega}},\
y=\dfrac{n(n+1)/2}{3n(n+1)/2+l_{0}/\tilde{\omega}}$, then $x+3y=1$. Thus, by
the Hölder inequality and by noticing that

$v_{m}=(dd^{c}\|z\|^{2})^{m}=2m\|z\|^{2m-1}\sigma_{m}\wedge d\|z\|,$

we obtain

$\displaystyle\int\limits_{B(1)}e^{a+a_{1}+a_{2}+a_{3}}dV$
$\displaystyle\leqslant
C\prod_{u=1}^{3}\left(\int\limits_{B(1)}\big{(}|\psi_{u}|\|f^{u}\|^{\sum_{i=1}^{q}\hat{\omega}_{i}-n-1}\big{)}^{\frac{t}{y}}v_{m}\right)^{y}\left(\int\limits_{B(1)}|z^{\beta}g|^{\frac{t}{x\tilde{\omega}}}v_{m}\right)^{x}$
$\displaystyle\leqslant
C\prod_{u=1}^{3}\bigl{(}2m\int\limits_{0}\limits^{1}r^{2m-1}\bigl{(}\int\limits_{S(r)}\big{(}|\psi_{u}|\|f^{u}\|^{\sum_{i=1}^{q}\hat{\omega}_{i}-n-1}\big{)}^{\frac{t}{y}}\sigma_{m}\bigl{)}dr\bigl{)}^{y}$
$\displaystyle\times\bigl{(}2m\int\limits_{0}\limits^{1}r^{2m-1}\bigl{(}\int\limits_{S(r)}|z^{\beta}g|^{\frac{t}{x\tilde{\omega}}}\sigma_{m}\bigl{)}dr\bigl{)}^{x}.$

We see from (2.13) that
$\dfrac{l_{0}t}{\tilde{\omega}x}=\big{(}\dfrac{3n(n+1)}{2}+\dfrac{l_{0}}{\tilde{\omega}}\big{)}t<1$
and

$\sum\limits_{s=0}^{n}|\alpha_{u,s}|\dfrac{t}{y}\leqslant\dfrac{n(n+1)}{2}\dfrac{t}{y}=\big{(}\dfrac{3n(n+1)}{2}+\dfrac{l_{0}}{\tilde{\omega}}\big{)}t<1.$

Then, we can choose a positive number $p$ such that
$\dfrac{l_{0}t}{\tilde{\omega}x}<p<1$ and
$\sum\limits_{s=0}^{n}|\alpha_{u,s}|\dfrac{t}{y}<p<1.$ Applying Lemma 2.7 to
$\hat{\omega}_{i}$, and from the property of $g$, we get

$\int\limits_{S(r)}\big{(}|\psi_{u}|\|f^{u}\|^{\sum_{i=1}^{q}\hat{\omega}_{i}-n-1}\big{)}^{\frac{t}{y}}\sigma_{m}\leqslant
K_{1}\left(\frac{R^{2m-1}}{R-r}T_{f}^{u}(R,r_{0})\right)^{p}$

and

$\int\limits_{S(r)}|z^{\beta}g|^{\frac{t}{\tilde{\omega}x}}\sigma_{m}\leqslant
K\left(\frac{R^{2m-1}}{R-r}T_{g}(R,r_{0})\right)^{p}$

outside a subset $E\subset[0,1]$ such that
$\displaystyle\int\limits_{E}\dfrac{1}{1-r}dr\leqslant+\infty.$ Choosing
$R=r+\dfrac{1-r}{eT_{f^{u}}(r,r_{0})},$ we have

$T_{f^{u}}(R,r_{0})\leqslant 2T_{f^{u}}(r,r_{0}),$

Hence, the above inequality implies that

$\int\limits_{S(r)}\big{(}|\psi_{u}|\|f^{u}\|^{\sum_{i=1}^{q}\hat{\omega}_{i}-n-1}\big{)}^{\frac{t}{y}}\sigma_{m}\leqslant\frac{K_{2}}{(1-r)^{p}}(T_{f^{u}}(r,r_{0}))^{2p}\leqslant\frac{K_{2}}{(1-r)^{p}}(\log\frac{1}{1-r})^{2p},$

since $\lim\limits_{r\to
R_{0}}\sup\dfrac{T_{f^{1}}(r,r_{0})+T_{f^{2}}(r,r_{0})+T_{f^{2}}(r,r_{0})}{\log(1/(R_{0}-r))}<\infty.$
It implies that

$\int\limits_{0}\limits^{1}r^{2m-1}\left(\int\limits_{S(r)}\big{(}|\psi_{u}|\|f^{u}\|^{\sum_{i=1}^{q}\hat{\omega}_{i}-n-1}\big{)}\sigma_{m}\right)dr\leqslant\int\limits_{0}\limits^{1}r^{2m-1}\frac{K_{2}}{(1-r)^{p}}\left(\log\frac{1}{1-r}\right)^{2p}dr<\infty.$

Similarly,

$\int\limits_{0}\limits^{1}r^{2m-1}\left(\int\limits_{S(r)}|z^{\beta}g|^{\frac{t}{\tilde{\omega}x}}\sigma_{m}\right)dr\leqslant\int\limits_{0}\limits^{1}r^{2m-1}\frac{K_{2}}{(1-r)^{p}}\left(\log\frac{1}{1-r}\right)^{2p}dr<\infty.$

Hence, we conclude that $\int\limits_{B(1)}e^{a+a_{1}+a_{2}+a_{3}}dV<\infty,$
which contradicts Yau’s result [23] and Karp’s result [7]. The proof of Lemma
2.8 is complete. ∎

##  3\. Proof of Theorem 1.1

######  Lemma 3.1 (see [22], Lemma 3.1).

If $q>2N+1+\sum_{v=1}^{q}\frac{n}{k_{v}+1}+\rho n(2N-n+1)$, then every
$g\in\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$ is linearly nondegenerate.

######  Lemma 3.2 (see [10], Lemma 12).

Let $q,N$ be two integers satisfying $q\geq 2N+2$, $N\geq 2$ and $q$ be even.
Let $\\{a_{1},a_{2},\ldots,a_{q}\\}$ be a family of vectors in a 3-dimensional
vector space such that $\text{rank}\\{a_{j}\\}_{j\in R}=2$ for any subset
${R}\subset Q=\\{1,\ldots,q\\}$ with cardinality $|R|=N+1$. Then there exists
a partition $\bigcup_{j=1}^{q/2}I_{j}$ of $\\{1,\ldots,q\\}$ satisfying
$|I_{j}|=2$ and $\text{rank}\\{a_{i}\\}_{i\in I_{j}}=2$ for all
$j=1,\ldots,q/2.$

We need the following result which slightly improves [22, Theorem 1.3].

######  Lemma 3.3.

Let $k$ be the largest integer number not exceeding $\dfrac{q-2N-2}{2}$. If
$n\geqslant 2$ then $f^{1}\wedge f^{2}\wedge f^{3}\equiv 0$ for every
$f^{1},f^{2},f^{3}\in\mathcal{(}f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$ provided

$q>2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho
n(2N-n+1)+\frac{3nq}{2\big{(}q+(n-1)\frac{l+1}{l}\big{)}},$

where $l$ is the smallest integer number not less than $\dfrac{2N+2+2k}{k+2}$
if $k>0$ or $l=2N+1$ if $k=0.$

###### Proof.

We consider $\mathcal{M}^{3}$ as a vector space over the field $\mathcal{M}$
and denote $Q=\\{1,\ldots,q\\}$. For each $i\in Q$, we set

$V_{i}=\left((f^{1},H_{i}),(f^{2},H_{i}),(f^{3},H_{i})\right)\in\mathcal{M}^{3}.$

By Lemma 3.1, $f^{1},f^{2},f^{3}$ are linearly nondegenerate. Suppose that
$f^{1}\wedge f^{2}\wedge f^{3}\not\equiv 0$. Since the family of hyperplanes
$\\{H_{1},H_{2},\ldots,H_{q}\\}$ are in $N$-subgeneral position, for each
subset $R\subset Q$ with cardinality $|R|=N+1$, there exist three indices
$l,t,s\in R$ such that the vectors $V_{l},V_{t}$ and $V_{s}$ are linearly
independent. This means that

$P_{I}:=\det\left(\begin{array}[]{ccc}(f^{1},H_{l})&(f^{1},H_{t})&(f^{1},H_{s})\\\
(f^{2},H_{l})&(f^{2},H_{t})&(f^{2},H_{s})\\\
(f^{3},H_{l})&(f^{3},H_{t})&(f^{3},H_{s})\end{array}\right)\not\equiv 0,$

where $I:=\\{l,t,s\\}.$ We separate into the following cases.

$\bullet$ Case 1: $q\mod 2=0$

By the assumption, we have $q=2N+2+2k$ $(k\geq 0)$. Applying Lemma 3.2, we can
find a partition $\\{J_{1},\ldots,J_{q/2}\\}$ of $Q$ satisfying $|J_{j}|=2$
and $\text{rank}\\{V_{v}\\}_{v\in J_{j}}=2$ for all $j=1,2,\ldots,q/2.$ Take a
fixed subset $S_{j}=\\{j_{1},\ldots,j_{k+2}\\}\subset\\{1,\ldots,q\\}$. We
claim that:

There exists a partition $J^{j}_{1},\ldots,J^{j}_{N+1+k}$ with $k+2$ indices
${r^{j}_{1},\ldots,r^{j}_{k+2}}\in\\{1,\ldots,N+1+k\\}$ satisfying
$\text{rank}\\{V_{v},V_{j_{i}}\\}_{v\in J^{j}_{r^{j}_{i}}}=3$ for all
$1\leqslant i\leqslant k+2$.

Indeed, consider $N$ sets $J_{1},\ldots,J_{N}$ and $j_{1}$. Assume that
$\text{rank}\\{V_{j_{1}},V_{t_{2}}\ldots,V_{t_{u}}\\}=1$ where $u$ is maximal.
By the assumption, we have $1\leqslant u\leqslant N-1.$ It follows that there
exist $N-u$ pairs, for instance $\\{V_{v}\\}_{v\in
J_{1}},\ldots,\\{V_{v}\\}_{v\in J_{N-u}}$ which do not contain $V_{j_{1}}$ or
$V_{t_{i}}$ with $2\leqslant i\leqslant u$. Obviously, $N-u\geq 1$. Without
loss of generality, we can assume that $V_{j_{1}}\in\\{V_{v}\\}_{v\in J_{N}}$.

If $u=N-1$ then obviously, $\text{rank}\\{V_{v},V_{j_{1}}\\}_{v\in J_{1}}=3$
since
$\sharp(\\{V_{j_{1}},V_{t_{2}},\ldots,V_{t_{N-1}}\\}\cup\\{V_{v}\\}_{v\in
J_{1}})=N+1.$

If $u\leqslant N-2$, there are at least two pairs vectors, which do not
contain $V_{j_{1}}$ or $V_{t_{i}}$ with $2\leqslant i\leqslant u$. Assume that
$V_{j_{1}}\in$ span$\\{V_{v}\\}_{v\in J_{r_{1}}}$ with some
$r_{1}\in\\{1,\ldots,N-u\\}$, there exists at least one pair, for instance
$\\{V_{v}\\}_{v\in J_{j_{0}}}$ with $j_{0}\in\\{1,\ldots,N-u\\}$ such that
$\text{rank}\\{V_{v}\\}_{v\in(J_{r_{1}}\cup J_{j_{0}})}=3$. Indeed, otherwise
$\text{rank}\\{V_{v}\\}_{v\in(\cup_{i=1}^{N-u}J_{i})\cup\\{j_{1},t_{2}\ldots,t_{u}\\}}=\text{rank}\\{V_{v}\\}_{v\in
J_{r_{1}}}=2$. This is impossible since
$\\{V_{v}\\}_{v\in(\cup_{i=1}^{N-u}J_{i})\cup\\{j_{1},t_{2}\ldots,t_{u}\\}}$
has at least $N+2$ vectors. From sets $\\{V_{v}\\}_{v\in J_{r_{1}}}$ and
$\\{V_{v}\\}_{v\in J_{j_{0}}}$, we can rebuild two linearly independent pairs
$\\{V_{i_{1}},V_{i_{2}}\\}$ and $\\{V_{i_{3}},V_{i_{4}}\\}$ such that
$\text{rank}\\{V_{i_{1}},V_{i_{2}},V_{j_{1}}\\}=3$, where
$\\{i_{1},i_{2},i_{3},i_{4}\\}=J_{r_{1}}\cup J_{j_{0}}.$ We redenote by
$J_{r_{1}}=\\{i_{1},i_{2}\\}$ and $J_{j_{0}}=\\{i_{3},i_{4}\\}$.

Therefore, we obtain a partition still denoted by $J_{1},\ldots,J_{N+1+k}$
such that there exists an index ${r^{j}_{1}}\in\\{1,\ldots,N\\}$ satisfying
$\text{rank}\\{V_{v},V_{j_{1}}\\}_{v\in J_{r^{j}_{1}}}=3$.

Next, we consider $N$ sets
$J_{1},\ldots,J_{r^{j}_{1}-1},J_{r^{j}_{1}+1},\ldots,J_{N+1}$ and $j_{2}$.
Repeating the above argument, we get a partition still denoted by
$J_{1},\ldots,J_{q/2}$ such that there exists an index
${r^{j}_{2}}\in\\{1,\ldots,{r^{j}_{1}-1},{r^{j}_{1}+1},\ldots,N+1\\}$
satisfying $\text{rank}\\{V_{v},V_{j_{2}}\\}_{v\in J_{r^{j}_{2}}}=3$. Of
course, this partition still satisfies $\text{rank}\\{V_{v},V_{j_{1}}\\}_{v\in
J_{r^{j}_{1}}}=3$.

Continue to the process, after $k+2$ times, we will obtain a new partition
denoted by $J^{j}_{1},\ldots,J^{j}_{N+1+k}$ such that there exists $k+2$
indices ${r^{j}_{1},\ldots,r^{j}_{k+2}}\in\\{1,\ldots,N+1+k\\}$ satisfying
$\text{rank}\\{V_{v},V_{j_{i}}\\}_{v\in J^{j}_{r^{j}_{i}}}=3$ for all
$1\leqslant i\leqslant k+2$. The claim is proved.

Put $I^{j}_{r^{j}_{i}}=J^{j}_{r^{j}_{i}}\cup\\{j_{i}\\}$, then
$P_{{I^{j}_{r^{j}_{i}}}}\not\equiv 0$ for all $1\leqslant i\leqslant k+2$.

For each remained index
$i\in\\{1,\ldots,N+1+k\\}\setminus\\{r^{j}_{1},\ldots,r^{j}_{k+2}\\}$, we
choose a vector $V_{s_{i}}$ such that $\text{rank}\\{V_{v}\\}_{v\in
J^{j}_{i}\cup\\{s_{i}\\}}=3.$ Put $I^{j}_{i}=J^{j}_{i}\cup\\{s_{i}\\}$, then
$P_{{I^{j}_{i}}}\not\equiv 0$ for all $i.$

$\bullet$ If $k=0$ then $l=2N+1$ and $q=2N+2$. Put
$S_{1}=\\{1\\},S_{2}=\\{2\\},\ldots,S_{l-1}=\\{2N\\},S_{l}=\\{2N+1,2N+2\\}.$

$\bullet$ If $k>0$ then $q=(k+2)(l-1)+t$ with $0<t\leqslant k+2.$ Put
$S_{1}=\\{1,\ldots,k+2\\},S_{2}=\\{(k+2)+1,\ldots,2(k+2)\\},\ldots,S_{l-1}=\\{(k+2)(l-2)+1,\ldots,(k+2)(l-1)\\},S_{l}=\\{(k+2)(l-1)+1,\ldots,2N+2+2k\\}.$

Applying the claim to each set $S_{j}$ $(1\leqslant j\leqslant l)$, we get a
partition $J^{j}_{1},\ldots,J^{j}_{N+1+k}$ with $s_{j}=\sharp S_{j}$ indices
${r^{j}_{1},\ldots,r^{j}_{s_{j}}}\in\\{1,\ldots,N+1+k\\}$ satisfying
$\text{rank}\\{V_{v},V_{u}\\}_{v\in J^{j}_{r^{j}_{i}},u\in S_{j}}=3$ for all
$1\leqslant i\leqslant s_{j}$.

We put

$P_{Q}=\prod_{j=1}^{l}\prod_{i=1}^{N+1+k}P_{I^{j}_{i}},$

where $I^{j}_{i}$ is defined as in the above.

Since $(\min\\{a,b,c\\}-1)\geq\min\\{a,n\\}+\min\\{b,n\\}+\min\\{c,n\\}-2n-1$
for any positive integers $a,b,c$, we have

$\displaystyle\min_{1\leqslant u\leqslant 3}\\{\nu_{(f^{u},H_{v}),\leqslant
k_{v}}(z)\\}-\nu^{[1]}_{(f^{k},H_{v}),\leqslant k_{v}}(z)$
$\displaystyle\geq\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z)-(2n+1)\nu^{[1]}_{(f^{k},H_{v}),\leqslant k_{v}}(z),$

for all $z\in\mathrm{Supp}\,\nu_{(f^{k},H_{v}),\leqslant k_{v}}$.

Putting $\nu_{v}(z)=\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z)-(2n+1)\nu^{[1]}_{(f^{k},H_{v}),\leqslant k_{v}}(z)\ (1\leqslant
k\leqslant 3,\ v\in Q),$ from Lemma 2.6, we have

$\displaystyle\nu_{P_{{I}^{j}_{i}}}(z)\geq\sum_{v\in{I}^{j}_{i}}\nu_{v}(z)+2\sum_{v=1}^{q}\nu^{[1]}_{(f^{k},H_{v}),\leqslant
k_{v}}(z)$

and

$\displaystyle\nu_{P_{{I}^{j}_{i}}}(z)\geq\sum_{v\in{J}^{j}_{i}}\nu_{v}(z)+2\sum_{v=1}^{q}\nu^{[1]}_{(f^{k},H_{v}),\leqslant
k_{v}}(z).$

Note that for $k=0$ then $l(q-2N-1)-(2N+1)=0$. For $k>0$ then
$2N+1\leqslant\frac{q}{k+2}(2k+1)\leqslant l(2k+1)=l(q-2N-1)$. Therefore, we
always have $l(q-2N-1)-(2N+1)\geqslant 0.$ It implies that
$l(q-2n-1)-(2n+1)\geqslant 0$ since $N\geq n.$ Then, for all
$z\not\in\mathcal{S}$, we obtain

$\displaystyle\nu_{P_{Q}}(z)$ $\displaystyle\geq
l\sum_{v=1}^{q}\nu_{v}(z)+\sum_{v=1}^{q}\nu_{v}(z)+lq\sum_{v=1}^{q}\nu^{[1]}_{(f^{k},H_{v}),\leqslant
k_{v}}(z)$
$\displaystyle=(l+1)\sum_{v=1}^{q}(\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z)-(2n+1)\nu^{[1]}_{(f^{k},H_{v}),\leqslant
k_{v}}(z))+lq\sum_{v=1}^{q}\nu^{[1]}_{(f^{k},H_{v}),\leqslant k_{v}}(z)$
$\displaystyle=(l+1)\sum_{v=1}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z)+\big{(}l(q-2n-1)-(2n+1)\big{)}\sum_{v=1}^{q}\nu^{[1]}_{(f^{k},H_{v}),\leqslant
k_{v}}(z)$
$\displaystyle\geq\left(l+1+\frac{l(q-2n-1)-(2n+1)}{3n}\right)\sum_{v=1}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z)$
$\displaystyle\geq\frac{l(q+n-1)+n-1}{3n}\sum_{v=1}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z).$

We put $P:=P_{Q}$. The above inequality implies that

$\displaystyle\sum_{v=1}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z)\leqslant\frac{3n}{l(q+n-1)+n-1}\nu_{P}(z),\forall
z\not\in\mathcal{S}.$

Define $\beta:=\dfrac{3n}{l(q+n-1)+n-1}$ and $\gamma:=\dfrac{lq}{2}$.

$\bullet$ Case 2: $q\mod 2=1$.

By the assumption, we have $q-1=2N+2+2k.$ We consider any subset
${R}=\\{j_{1},\ldots,j_{q-1}\\}$ of $\\{1,\ldots,q\\}$. By the same argument
as in Case 1 for $R$, we get

$\displaystyle\nu_{P_{R}}(z)$
$\displaystyle\geq(l+1)\sum_{v=1}^{q-1}\nu_{j_{v}}(z)+l(q-1)\sum_{v=1}^{q}\nu^{[1]}_{(f^{k},H_{v}),\leqslant
k_{v}}(z),\forall z\not\in\mathcal{S}.$

We now define $P:=\prod_{|R|=q-1}P_{R},$ so we obtain

$\displaystyle\nu_{P}(z)$ $\displaystyle=\sum_{|R|=q-1}\nu_{P_{R}}$
$\displaystyle\geq(q-1)(l+1)\sum_{v=1}^{q}\nu_{v}(z)+ql(q-1)\sum_{v=1}^{q}\nu^{[1]}_{(f^{k},H_{v}),\leqslant
k_{v}}(z)$
$\displaystyle\geq(q-1)\frac{l(q+n-1)+n-1}{3n}\sum_{v=1}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z).$

Hence, we have

$\displaystyle\sum_{v=1}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{v}),\leqslant
k_{v}}(z)\leqslant\frac{3n}{(l(q+n-1)+n-1)(q-1)}\nu_{P}(z),\forall
z\not\in\mathcal{S}.$

Define $\beta:=\dfrac{3n}{\big{(}l(q+n-1)+n-1\big{)}(q-1)}$ and
$\gamma:=\dfrac{(q-1)lq}{2}.$ Then, from all the above cases, we always get

$\alpha:=\beta\gamma=\dfrac{3nlq}{2(l(q+n-1)+n-1)}=\frac{3nq}{2\big{(}q+(n-1)\frac{l+1}{l}\big{)}},$

and

$\displaystyle\sum_{u=1}^{3}\sum_{v=1}^{q}\nu_{(f^{u},H_{v}),\leqslant
k_{v}}^{[n]}(z)\leqslant\beta\nu_{P}(z),\forall z\not\in\mathcal{S}.$

It is easy to see that $|P|^{\beta}\leqslant
C(\|f^{1}\|\|f^{2}\|\|f^{3}\|)^{\beta\gamma}=C(\|f^{1}\|\|f^{2}\|\|f^{3}\|)^{\alpha}$,
where $C$ is some positive constant. This means that $P^{\beta}\in
B(\alpha,0;f^{1},f^{2},f^{3})$. Applying Lemma 2.8, we obtain

$\displaystyle q$ $\displaystyle\leqslant
2N-n+1+\sum_{j=1}^{q}\frac{n}{k_{j}+1}+\rho n(2N-n+1)+\alpha$
$\displaystyle=2N-n+1+\sum_{j=1}^{q}\frac{n}{k_{j}+1}+\rho
n(2N-n+1)+\frac{3nq}{2\big{(}q+(n-1)\frac{l+1}{l}\big{)}},$

which contradicts the assumption. Therefore, $f^{1}\wedge f^{2}\wedge
f^{3}\equiv 0$ on $M$. The proof of Lemma 3.3 is complete. ∎

By basing on the proofs of Quang [18, Lemma 3.3, 3.4, 3.5, 3.6] or [19, Lemma
4.4, 4.5, 4.6, 4.8], we obtain the following Lemmas which are necessary for
the proof of our theorem.

The first, for three mappings
$f^{1},f^{2},f^{3}\in\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$, we define

$\bullet F^{ij}_{k}=\frac{(f^{k},H_{i})}{(f^{k},H_{i})},\ \ 0\leqslant
k\leqslant 2,\ 1\leqslant i,j\leqslant q,$

$\bullet
V_{i}=((f^{1},H_{i}),(f^{2},H_{i}),(f^{3},H_{i}))\in\mathcal{M}^{3}_{m},$

$\bullet\nu_{i}:\text{ the divisor whose support is the closure of the set }$
$\\{z:\nu_{(f^{u},H_{i}),\leqslant
k_{i}}(z)\geqslant\nu_{(f^{v},H_{i}),\leqslant
k_{i}}(z)=\nu_{(f^{t},H_{i}),\leqslant k_{i}}(z)\text{ for a permutation
}(u,v,t)\text{ of }(1,2,3)\\}.$

We write $V_{i}\cong V_{j}$ if $V_{i}\wedge V_{j}\equiv 0$, otherwise we write
$V_{i}\not\cong V_{j}$. For $V_{i}\not\cong V_{j}$, we write $V_{i}\sim V_{j}$
if there exist $1\leqslant u<v\leqslant 3$ such that $F_{u}^{ij}=F_{v}^{ij}$,
otherwise we write $V_{i}\not\sim V_{j}.$

######  Lemma 3.4.

[18, Lemma 3.3] or [19, Lemma 4.4] With the assumption of Theorem 1.1, let $h$
and $g$ be two elements of the family
$\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$. If there exists a constant
$\lambda$ and two indices $i,j$ such that
$\frac{(h,H_{i})}{(h,H_{j})}=\lambda\frac{(g,H_{i})}{(g,H_{j})},$ then
$\lambda=1.$

######  Lemma 3.5.

[18, Lemma 3.4] or [19, Lemma 4.5] Let $f^{1},f^{2},f^{3}$ be three elements
of $\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$. Suppose that $f^{1}\wedge
f^{2}\wedge f^{3}\equiv 0$ and $V_{i}\sim V_{j}$ for some distinct indices $i$
and $j$. Then $f^{1},f^{2},f^{3}$ are not distinct.

######  Lemma 3.6.

[18, Lemma 3.5] or [19, Lemma 4.6] With the assumption of Theorem 1.1, let
$f^{1},f^{2},f^{3}$ be three maps in
$\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$. Suppose that
$f^{1},f^{2},f^{3}$ are distinct and there are two indices
$i,j\in\\{1,2,\ldots,q\\}\ (i\not=j)$ such that $V_{i}\not\cong V_{j}$ and

$\Phi^{\alpha}_{ij}:=\Phi^{\alpha}(F_{1}^{ij},F_{2}^{ij},F_{3}^{ij})\equiv 0$

for every $\alpha=(\alpha_{1},\ldots,\alpha_{m})\in\mathbb{Z}^{m}_{+}$ with
$|\alpha|=1.$ Then for every $t\in\\{1,\ldots,q\\}\setminus\\{i\\}$, the
following assertions hold:

(i) $\Phi^{\alpha}_{it}\equiv 0$ for all $|\alpha|\leqslant 1,$

(ii) if $V_{i}\not\cong V_{t}$, then $F^{ti}_{1},F^{ti}_{2},F^{ti}_{3}$ are
distinct and there exists a meromorphic function $h_{it}\in
B(0,1;f^{1},f^{2},f^{3})$ such that

$\displaystyle\nu_{h_{ti}}\geq-\nu^{[1]}_{(f,H_{i}),\leqslant
k_{i}}-\nu^{[1]}_{(f,H_{t}),\leqslant
k_{t}}+\sum_{j\not=i,t}\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}.$

######  Lemma 3.7.

[18, Lemma 3.6] or [19, Lemma 4.8] With the assumption of Theorem 1.1, let
$f^{1},f^{2},f^{3}$ be three maps in
$\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$. Assume that there exist
$i,j\in\\{1,2,\ldots,q\\}\ (i\not=j)$ and $\alpha\in\mathbb{Z}^{m}_{+}$ with
$|\alpha|=1$ such that $\Phi^{\alpha}_{ij}\not\equiv 0$. Then there exists a
holomorphic function $g_{ij}\in B(1,1;f^{1},f^{2},f^{3})$ such that

$\displaystyle\nu_{g_{ij}}$
$\displaystyle\geq\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}+\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{j}),\leqslant
k_{j}}+2\sum_{t=1,t\not=i,j}\nu^{[1]}_{(f,H_{t}),\leqslant
k_{t}}-(2n+1)\nu^{[1]}_{(f,H_{i}),\leqslant k_{i}}$
$\displaystyle-(n+1)\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}+\nu_{j}.$

We now prove Theorem 1.1.

Suppose that there exist three distinct meromorphic mappings
$f^{1},f^{2},f^{3}$ belonging to
$\mathcal{F}(f,\\{H_{i},k_{i}\\}_{i=1}^{q},1)$. By Lemma 3.3, we get
$f^{1}\wedge f^{2}\wedge f^{3}\equiv 0.$ We may assume that

${\mathrel{\mathop{{\underbrace{{V_{1}\cong\cdots\cong
V_{l_{1}}}}}}\limits_{{group\
1}}}}\not\cong{\mathrel{\mathop{{\underbrace{V_{l_{1}+1}\cong\cdots\cong
V_{l_{2}}}}}\limits_{{group\
2}}}}\not\cong{\mathrel{\mathop{{\underbrace{V_{l_{2}+1}\cong\cdots\cong
V_{l_{3}}}}}\limits_{{group\
3}}}}\not\cong\cdots\not\cong{\mathrel{\mathop{{\underbrace{V_{l_{s-1}+1}\cong\cdots\cong
V_{l_{s}}}}}\limits_{{group\ s}}}},$

where $l_{s}=q.$

Denote by $P$ the set of all $i\in\\{1,\ldots,q\\}$ satisfying that there
exists $j\in\\{1,\ldots,q\\}\setminus\\{i\\}$ such that $V_{i}\not\cong V_{j}$
and $\Phi^{\alpha}_{ij}\equiv 0$ for all $\alpha\in\mathbb{Z}^{m}_{+}$ with
$|\alpha|\leqslant 1.$ We separate into three cases.

$\bullet$ Case 1: $\sharp P\geq 2.$ It follows that $P$ contains two elements
$i,j.$ We get $\Phi^{\alpha}_{ij}=\Phi^{\alpha}_{ji}=0$ for all
$\alpha\in\mathbb{Z}^{m}_{+}$ with $|\alpha|\leqslant 1.$ By Lemma 2.1, there
exist two functions, for instance $F_{1}^{ij}$ and $F_{2}^{ij}$, and a
constant $\lambda$ such that $F_{1}^{ij}=\lambda F_{2}^{ij}.$ Applying Lemma
3.4, we have $F_{1}^{ij}=F_{2}^{ij}$. Hence, since Lemma 3.6 (ii), we can see
that $V_{i}\cong V_{j}$, i.e., $V_{i}$ and $V_{j}$ belong to the same group in
the partition. We may assume that $i=1$ and $j=2.$ Since our assumption
$f^{1},f^{2},f^{3}$ are distinct, the number of each group in the partition is
less than $N+1.$ Thus, we get $V_{1}\cong V_{2}\not\cong V_{t}$ for all
$t\in\\{N+1,\ldots,q\\}.$ By Lemma 3.6 (ii), we obtain

$\displaystyle\nu_{h_{1t}}\geq-\nu^{[1]}_{(f,H_{1}),\leqslant
k_{1}}-\nu^{[1]}_{(f,H_{t}),\leqslant
k_{t}}+\sum_{s\not=1,t}\nu^{[1]}_{(f,H_{s}),\leqslant k_{s}},$

and

$\displaystyle\nu_{h_{2t}}\geq-\nu^{[1]}_{(f,H_{2}),\leqslant
k_{2}}-\nu^{[1]}_{(f,H_{t}),\leqslant
k_{t}}+\sum_{s\not=2,t}\nu^{[1]}_{(f,H_{s}),\leqslant k_{s}}.$

By summing up both sides of the above two inequalities, we have

$\displaystyle\nu_{h_{1t}}+\nu_{h_{2t}}\geq-2\nu^{[1]}_{(f,H_{t})\leqslant
k_{t}}+\sum_{s\not=1,2,t}\nu^{[1]}_{(f,H_{s}),\leqslant k_{s}}.$

Summing up both sides of the above inequalities over all
$t\in\\{N+1,\ldots,q\\},$ we obtain

$\displaystyle\sum_{t=N+1}^{q}(\nu_{h_{1t}}+\nu_{h_{2t}})$
$\displaystyle\geq(q-N)\sum_{t=3}^{N}\nu^{[1]}_{(f,H_{t})\leqslant
k_{t}}+(q-N-3)\sum_{t=N+1}^{q}\nu^{[1]}_{(f,H_{t})\leqslant k_{t}}$
$\displaystyle\geq(q-N-3)\sum_{t=3}^{q}\nu^{[1]}_{(f,H_{t})\leqslant
k_{t}}\geq\frac{q-N-3}{3n}\sum_{u=1}^{3}\sum_{t=3}^{q}\nu^{[n]}_{(f,H_{t})\leqslant
k_{t}}.$

Hence, we get

$\displaystyle\sum_{u=1}^{3}\sum_{t=3}^{q}\nu^{[n]}_{(f,H_{t})\leqslant
k_{t}}\leqslant\frac{3n}{q-N-3}\nu_{\prod_{t=N+1}^{q}}(h_{1t}h_{2t}).$

Since $({\prod_{t=N+1}^{q}}(h_{1t}h_{2t}))^{\frac{3n}{q-N-3}}\in
B(0,2(q-N)\frac{3n}{q-N-3};f^{1},f^{2},f^{3})$, applying Lemma 2.8, we obtain

$q-2\leqslant
2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+4(q-N)\frac{n}{q-N-3}\big{)}.$

From the definition of $l$ and the condition of $q$, it is easy to see that
$l\geq 3.$ It is easy to see that

$2\leqslant\frac{3nq}{2\big{(}q+n-1+\frac{n-1}{3}\big{)}}\leqslant\frac{3nq}{2\big{(}q+n-1+\frac{n-1}{l}\big{)}},$

and

$4(q-N)\frac{n}{q-N-3}\leqslant\frac{4(q-n)n}{n-1}.$

These inequalities imply that

$q\leqslant
2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{4(q-n)n}{n-1}\big{)}+\frac{3nq}{2\big{(}q+n-1+\frac{n-1}{l}\big{)}},$

which is a contradiction.

$\bullet$ Case 2: $\sharp P=1$. We assume that $P=\\{1\\}.$ It is easy to see
that $V_{1}\not\cong V_{i}$ for all $i=2,\ldots,q$. By Lemma 3.6 (ii), we
obtain

$\displaystyle\nu_{h_{1i}}\geq-\nu^{[1]}_{(f,H_{1})\leqslant
k_{1}}-\nu^{[1]}_{(f,H_{i})\leqslant
k_{i}}+\sum_{s\not=1,t}\nu^{[1]}_{(f,H_{s})\leqslant k_{s}}.$

Summing up both sides of the above inequalities over all $i=2,\ldots,q,$ we
have

(3.8)
$\displaystyle\sum_{i=2}^{q}\nu_{h_{1i}}\geq(q-3)\sum_{i=2}^{q}\nu^{[1]}_{(f,H_{i})\leqslant
k_{i}}-(q-1)\nu^{[1]}_{(f,H_{1})\leqslant k_{1}}.$

Obviously, $i\not\in P$ for all $i=2,\ldots,q.$ Now put

$\sigma(i)=\begin{cases}i+N,&\text{ if }i+N\leqslant q\\\ i-N-q+1,&\text{ if
}i+N>q,\end{cases}$

then $i$ and $\sigma(i)$ belong to distinct groups, i.e., $V_{i}\not\cong
V_{\sigma(i)}$ for all $i=2,\ldots,q$ and hence
$\Phi^{\alpha}_{i\sigma(i)}\not\equiv 0$ for some
$\alpha\in\mathbb{Z}^{m}_{+}$ with $|\alpha|\leqslant 1.$ By Lemma 3.7, we get

$\displaystyle\nu_{g_{i\sigma(i)}}$
$\displaystyle\geq\sum_{u=1}^{3}\sum_{t=i,\sigma(i)}\nu^{[n]}_{(f^{u},H_{t})\leqslant
k_{t}}-(2n+1)\nu^{[1]}_{(f,H_{i})\leqslant
k_{i}}-(n+1)\nu^{[1]}_{(f,H_{\sigma(i)})\leqslant k_{\sigma(i)}}$
$\displaystyle+2\sum_{t=1,t\not=i,\sigma(i)}\nu^{[1]}_{(f,H_{t})\leqslant
k_{t}}.$

Summing up both sides of this inequality over all $i\in\\{2,\ldots,q\\}$ and
using (3.8), we obtain

$\displaystyle\sum_{i=2}^{q}\nu_{g_{i\sigma(i)}}$ $\displaystyle\geq
2\sum_{i=2}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}+(2q-3n-8)\sum_{i=2}^{q}\nu^{[1]}_{(f,H_{i}),\leqslant
k_{i}})+2(q-1)\nu^{[1]}_{(f,H_{1}),\leqslant k_{1}}$ $\displaystyle\geq
2\sum_{i=2}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}+\frac{4q-3n-14}{3}\sum_{u=1}^{3}\sum_{i=2}^{q}\nu^{[1]}_{(f^{u},H_{i}),\leqslant
k_{i}})-2\sum_{i=2}^{q}\nu_{h_{1i}}$
$\displaystyle\geq\frac{4q+3n-14}{3n}\sum_{i=2}^{q}\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}-2\sum_{i=2}^{q}\nu_{h_{1t}}.$

It implies that

$\sum_{u=1}^{3}\sum_{i=2}^{q}\nu^{[n]}_{(f^{u},H_{i})}\leqslant\frac{3n}{4q+3n-14}\nu_{\prod_{i=2}^{q}(g_{i\sigma(i)}h^{2}_{1i})}.$

Obviously, $\prod_{i=2}^{q}(g_{i\sigma(i)}h^{2}_{1i})\in
B(q-1,3(q-1);f^{1},f^{2},f^{3})$. Applying Lemma 2.8, we obtain

$q-1\leqslant
2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{6n(q-1)}{4q+3n-14}\big{)}+\frac{3n(q-1)}{4q+3n-14}.$

Since $q\geq 2n+2$ and by the simple calculation, we have

$\frac{6n(q-1)}{4q+3n-14}\leqslant\frac{6n(q-1)}{11n-6}<\frac{4(q-n)n}{n-1}.$

It implies that

$q\leqslant
2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{4(q-n)n}{n-1}\big{)}+\frac{4q+3nq-14}{4q+3n-14},$

which is a contradiction.

$\bullet$ Case 3: $\sharp P=0$. By Lemma 3.7, for all $i\not=j$, we get

$\displaystyle\nu_{g_{ij}}$
$\displaystyle\geq\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}+\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{j}),\leqslant
k_{j}}+2\sum_{t=1,t\not=i,j}\nu^{[1]}_{(f,H_{t}),\leqslant
k_{t}}-(2n+1)\nu^{[1]}_{(f,H_{i}),\leqslant k_{i}}$
$\displaystyle-(n+1)\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}+\nu_{j}.$

Put

$\gamma(i)=\begin{cases}i+N&\text{ if }i\leqslant q-N\\\ i+N-q&\text{ if
}i>q-N.\end{cases}$

By summing up both sides of the above inequality over all pairs
$(i,\gamma(i)),$ we obtain

(3.9) $\displaystyle\sum_{i=1}^{q}\nu_{g_{i\gamma(i)}}\geq
2\sum_{u=1}^{3}\sum_{i=1}^{q}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}+(2q-3n-6)\sum_{t=1}^{q}\nu^{[1]}_{(f,H_{t}),\leqslant
k_{t}}+\sum_{t=1}^{q}\nu_{t}.$

By Lemma 3.5, we can see that $V_{j}\not\sim V_{l}$ for all $j\not=l.$ Thus,
we have

$P^{i\gamma(i)}_{st}:=(f^{s},H_{i})(f^{t},H_{\gamma(i)})-(f^{t},H_{\gamma(i)})(f^{s},H_{i})\not\equiv
0,\ s\not=t,1\leqslant i\leqslant q.$

We claim that: With $i\not=j\not=\gamma(i)$, for every $z\in f^{-1}(H_{j})$,
we have

$\sum_{1\leqslant s<t\leqslant 3}\nu_{P^{i\gamma(i)}_{st}}(z)\geq
4\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}-\nu_{j}(z).$

Indeed, for $z\in f^{-1}(H_{j})\cap\mathrm{Supp}\,{\nu_{j}},$ we have

$4\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}(z)-\nu_{j}(z)\leqslant
4-1=3\leqslant\sum_{1\leqslant s<t\leqslant 3}\nu_{P^{i\gamma(i)}_{st}}.$

For $z\in f^{-1}(H_{j})\setminus\mathrm{Supp}\,\nu_{j}$, we assume that
$\nu_{(f^{1},H_{j}),\leqslant k_{j}}(z)<\nu_{(f^{2},H_{j}),\leqslant
k_{j}}(z)\leqslant\nu_{(f^{3},H_{j}),\leqslant k_{j}}(z).$ Since $f^{1}\wedge
f^{2}\wedge f^{3}\equiv 0,$ we have $\det(V_{i},V_{\gamma(i)},V_{j})\equiv 0,$
and hence

$(f^{1},H_{j})P^{i\gamma(i)}_{23}=(f^{2},H_{j})P^{i\gamma(i)}_{13}-(f^{3},H_{j})P^{i\gamma(i)}_{12}.$

It implies that $\nu_{P^{i\gamma(i)}_{23}}\geq 2$ and so

$\sum_{1\leqslant s<t\leqslant 3}\nu_{P^{i\gamma(i)}_{st}}(z)\geq
4=4\nu^{[1]}_{(f,H_{i}),\leqslant k_{i}}(z)-\nu_{j}(z).$

The claim is proved.

On the other hand, with $j=i$ or $j=\sigma(i)$, for every $z\in
f^{-1}(H_{j})$, we see that

$\displaystyle\nu_{P^{i\gamma(i)}_{st}}(z)$
$\displaystyle\geq\min\\{\nu_{(f^{s},H_{j}),\leqslant
k_{j}}(z),\nu_{(f^{t},H_{j}),\leqslant k_{j}}(z)\\}$
$\displaystyle\geq\nu^{[n]}_{(f^{s},H_{j}),\leqslant
k_{j}}(z)+\nu^{[n]}_{(f^{t},H_{j}),\leqslant
k_{j}}(z)-n\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}(z).$

Hence, $\sum_{1\leqslant s<t\leqslant 3}\nu_{P^{i\gamma(i)}_{st}}(z)\geq
2\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{j}),\leqslant
k_{j}}(z)-3n\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}(z).$ Together this
inequality with the above claim, we obtain

$\displaystyle\sum_{1\leqslant s<t\leqslant 3}\nu_{P^{i\gamma(i)}_{st}}(z)$
$\displaystyle\geq\sum_{j=i,\gamma(i)}\big{(}2\sum_{u=1}^{3}\nu^{[n]}_{(f^{u},H_{j}),\leqslant
k_{j}}(z)-3n\nu^{[1]}_{(f,H_{j}),\leqslant k_{j}}(z)\big{)}$
$\displaystyle+\sum_{j=1,j\not=i,\gamma(i)}(4\nu^{[1]}_{(f,H_{j}),\leqslant
k_{j}}(z)-\nu_{j}(z)).$

On the other hand, it is easy to see that $\prod_{1\leqslant s<t\leqslant
3}P^{i\gamma(i)}_{st}\in B(2,0;f^{1},f^{2},f^{3}).$ Summing up both sides of
the above inequality over all $i,$ we obtain

$\displaystyle\sum_{i=1}^{q}\sum_{1\leqslant s<t\leqslant
3}\nu_{P^{i\gamma(i)}_{st}}\geq
4\sum_{u=1}^{3}\sum_{i=1}^{q}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}+(4q-6n-8)\sum_{i=1}^{q}\nu^{[1]}_{(f,H_{i}),\leqslant
k_{i}}-(q-2)\sum_{i=1}^{q}\nu_{i}.$

Thus,

$\sum_{i=1}^{q}\nu_{i}+\frac{1}{q-2}\sum_{i=1}^{q}\sum_{1\leqslant
s<t\leqslant
3}\nu_{P^{i\gamma(i)}_{st}}\geq\frac{4}{q-2}\sum_{u=1}^{3}\sum_{i=1}^{q}\nu^{[n]}_{(f^{u},H_{i}),\leqslant
k_{i}}+\frac{4q-6n-8}{q-2}\sum_{i=1}^{q}\nu^{[1]}_{(f,H_{i}),\leqslant
k_{i}}.$

Using this inequality and (3.9), we have

$\displaystyle\sum_{i=1}^{q}\nu_{g_{i\gamma(i)}}$
$\displaystyle+\frac{1}{q-2}\sum_{i=1}^{q}\sum_{1\leqslant s<t\leqslant
3}\nu_{P^{i\gamma(i)}_{st}}$
$\displaystyle\geq\big{(}2+\frac{4}{q-2}\big{)}\sum_{u=1}^{q}\sum_{t=1}^{q}\nu^{[n]}_{(f^{u},H_{t}),\leqslant
k_{t}}+\big{(}n-2+\frac{4q-6n-8}{q-2}\big{)}\sum_{i=1}^{q}\nu^{[1]}_{(f^{u},H_{i}),\leqslant
k_{i}}$
$\displaystyle\geq\big{(}2+\frac{4}{q-2}+\frac{n-2}{3n}+\frac{4q-6n-8}{3n(q-2)}\big{)}\sum_{u=1}^{q}\sum_{t=1}^{q}\nu^{[n]}_{(f^{u},H_{t}),\leqslant
k_{t}}.$

It implies that

$\sum_{u=1}^{q}\sum_{t=1}^{q}\nu^{[n]}_{(f^{u},H_{t}),\leqslant
k_{t}}\leqslant\frac{3n}{6nq+(n-2)(q-2)+4q-6n-8}\nu_{\prod_{i=1}^{q}(g^{q-2}_{i\gamma(i)}P^{i\gamma(i)}_{12}P^{i\gamma(i)}_{13}P^{i\gamma(i)}_{23})}.$

Observe that
$\prod_{i=1}^{q}g^{q-2}_{i\gamma(i)}P^{i\gamma(i)}_{12}P^{i\gamma(i)}_{13}P^{i\gamma(i)}_{23}\in
B(q^{2},q(q-2);f^{1},f^{2},f^{3})$, hence applying Lemma 2.8, we obtain

$\displaystyle q$ $\displaystyle\leqslant
2N-n+1+\sum_{i=1}^{q}\frac{n}{k_{i}+1}+\rho\big{(}n(2N-n+1)+\frac{2nq(q-2)}{6nq+(n-2)(q-2)+4q-6n-8}\big{)}$
$\displaystyle+\frac{3nq^{2}}{6nq+(n-2)(q-2)+4q-6n-8},$

which is impossiple since

$\frac{2nq(q-2)}{6nq+(n-2)(q-2)+4q-6n-8}<\frac{2nq(q-2)}{6nq+q-2}=\frac{2n(q-2)}{6n+1}\leqslant\frac{4(q-n)n}{n-1}.$

The proof of Theorem 1.1 is complete. $\square$

Acknowledgement: This work was done while the first author was staying at the
Vietnam Institude for Advanced Study in Mathematics (VIASM). He would like to
thank VIASM for the support.

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