1. Introduction
Suppose that K is a complete closed field of characteristic 0 and
is a transcendental general meromorphic function in K. Let
be the set of power series with coefficients converging in all K, and let
be a general meromorphic function in K, and if
we denote by
the disk
. For meromorphic function in a first order system and factorization of p-adic meromorphic functions, see [1] [2] [3].
Definition 1. Given a meromorphic function in
, we call exceptional value of f (or Picard value of f) a value
such that
has no zero. And, if f is transcendental, we call quasi-exceptional value a value
such that
has finitely many zeros (see [4] ). Also see [5] [6] [7] for meromorphic function with doubly periodic phase and with the uniqueness sharing a value.
Let
be the set of power series in
with coefficient in K whose radius of convergence is
and
be the field of fraction of
for more details (see [4] [8] [9] [10] ). So, the function
is an entire function admitting as zeros the distinct zeros of
, all with order 1. We can then set
where the function
is an entire function admitting for zeros the multiple zeros of
, each with order
when it is a zero of
of order q. Particularly, if
is constant, we set
and
.
According to the p-adic Hayman conjecture, for every
,
takes every non-zero value infinitely many times (see [8] [9] [10] [11] [12] ).
Now,
is a power series of infinite radius of convergence. According to classical notation [13], we set
.
We know that
That notation defines an absolute value on
and has continuation to
as
with
. In the paper [11], the Theorem 1 is proven. In this paper, we use information from related literature and formulate the method of Bezivin, J., Boussaf, K. and Escassut, A. [4] by using a general meromorphic function to show that for every
has infinitely many zeros and
has no practically exceptional value.
2. Theorems and Lemmas
Theorem 1. Let
be a transcendental general meromorphic function on
having finitely many multiple poles. Then
takes every value infinitely many times.
That has suggested the following conjecture:
Conjecture 1. Let
be a general meromorphic function on
such that
has finitely many zeros. Then
is a rational function.
Now we will define new expressions:
Let
. For each
, we denote by
the number of multiple zeros of
in
, each counted with its multiplicity and we set
.
Similarly, we denote by
the number of zeros of
in
, taking multiplicity into account and set
.
We need several lemmas:
Lemma 1. Let
have no common zero and let
. If
has finitely many zeros, there exists a polynomial
such that
.
Proof. If V is a constant, the statement is obvious. So, we assume that V is not a constant. Now
divides
and hence
factorizes in the way
with
. Then no zero of Y can be a zero of V. Consequently, we have
The two functions
and
have no common zero since neither have U and V. Consequently, the zeros of
are those of
which therefore has finitely many zeros and consequently is a polynomial.
Lemma 2 is known as the p-adic Schwarz Lemma (Lemma 23.12 [14] ). Lemmas 3 and 4 are immediate corollaries:
Lemma 2. Let
be such that
and let
admits zeros and t poles in
and no zero and no pole in
. Then
.
Lemma 3. Let
be such that
and let
have q zeros in
. Then
.
Lemma 4. Let
. Then
is a polynomial of degree q if and only if there exists a constant c such that
.
Let
be the disc
. We denote by
the
-algebra of analytic functions in
, i.e. the set of power series in
with coefficients in
whose radius of convergence is
and we denote by
the field of general meromorphic functions in
, i.e. the field of fraction of
.
Lemma 5. Let
. For each
, and
, we have
.
Proof. Suppose first f belongs to
and set
.
Then
.
The statement then is immediate. Consider now the general case and set
with
. The stated inequality is obvious when
. So, we assume it holds for
and consider
. Writing
, by Leibniz Theorem we have
and hence
Now,
and for each
, we have
and
.
Therefore, we can derive that terms on the right hand side are upper bounded by
and hence the conclusion holds for
.
Lemma 6. Let
and let
. For all
with
and
, we have the inequality:
Proof. By Taylor’s formula at the point x, we have
Now,
.
But we have
, hence
And we notice that
. Consequently, we can define
and we have
We can check that the function h defined in
as
reaches it maximum at the point
.
Consequently,
and therefore
Theorem 2. Let f be a meromorphic function on
such that, for some
satisfies
in
. If
has finitely many zeros, then f is a rational function.
Proof. Suppose
has finitely many zeros. If V is a constant, the statement is immediate. So, we suppose V is not a constant and hence it admits at least one zero a. By Lemma1 there exists a polynomial
such that
. Next, we take
such that
and
. By Lemma 6 we have
Notice that
because U and V have no common zero. Now set
and take
. Setting
, we have
Then taking the supremum of
inside the disc
, we can derive
(1)
Let us apply Lemma 3, by taking
, after noticing that the number of zeros of
is bounded by
. So, we have
(2)
Now, due to the hypothesis:
in
, we have
(3)
The function
is continuous on
and equivalent to
when r tends to
. Consequently, it is bounded on
.
Therefore, by (2) and (3) there exists a constant
such that, for all
by (3) we obtain
(4)
On the other hand,
clearly satisfies an inequality of the form
in
with
. Moreover, we can obviously find positive constants
such that
.
Consequently, by (1) and (4) we can find positive constants
such that
. Thus, writing again
, we have
and hence
, consequently, by Lemma 4,
is a polynomial of degree
and hence it has finitely many zeros and so does. And then, by Theorem 1, f must be a rational function.
3. Main Results
The main generalized meromorphic results are the following corollaries and theorem.
Corollary 1. Let
. For each
, and
, we have
.
Proof. Suppose first
belongs to
and set
then
The statement then is immediate. Consider now the general case and set
with
. The stated inequality is obvious when
. So, we assume it holds for
and consider
.
Writing
, by Leibniz Theorem we have
and hence
Now,
and for each
, we have
and
.
Therefore, we can derive that terms on the right hand side are upper bounded by
and hence the conclusion holds for
.
Corollary 2. Let
and let
. For all
with
and
, we have the inequality:
Proof. By Taylor’s formula at the point
, we have
Now,
.
But we have
, hence
And we notice that
. Consequently, we can define
and we have
We can check that the function h defined in
as
reaches it maximum at the point
.
Consequently,
and therefore
Theorem 3. Let
be a general meromorphic function on
such that, for some
,
satisfies
in
. If
has finitely many zeros, then
is a rational function.
Proof. Suppose
has finitely many zeros. If
is a constant, the statement is immediate. So, we suppose
is not a constant and hence it admits at least one zero . By Lemma 4, there exists a polynomial
such that
. Next, we take
such that
and
. By Lemma 6 we have
Notice that
because
and
have no common zero. Now set
and take
. Setting
, we have
Then taking the supremum of
inside the disc
, we can derive
(5)
Let us apply Lemma 3, by taking
, after noticing that the number of zeros of
is bounded by
. So, we have
(6)
Now, due to the hypothesis:
in
, we have
(7)
The function
is continuous on
and equivalent to
when
tends to
. Consequently, it is bounded on
. Therefore, by (5) and (6) there exists a constant
such that, for all
by (6) we obtain
(8)
On the other hand,
clearly satisfies an inequality of the form
in
with
. Moreover, we can obviously find positive constants
such that
Consequently, by (5) and (6) we can find positive constants
such that
Thus, writing again
, we have
and hence
, consequently, by Lemma 4,
is a polynomial of degree
and hence it has finitely many zeros and so does. And then, by Theorem 1,
must be a rational function.
Corollary 3. Let
be a general meromorphic function on
. Suppose that there exist
, such that
.
If
has finitely many zeros for some
, with
then
is a rational function.
Proof. Suppose
is transcendental. Due to hypothesis,
satisfies
hence by Theorem 3,
has no practically exceptional value.
Corollary 4. Let
be a transcendental general meromorphic function on
such that, for some
, we have
in
. Then for every
,
,
has infinitely many zeros.
Proof. Suppose
admits a practically exceptional value
.
Then
is of the form
with
and h a transcendental entire function.
Consequently there exists
such that
and hence
. Then by Lemma 3, the numbers of zeros and poles of
in disks
are equal when
. So, there exists
such that for every
we have
(9)
On the other hand, of course we have
, hence by (9) and by hypothesis of corollary 4, we have
. Therefore by Theorem 2,
has no practically exceptional value, a contradiction.