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BY-NC-ND 3.0 license Open Access Published by De Gruyter Open Access February 24, 2016

Results on the deficiencies of some differential-difference polynomials of meromorphic functions

Xiu-Min Zheng and Hong-Yan Xu
From the journal Open Mathematics

Abstract

In this paper, we study the relation between the deficiencies concerning a meromorphic function f(z), its derivative f′(z) and differential-difference monomials f(z)mf(z+c)f′(z), f(z+c)nf′(z), f(z)mf(z+c). The main results of this paper are listed as follows: Let f(z) be a meromorphic function of finite order satisfying

limsupr+T(r,f)T(r,f)<+,

and c be a non-zero complex constant, then δ(∞, f(z)mf(z+c)f′(z))≥δ(∞, f′) and δ(∞,f(z+c)nf′(z))≥ δ(∞, f′). We also investigate the value distribution of some differential-difference polynomials taking small function a(z) with respect to f(z).

MSC 2010: 30D35; 39A10

1 Introduction and main results

The fundamental theorems and the standard notations of the Nevanlinna value distribution theory of meromorphic functions will be used (see Hayman [1], Yang [2] and Yi-Yang [3]). In addition, for a meromorphic function f(z), we use δ (a, f) to denote the Nevanlinna deficiency of a ∊ ℂ̃ = ℂ⋃{} where

δ(a,f)=liminfr+m(r,1fa)T(r,f)=1limsupr+m(r,1fa)T(r,f).

We also use S(r, f) to denote any quantity satisfying S(r, f) = o(T(r, f)) for all r outside a possible exceptional set E of finite logarithmic measure limr[1,r)Edtt<

Throughout this paper, we assume m; n; k; t are positive integers.

Many people were interested in the value distribution of different expressions of meromorphic functions and obtained lots of important theorems (see [1, 46]).

Recently, the topic of complex differences has attracted the interest of many mathematicians, and a number of papers have focused on the value distribution of complex differences and difference analogues of Nevanlinna theory (including [711]). By combining complex differentiates and complex differences, we proceed in this way in this paper.

Firstly, we study the Nevanlinna deficiencies related to a meromorphic function f(z), its derivative f′(z) and its differential-difference monomials

F1(z)=f(z)mf(z+c)f(z)andF2(z)=f(z+c)nf)f(z),

and obtain the following theorems.

Theorem 1.1

Let f(z) be a meromorphic function of finite order satisfying

(1)limsupr+T(r,f)T(r,f)<+,
and c be a non-zero complex constant. Then
δ(,F1)δ(,f).
Theorem 1.2

Let f(z) be a meromorphic function of finite order satisfying (1), and c be a non-zero complex constant. Then

δ(,F2)δ(,f).
Example 1.3.

It is easy to find meromorphic functions to make the inequalities in Theorems1.1and1.2hold. For example, let f1(z) = ez, then

δ(,f1')=δ(,F1)=δ(,F2)=δ(,f1)=1,
showing the equalities in Theorems 1.1and1.2may hold.

Letf2(z)=z3+2zand c = 1 = m = n, then

f2'(z)=2z32z2andf2(z+1)=z3+3z2+3z+3z+1.

It follows that

F1(z)=f2(z)f2(z+1)f2'(z)=2z9P1(z)z4+z3,F2(z)=f2(z+1)f2'(z)=2z6P2(z)z3+z2,
where P1(z); P2(z) are polynomials in z with deg P1(z) ≥ 8 and deg P2(z) ≥ 5. Clearly, f2(z) satisfies
limsupr+T(r,f2)T(r,f2')=limsupr+3logr3logr=1<+.

Thus, we have

δ(,F1)=59>δ(,f2')=13,
and
δ(,F2)=12>δ(,f2')=13,
showing the inequalities in Theorems 1.1 and 1.2 may hold.

Thus, Theorems 1.1 and 1.2 are sharp.

In addition, from the above examples, we can find thatδ(,F1)=59<δ(,f2)=23 and δ(,F2)=12<δ(,f2)=23.. So, we give the following question: Under the conditions of Theorems 1.1 and 1.2, do F1; F2satisfy

δ(,f)δ(,F1)δ(,f),
and
δ(,f)δ(,F2)δ(,f)?

We also get the following relations between δ(∞, f) and δ(∞, Fi), i = 1, 2.

Theorem 1.4

Let f(z) be a meromorphic function of finite order, and c be a non-zero complex constant. Ifδ=δ(,f)>8m+6,, then δ(∞, F1) > 0.

Theorem 1.5

Letf(z) be a meromorphic function of finite order, and c be a non-zero complex constant. Ifδ=δ(,f)>8m+5,, then δ(∞, F2) > 0.

Theorem 1.6

Let f(z) be a meromorphic function of finite order, and c be a non-zero complex constant. Set

F3(z)=f(z)mf(z+c).

Ifδ=δ(,f)>4m+3,, then δ(∞, F)> 0.

Remark 1.7

From the conclusions of Theorems 1.1, 1.4, 1.5 and 1.6, we see that there may exist some meromorphic function f(z) satisfying δ(∞, Fi) = 0, i = 1, 2, 3 as δ(∞, f) > 0. An interesting problem arises naturally: How can we find some meromorphic function f(z) to satisfy δ(∞, Fi) = 0, i = 1, 2, 3 as δ(∞, f) > 0?

The following ideas derive from Hayman [5], Laine-Yang [12], Zheng-Chen [13]. In 1959, Hayman [5] studied the value distribution of meromorphic functions and their derivatives, and obtained the following famous theorems.

Theorem 1.8

([5]). Let f(z) be a transcendental entire function. Then

  1. for n = 3 and a ≠ 0, Ψ(z) = f′(z) − af(z)nassumes all finite values infinitely often.

  2. for n = 2, Φ(z) = f′(z) f(z)nassumes all finite values except possibly zero infinitely often.

Recently, some authors studied the zeros of f(z+c) f(z)na and f(z+c) − af(z)nb, where a(≠0), b are complex constants or small functions. Some related results can be found in [1217]. Especially, Laine-Yang [12] and Zheng-Chen [13] proved the following result, which is regarded as a difference counterpart of Theorem1. 8.

Theorem 1.9

([12, 13]). Letf(z) be a transcendental entire function of finite order, and c be a non-zero complex constant. Then

  1. for n ≥ 2, Φ1(z) = f(z+c) f(z)nassumes every a ∈ ℂ\{0} infinitely often.

  2. for n ≥ 3, a ≠0, Ψ1(z) = f(z+c)− af(z)nassumes every b ∈ ℂ infinitely often.

In the following, we investigate the zeros of some differential-difference polynomials of a meromorphic function f(z) taking small function a(z) with respect to f(z), where and in the following a(z) is a non-zero small function of growth S(r, f), and obtain some theorems as follows.

Theorem 1.10

Letf(z) be a transcendental meromorphic function of finite order, and c be a non-zero complex constant. Set

G1(z)=f(z)mf(z+c)nf(z).

If mn + 8 or nm + 8, then G1(z) − a(z) has infinitely many zeros.

Example 1.11.

An example shows that the conclusion can not hold if f(z) is of infinite order. Let f(z) = 2eez, a(z) = ez, m = 9, n = 1 and ec = −10, then

G1(z)a(z)=(2m+n+1e[(m+1)+nec]ez1)ez=(2111)ez,
then G1(z) − a(z) has finite many zeros.
Theorem 1.12

Let f(z) be a transcendental meromorphic function of finite order, andc1, c2, … cn be non-zero complex constants. Set

G2(z)=f(z)mj=1nf(z+cj)sji=1kf(i)(z).

If m ≥ σ1 + 2n + k(k + 3) + 4, where σ1 = s1 + s2 + … + sn, then G2(z) − a(z) has infinitely many zeros.

Let Pn(z) = anzn + an−1zn−1 + … + a1z + a0 be a non-zero polynomial, where a0a1an are complex constants and t is the number of the distinct zeros of Pn(z). Then we further obtain the following results.

Theorem 1.13

Let f(z) be a transcendental meromorphic function of finite order, and c be a non-zero complex constant. Set

G3(z)=f(z)mPn(f(z+c))i=1kf(i)(z).

If mn + t + k(k + 3) + 4, then G3(z) − a(z) has infinitely many zeros.

Theorem 1.14

Let f(z) be a transcendental meromorphic function of finite order, and c be a non-zero complex constant. Set

G4(z)=Pm(f(z))f(z+c)ni=1kf(i)(z).

If mn + t + k(k + 3) + 4, then G4(z) − a(z) has infinitely many zeros.

2 Some lemmas

To prove the above theorems, we will require some lemmas as follows.

Lemma 2.1

([7, 10]).Let f(z) be a meromorphic function of finite order ρ and c be a fixed non-zero complex number, then we have

m(r,f(z+c)f(z))+m(r,f(z)f(z+c))=S(r,f).

By [18], [19, p.66] and [20], we immediately deduce the following lemma.

Lemma 2.2

Let f(z) be a meromorphic function of finite order, and c be a non-zero complex constant. Then

T(r,f(z+c))=T(r,f)+S(r,f),N(r,f(z+c))=N(r,f)+S(r,f),N(r,1f(z+c))=N(r,1f)+S(r,f).
Lemma 2.3

([3, p.37]).Let f(z) be a nonconstant meromorphic function in the complex plane and l be a positive integer. Then

N(r,f(i))=N(r,f)+lN¯(r,f),T(r,f(l))T(r,f)+lN¯(r,f)+S(r,f).
Lemma 2.4.

Let f(z) be a transcendental meromorphic function of finite order, and G1(z) = f(z)mf(z + c)n .f(z). Then we have

(2)(|mn|1)T(r,f)+S(r,f)T(r,G1)(n+m+2)T(r,f)+S(r,f).
Proof.

From Lemmas 2.2 and 2.3, we have

T(r,G1)T(r,fm)+T(r,f(z+c)n)+T(r,f)(n+m+2)T(r,f)+S(r,f).

On the other hand, from Lemma 2.2 again, we have

(n+m+1)T(r,f)=T(r,fn+m+1)=T(r,f(z)n+1G1(z)f(z+c)nf(z))T(r,G1)+T(r,ff)+T(r,f(z)nf(z+c)n)T(r,G1)+(2n+2)T(r,f)+S(r,f),

where we assume mn without loss of generality. Thus, (2) is proved. □

Using the similar method as in Lemma 2.4, we get the following lemmas.

Lemma 2.5.

Let f(z) be a transcendental meromorphic function of finite order, andG2(z)=f(z)mj=1nf(z+cj)Sji=1kf(i)(z), then we have

(mσ1k(k+3)2)T(r,f)+S(r,f)(r,G2)(m+σ1+k(k+3)2)T(r,f)+S(r,f).
Lemma 2.6

Let f(z)be a transcendental meromorphic function of finite order, andG3(z)=f(z)mPn(f(z+c))i=1kf(i)(z), then we have

(mnk(k+3)2)T(r,f)+S(r,f)T(r,G3)(m+n+k(k+3)2)T(r,f)+S(r,f).
Lemma 2.7

Let f(z) be a transcendental meromorphic function of finite order, andG4(z)=Pm(f(z))f(z+c)ni=1kf(i)(z), then we have

(mnk(k+3)2)T(r,f)+S(r,f)T(r,G4)(m+n+k(k+3)2)T(r,f)+S(r,f).

3 Proofs of Theorems 1.1 and 1.2

3.1 Proof of Theorem 1.1

We firstly give the following elementary inequalities

(3)αα+βα1α1+β,αα+βα+γa+β+γ,

for α, β, γ ≥ 0 and αα1.

Since

f(z)m+2=F1(z)f(z)m+1f(z)m+1f(z)f(z+c),

it follows that

(m+2)m(r,f)m(r,F1)+(m+1)m(r,ff)+m(r,f(z)f(z+c),

Then by Lemma 2.1, we have

(4)m(r,F1)(m+2)m(r,f)+S(r,f).

Since N(r, f′) = N(r, f) = N̅ (r, f), it follows by Lemma 2.2 that

(5)N(r,F1)(m+2)N(r,f)+N¯(r,f)(m+2)N(r,f).

From (1), we have lim sup

(6)limr+S(r,f)T(r,f)=limr+S(r,f)T(r,f)T(r,f)T(r,f)=0

Then, from (3)- (6), we have

N(r,F1)T(r,F1)(m+2)N(r,f)(m+2)N(r,f)+(m+2)m(r,f)+S(r,f)N(r,f)T(r,f)+S(r,f)=N(r,f)(1+ο(1))T(r,f).

It follows that δ(∞, f′) ≤ δ(∞, F′).

Thus, we complete the proof of Theorem 1.1.

3.2 Proof of Theorem 1.2

Since

f(z)n+1=F2f(z)nf(z)nf(z)nf(z+c)n,

then by using the similar method as in the proof of Theorem 1.1, we can prove Theorem 1.2 easily.

4 Proofs of Theorems 1.4, 1.5 and 1.6

4.1 Proof of Theorem 1.4

Let F4(z) = f(z)m + 2 then we have

N(r,F4)=(m+2)N(r,f)andT(r,F4)=(m+2)T(r,f)

It follows that δ(∞, F4) = δ(∞, f′) = δ. Since F4(z) = f(z)m+2, we have

(7)N¯(r,f)N(r,f)1m+2N(r,F4)1δm+2T(r,F4)+S(r,F4),
(8)N¯(r,1f)N(r,1f)1m+2N(r,1F4)1m+2T(r,F4)+ο(1).

From (7), (8) and Lemmas 2.1 and 2.2, we have

T(r,F4)=T(r,F1(z)f(z)f(z+c)f(z)f(z))T(r,F1)+N(r,1f)+N¯(r,1f)+N(r,f)+N¯(r,f)+S(r,f)T(r,F1)+42δm+2T(r,F4)+S(r,f),

that is

(9)T(r,F1)(m2(1δ)m+2+ο(1))T(r,F4).

From (7) and (8) again, we have

N(r,F1)N(r,F4)+N(r,f(z+c)f(z)+N(r,ff)N(r,F4)+N(r,1f)+N¯(r,1f)+N(r,f)+N¯(r,f)+S(r,f)(m+4)(1δ)+2m+2T(r,F4)+S(r,f),

that is,

(10)N(r,F1)((m+4)(1δ)+2m+2+ο(1))T(r,F4).

Thus, from (9), (10) and δ=δ(,f)=δ(,F4)>8m+6, it follows that that is,

limr+N(r,F1)T(r,F1)(m+4)(1δ)+2m2(1δ)<1,δ(,F4)=1limr+supN(r,F1)T(r,F1)>0.

This completes the proof of Theorem 1.4.

4.2 Proofs of Theorems 1.5 and 1.6

Using the similar method as in the proof of Theorem 1.4, we can prove Theorems 1.5 and 1.6 easily.

5 Proofs of Theorems 1.10, 1.12, 1.13 and 1.14

5.1 Proof of Theorem 1.10

Suppose that f(z) is a transcendental meromorphic function of finite order. Since mn + 8 or nm + 8, then by Lemma 2.4, we have S(r, f) = S(r, G1). Thus, by using the second fundamental theorem and Lemmas 2.2 and 2.4, we have

(|mn|1)T(r,f)T(r,G1)+S(r,f)N¯(r,G1)+N¯(r,1G1)+N¯(r,1G1(z)a(z))+S(r,G1)+N¯(r,1f(z+c))+N¯(r,1f)+N¯(r,1G1(z)a(z))+S(r,G1)6T(r,f)+N¯(r,1G1(z)a(z))+S(r,G1),

that is

|mn|7n+m+2T(r,G1)+S(r,G1)(|mn|7)T(r,f)N¯(r,1G1(z)a(z))+S(r,G1).

Thus, we have from mn + 8 or nm + 8 that

δ(a,G1)1|mn|7n+m+2<1.

Consequently, G1(z)− a(z) has infinitely many zeros.

This completes the proof of Theorem 1.10.

5.2 Proof of Theorem 1.14

If f(z) is a transcendental meromorphic function of finite order, then by Lemma 2.7, we have S(r, f) = S(r. G4). Thus, by using the second fundamental theorem and Lemmas 2.2 and 2.7 again,

(mnk(k+3)2)T(r,f)T(r,G4)+S(r,f)N¯(r,G4)+N¯(r,1G4)+N¯(r,1G4(z)a(z))+S(r,G4)N¯(r,f)+N¯(r,f(z+c))+i=1tN¯(r,1fγi)+N¯(r,1f(z+c))+j=1kN¯(r,1f(j))+N¯(r,1G4(z)a(z))+S(r,G4)(3+t+k(k+1)2+k)T(r,f)+N¯(r,1G4(z)a(z))+S(r,G4),

that is,

mntk(k+3)3m+n+k(k+3)2T(r,G4)+S(r,G4)(mntk(k+3)3)T(r,f)N¯(r,1G4(z)a(z))+S(r,G4),

where γ1, γ2,.....,γt are distinct zeros of Pm(z). Since mn + t + k(k + 3) + 4, we have

δ(a,G4)1mntk(k+3)m+n+k(k+3)2<1.

Consequently, G4(z) − a(z) has infinitely many zeros.

This completes the proof of Theorem 1.14.

5.3 Proofs of Theorems 1.12 and 1.13

Using the similar method as in the proofs of Theorems 1.10 and 1.14 and combining Lemmas 2.5 and 2.6, we can prove Theorems 1.12 and 1.13 easily.

Acknowledgement

The authors are grateful to the referees and editors for their valuable comments which lead to the improvement of this paper.

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Received: 2015-8-28
Accepted: 2015-11-24
Published Online: 2016-2-24
Published in Print: 2016-1-1

© 2016 Zheng and Xu, published by De Gruyter Open.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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