On Picard value problem of some difference polynomials
On Picard value problem of some difference polynomials
Zinelâabidine Latreuch
Benharrat Belaïdi
Mathematics Subject Classification
In this paper, we study the value distribution of zeros of certain nonlinear difference polynomials of entire functions of finite order. Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna's value distribution theory [10,13]. In addition, we will use ρ ( f ) to denote the order of growth of f , we say that a meromorphic function a (z) is a small function of f (z) if T (r, a) = S (r, f ) , where S (r, f ) = o (T (r, f )) , as r → +∞ outside of a possible exceptional set of finite logarithmic measure, we use S ( f ) to denote the family of all small functions with respect to f (z). For a meromorphic function f (z) , we define its shift by fc (z) = f (z + c) . In 1959, Hayman proved in [11] that if f is a transcendental entire function, then f n f assume every nonzero complex number infinitely many times, provided that n ≥ 3. Later, Hayman [12] conjectured that this result remains to be valid when n = 1 and n = 2. Then Mues [18] confirmed the case when n = 2 and BergweilerEremenko [2] and ChenFang [3] confirmed the case when n = 1, independently. Since then, there are many research publications (see [17]) regarding this type of Picard value problem. In 1997, Bergweiler obtained the following result. Theorem 1.1 [1] If f is a transcendental meromorphic function of finite order and q is a not identically zero polynomial, then f f − q has infinitely many zeros.

In the same paper, Laine and Yang showed that Theorem 1.2 does not remain valid for the case n = 1. Indeed,
take f (z) = ez + 1. Then
f (z) f (z + π i ) − 1 = 1 + ez
1 − ez
− 1 = −e2z .
After their, a stream of studies on the value distribution of nonlinear difference polynomials in f has been
launched and many related results have been obtained, see, e.g. [
5,14–16
] . For example, Liu and Yang improved
the previous result and obtained the following.
Theorem 1.3 [
15
] Let f (z) be a transcendental entire function of finite order, and c be a nonzero complex
constant. Then for n ≥ 2, f n (z) f (z + c) − p (z) has infinitely many zeros, where p (z) ≡ 0 is a polynomial
in z.
Hence, it is natural to ask: What can be said about the value distribution of f (z) f (z + c) − q(z); when f
is a transcendental meromorphic function and q be a not identically zero small function of f? In this paper, as
an attempt in resolving this question, we obtain the following results.
Theorem 1.4 Let f be a transcendental entire function of finite order, let c1, c2 be two nonzero complex
numbers such that f (z + c1) ≡ f (z + c2) and q be not identically zero polynomial. Then f (z) f (z + c1) −
q (z) and f (z) f (z + c2) − q (z) at least one of them has infinitely many zeros.
The following corollary arises directly from Theorems 1.4 and 1.3.
Corollary 1.5 Let n ≥ 1 be an integer and let c1, c2 (c1c2 = 0) be two distinct complex numbers. Let α, β,
p1, p2 and q (≡ 0) be nonconstant polynomials. If f is a finite order transcendental entire solution of
f n (z) f (z + c1) − q (z) = p1 (z) eα(z)
f n (z) f (z + c2) − q (z) = p2 (z) eβ(z) ,
then n = 1 and f must be a periodic function of period c1 − c2.
2 Some lemmas
The following lemma is an extension of the difference analogue of the Clunie lemma obtained by Halburd and
Korhonen [8].
Lemma 2.1 [
4
] Let f (z) be a nonconstant, finite order meromorphic solution of
f n P (z, f ) = Q (z, f ) ,
where P (z, f ) , Q (z, f ) are the difference polynomials in f (z) with meromorphic coefficients a j (z)
( j = 1, . . . , s) , and let δ < 1. If the degree of Q (z, f ) as a polynomial in f (z) and its shifts is at most
n, then
m (r, P (z, f )) = o
T (r + c , f )
r δ
+ o (T (r, f )) + O ⎝
⎛
s
j=1
⎞
m r, a j ⎠ ,
for all r outside an exceptional set of finite logarithmic measure.
Lemma 2.2 [
6
] Let f (z) be a nonconstant, finite order meromorphic function and let c = 0 be an arbitrary
complex number. Then
T (r, f (z + c)) = T (r, f (z)) + S (r, f ) .
Lemma 2.3 [
7
] Let f (z) be a transcendental meromorphic function of finite order ρ , and let ε > 0 be a
given constant. Then, there exists a set E0 ⊂ (1, +∞) that has finite logarithmic measure, such that for all z
satisfying z ∈/ E0 ∪ [
0, 1
] , and for all k, j, 0 ≤ j < k, we have
f (k) (z)
f ( j) (z)
≤ z(k− j)(ρ−1+ε) .
The following lemma is the lemma of the logarithmic derivative.
Lemma 2.6 Let f (z) be a transcendental meromorphic solution of the system
m r,
f (z + η1)
f (z + η2)
= O r σ −1+ε .
f (z) f (z + c1) − q (z) = p1 (z) eα(z),
f (z) f (z + c2) − q (z) = p2 (z) eβ(z),
where α, β are polynomials and p1, p2, q are not identically zero rational functions. If N (r, f ) = S (r, f ) ,
then
deg α = deg β = deg (α + β) = ρ ( f ) > 0.
Proof First, we prove that deg α = ρ ( f ) and by the same we can deduce that deg β = ρ ( f ) . It is clear from
(2.1) that deg α ≤ ρ ( f ) . Suppose that deg α < ρ ( f ) , this means that
f (z) f (z + c1) := F = q (z) + p1 (z) eα(z) ∈ S ( f ) .
Applying Lemmas 2.1 and 2.2 into (2.2) , we obtain T (r, fc) = T (r, f ) = S (r, f ) which is a contradiction.
Assume now that deg (α + β) < ρ ( f ) , this leads to p1 p2eα+β ∈ S ( f ) . From this and (2.1), we have
Lemma 2.4 [
10
] Let f be a meromorphic function and let k ∈ N. Then
f (k)
f
f (k)
f
m r,
= S (r, f ) ,
m r,
= O (log r ) .
where S (r, f ) = O (log T (r, f ) + log r ) , possibly outside a set E1 ⊂ [0, +∞) of a finite linear measure. If
f is of finite order of growth, then
The following lemma is a difference analogue of the lemma of the logarithmic derivative for finite order
meromorphic functions.
Lemma 2.5 [
6,8,9
] Let η1, η2 be two arbitrary complex numbers such that η1 = η2 and let f (z) be a finite
order meromorphic function. Let σ be the order of f (z). Then for each ε > 0, we have
(2.1)
(2.2)
where
and
which leads to
It is clear that P (z, f ) ≡ 0, and using Lemma 2.1, we get
f 2 P (z, f ) = p1 p2eα+β + q2,
P (z, f ) = a (z) f 2 − b (z)
a =
fc1 fc2
f f
, b = q
fc1 fc2
f + f
.
m (r, P (z, f )) = S (r, f )
2T (r, f ) = m r,
b (z) + P (z, f )
a (z)
which is a contradiction. Hence, deg (α + β) = deg α = deg β. Finally, using Lemma 2.1, it is easy to see that
both of α and β are nonconstant polynomials.
3 Proof of Theorem 1.4
We shall prove this theorem by contradiction. Suppose contrary to our assertion that both of f (z) f (z + c1) −
q (z) and f (z) f (z + c2) − q (z) have finitely many zeros. Then, there exist four polynomials α, β, p1 and
p2 such that
and
By differentiating (3.1) and eliminating eα, we get
f (z) f (z + c1) − q (z) = p1 (z) eα(z)
f (z) f (z + c2) − q (z) = p2 (z) eβ(z).
p
where A1 = p11 + α , B1 =
p
p11 + α
A1 f fc1 − f fc1 − f fc1 = B1,
q − q . By Lemma 2.6, we have
deg α = deg β = deg (α + β) = ρ ( f ) > 0.
Now, we prove that A1 ≡ 0. To show this, we suppose the contrary. Then, there exists a constant A such that
A = p1 (z) eα, which implies the contradiction deg α = ρ ( f ) = 0. By the same, we can prove that B1 ≡ 0.
By the same arguments as above, (3.2) gives
A2 f fc2 − f fc2 − f fc2 = B2,
p
where A2 = p22 + β and B2 =
p
p22 + β
(3.3) and (3.4) by f 2, we get for each ε > 0
q − q . Obviously, A2 ≡ 0 and B2 ≡ 0. Dividing both sides of
2m r,
1
f
fci
f
≤ m r,
+ m r,
+ m r,
= O r ρ−1+ε
+ O (log r ) = S (r, f ) .
f fci
f f
fci fci
fci f
+ O (log r )
So, by the first fundamental theorem, we deduce that
(3.1)
(3.2)
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
(3.8)
(3.9)
T (r, f ) = N r,
It is clear from (3.3) and (3.4) that any multiple zero of f is a zero of Bi (i = 1, 2) . Hence,
where N(2 r, 1f
denotes the counting function of zeros of f whose multiplicities are not less than 2. It follows
by this and (3.5) that
T (r, f ) = N1) r,
and (3.4) , for every zero z0 such that f (z0) = 0 which is not zero or pole of B1 and B2, we have
where N1) r, 1f
and
By (3.7) and (3.8), we obtain
which means that the function B2 fc1 −B1 fc2 has at most a finite number of simple poles. We consider two cases:
f
Case 1. B2 fc1 − B1 fc2 ≡ 0. Set
h (z) =
B1 A2
B2
f fc2 + BB21 f fc2 + BB21 f fc2 = −B1.
A1h
B2 −
and
≡ 0. Then, by the definition of A1 and by simple integration,
A1h h
B2 − B2
Then, we obtain
h
p1eα = C1 B2 ,
where C1 is a nonzero constant. This implies that deg α = ρ ( f ) − 1, which is a contradiction. Hence,
≡ 0. Next, we shall prove AB1B21 − BB21 − BB1A22 ≡ 0. Suppose that AB1B21 − BB21 − BB1A22 ≡ 0.
pp21 eα−β = C2 BB21 := γ ,
where C2 is a nonzero constant and γ is a small function of f. From (3.1) and (3.2), we get
f fc1 − γ fc2 = (1 − γ ) q.
(3.10)
(3.11)
(3.12)
(3.13)
(3.15)
If γ ≡ 1, then by applying Clunie’s lemma to (3.15), we obtain
By this and (3.15), we have
,
N =
2h
B2
( A1 − A2) BB21 −
B1
B2
ϕ (z) = BB21 M + Bh2 ,
ψ = BB21 N .
Differentiation of (3.16) gives
fc2 = M f +
Substituting (3.16) and (3.18) into (3.4), we get
M + N
f + N f .
Differentiating (3.19), we get
M − A2 M f 2 +
N − A2 N + 2M f f + N
f 2 + f f
= −B2.
M − A2 M
+ 2N − A2 N + 2M f 2 + f f + N 3 f f + f f = −B2.
Suppose z0 is a simple zero of f and not a zero or pole of B2. Then from (3.19) and (3.20), we have
B2
N f + f
2N − A2 N + 2M f + 3N f
− B2 N f + 3B2 N f . Therefore, the function
satisfies T (r, H ) = S (r, f ) and
f = 3BH2 N f +
Substituting (3.21) into (3.19), we get
B1
B2
−
(3.16)
(3.17)
(3.18)
(3.19)
(3.20)
(3.21)
(3.22)
H
q1 = M − A2 M + 3B2 ,
q2 = 31 N + 31 BB22 − 2 A2 N + 43 M, q3 = N .
where c is a nonzero constant, this leads to the contradiction deg (α + β) < deg α = deg β. Hence, q2 ≡ 0.
Differentiating (3.22), we obtain
Therefore, z0 is a zero of B2 q2 + q3 − B2q3 f + 2B2q3 f . Hence, the function
where
We prove first q2 ≡ 0. Suppose the contrary. Then satisfies T (r, R) = S (r, f ) and
Substituting (3.25) into (3.24)
q2 R
q1 + 2B2q3
f 2 + 2q1 + q2 + 21 BB22 q2 − 2 q2 + q3 q3 + B2
1 q2 R
f f
+ BB2q23 f 2 = −B2.
Combining (3.26) and (3.22), we obtain
From (3.27), we deduce that
q2 R B2 q1 f + 2q1 + q2 − 21 BB22 q2 − 21 q2 + q3 qq23 + BR2
q1 + 2B2q3 − B2
f = 0.
q2 R B2 q1 = 0
q1 + 2B2q3 − B2
(3.23)
(3.24)
(3.25)
(3.26)
(3.27)
and
By eliminating R from the above two equations, we obtain
Thus, Eq. (3.25) can be rewritten as
Subcase 1. If 4q1q3 − q22 ≡ 0, then from (3.28), we have
f =
(3.28)
(3.29)
(3.30)
(3.31)
On the other hand,
Hence,
On the other hand,
By the definition of Ai (i = 1, 2) and simple integration, we deduce that
which is a contradiction.
Subcase 2. If 4q1q3 ≡ q22, then from (3.29) and (3.21), we have
deg (α + β) < deg α = deg β
B2 q1 q1 H
B2 q2 − q2 = 3B2 N
.
q1
q3 −
M − A2 M
N
H
= 3B2 N .
Combining (3.30) and (3.31), we obtain
function, we obtain
Dividing both sides of the above equation by (A1+A2)2 and since lim R(z) = 0 if R is a nonzero rational
R (z)
2 z→∞
A2 A1
( A1 + A2)
2
On the other hand, since ρ (h) ≤ ρ ( f ) − 1 and by Lemma 2.3
(3.32) and (3.33), we deduce
for all z satisfying z ∈/ E0 ∪ [
0, 1
] , where E0 ⊂ (1, ∞) is a set of finite logarithmic measure. By combining
h (z)
h (z)
≤ zρ( f )−2+ε ,
lim
z→∞
z∈/E0∪[
0,1
]
A2 A1
( A1 + A2)
2 =
lim
z→∞
z∈/E0∪[
0,1
]
α β
(α + β )
By setting α (z) = am zm + · · · + a0 and β (z) = bm zm + · · · + b0, we deduce
which implies that am 2
bm = 2 or 1 . We consider first the case abmm = 2
1 , we get from (3.1) and (3.17)
and
Therefore,
which implies the contradiction
T r, fc2 = T r, M f + N f = S (r, f ) .
Case 2. B2 fc1 − B1 fc2 ≡ 0, using the same arguments as in the proof of (3.14) , we obtain that
(3.35)
log+ ϕ r eiθ
f r eiθ
+ ψ
r eiθ
f
r eiθ
dθ + S (r, f ) ,
log+ ϕ r eiθ
f r eiθ
+ ψ
r eiθ
f
r eiθ
dθ
E1
E1
log+
f r eiθ
f r eiθ
log+ f
r eiθ
dθ + S (r, f )
dθ + S (r, f ) = S (r, f ) .
log+ ϕ r eiθ
f r eiθ
+ ψ
r eiθ
f
r eiθ
dθ
q r eiθ
f r eiθ
dθ
M r eiθ
B r eiθ
N r eiθ
+ B r eiθ
f r eiθ
f r eiθ
q r eiθ
− f 2 r eiθ
dθ + S (r, f ) = S (r, f ) .
which leads to
where k is a nonzero complex constant. By this (3.1) and (3.2), we have
p1 eα−β
B1
B2
= k
fc2
,
(3.36)
(3.37)
If k = 1, then by applying Clunie lemma to (3.37) , we deduce the contradiction T r, fci = S (r, f ) . Hence,
k = 1 and from the equation (3.36) , we conclude that fc1 ≡ fc2 which exclude the hypothesis of our theorem.
This shows that at least one of f (z) f (z + c1) − q (z) and f (z) f (z + c2) − q (z) has infinitely many zeros.
Acknowledgements The authors are grateful to the anonymous referee for his/her valuable comments and suggestions which
lead to the improvement of this paper.
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