# Results on Difference Analogues of Valiron-Mohon'ko Theorem

Authors: Ran Ran Zhang; Zhi Bo Huang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The classical Valiron-Mohon'ko theorem has many applications in the study of complex equations. In this paper, we investigate rational functions inf(z) and the shifts of f(z). We get some results on their characteristic functions. These results may be viewed as difference analogues of Valiron-Mohon'ko theorem.


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## Body

## 1. Introduction and Results

We use the basic notions of Nevanlinna’s theory in this work (see [1, 2]). Let f(z) be a meromorphic function. We say that a meromorphic function α(z) is a small function of f(z) if T(r,α)=S(r,f), where S(r,f)=o(T(r,f)) outside a possible exceptional set of finite logarithmic measure.The Valiron-Mohon’ko theorem has been proved to be an extremely useful tool in the study of meromorphic solutions of differential, difference, and functional equations. It is stated as follows.Theorem A (see [3, page 29]).
Letf be a meromorphic function. Then for all irreducible rational functions in f(1)R(z,f)=∑i=0pai(z)fi∑j=0qbj(z)fj
with meromorphic coefficients ai(z), bj(z) such that
(2)T(r,ai)=S(r,f),i=0,…,p,T(r,bj)=S(r,f),j=0,…,q,
the characteristic function of R(r,f(z)) satisfies
(3)T(r,R(z,f))=max{p,q}T(r,f)+S(r,f).Recently, a number of papers have focused on difference analogues of Nevanlinna’s theory; see, for instance, [4–12]. Among these papers, difference polynomials are investigated extensively (see [5, 9–11]). But the difference analogues of Valiron-Mohon’ko theorem have not been established. In this paper, we are devoted to this work.A difference polynomial off(z) is an expression of the form
(4)H(z,f)=∑λ∈J‍aλ(z)∏j=1τλ‍f(z+δλ,j)μλ,j,
where J is an index set, δλ,j are complex constants, and μλ,j are nonnegative integers. In what follows, we assume that the coefficients of difference polynomials are, unless otherwise stated, small functions. The maximal total degree of H(z,f) in f(z) and the shifts of f(z) is defined by
(5)degfH=maxλ∈J∑j=1τλ‍μλ,j.First, we investigate the rational function(6)R1(z,f)=P(z,f)d1(z)f(z+c)+d0(z),
where c is an arbitrary complex number, and d0(z) and d1(z) are small functions of f(z) with d0(z)≢0 or d1(z)≢0. Our result is stated as follows.Theorem 1.
Letf(z) be a meromorphic function of finite order such that N(r,f)=S(r,f). Suppose that P(z,f)≢0 is a difference polynomial in f(z) and that R1(z,f) is of the form (6). Then
(7)T(r,R1)≤(degfP)T(r,f)+S(r,f).In many papers (see, for instance, [7, 13, 14]), linear difference expressions often appear. Concerning their characteristic functions, we have the following corollary, which is obtained easily from Theorem 1.Corollary 2.
Letf(z) be a meromorphic function of finite order such that N(r,f)=S(r,f). Suppose that L(z,f)≢0 is a linear combination in f(z) and the shifts of f(z). Then
(8)T(r,L)≤T(r,f)+S(r,f).Next we consider the rational function(9)R2(z,f)=P(z,f)f(z+c1)⋯f(z+cn),
where c1,…,cn are different complex constants. We get the following result.Theorem 3.
Letf(z) be a meromorphic function of finite order such that N(r,f)=S(r,f). Suppose that P(z,f)≢0 is a difference polynomial in f(z) and that R2(z,f) is of the form (9). Then
(10)T(r,R2)≤max{degfP,n}T(r,f)+S(r,f).As for the general rational function inf(z) and the shifts of f(z),
(11)R3(z,f)=P(z,f)Q(z,f),
we get the following two results.Theorem 4.
Letf(z) be a meromorphic function of finite order such that N(r,f)=S(r,f). Suppose that P(z,f)≢0 and Q(z,f)≢0 are difference polynomials in f(z) and that R3(z,f) is of the form (11).(i)
IfdegfP≥degfQ and P(z,f) contains just one term of maximal total degree, then
(12)T(r,R3)≥(degfP-degfQ)T(r,f)+S(r,f).(ii)
IfdegfP≤degfQ and Q(z,f) contains just one term of maximal total degree, then
(13)T(r,R3)≥(degfQ-degfP)T(r,f)+S(r,f).Theorem 5.
Letf(z) be a meromorphic function of finite order such that N(r,f)+N(r,1/f)=S(r,f). Suppose that P(z,f)≢0 and Q(z,f)≢0 are difference polynomials in f(z) and that R3(z,f) is of the form (11). Then
(14)T(r,R3)≤max{degfP,degfQ}T(r,f)+S(r,f).The following two examples show that the results in Theorems1–5 are sharp; that is, “≤” and “≥” cannot be replaced by “<”, “>” or “=”.Example 6.
Letf(z)=ez and
(15)P(z,f)=f(z)2f(z+πi)+f(z)2+2f(z+πi)f(z)+2f(z)+f(z+πi)+1.
Let
(16)R11(z,f)=P(z,f)f(z+πi)+2,R12(z,f)=P(z,f)f(z+πi)+1.
Then R11(z,f)=(1+ez)2(1-ez)/(-ez+2) and R12(z,f)=(1+ez)2. Clearly,
(17)T(r,R11)=3T(r,f)+S(r,f),T(r,R12)=2T(r,f)+S(r,f).
Therefore,
(18)(degfP-1)T(r,f)+S(r,f)<T(r,R11)=(degfP)T(r,f)+S(r,f),(degfP-1)T(r,f)+S(r,f)=T(r,R12)<(degfP)T(r,f)+S(r,f).Example 7.
Letf(z)=sinz and
(19)P(z,f)=f(z+π2)2f(z)+f(z)2+f(z+π)f(z)-f(z).
Let
(20)R21(z,f)=P(z,f)f(z+π/2)3,R22(z,f)=P(z,f)f(z+π)2.
Then R21(z,f)=-tan3z and R22(z,f)=-sinz. Clearly,
(21)T(r,R21)=3T(r,f)+S(r,f),T(r,R22)=T(r,f)+S(r,f).
Therefore,
(22)(degfP-3)T(r,f)+S(r,f)<T(r,R21)=(degfP)T(r,f)+S(r,f),(degfP-2)T(r,f)+S(r,f)=T(r,R22)<(degfP)T(r,f)+S(r,f).


## 2. Proof of Theorem1

We need the following lemmas for the proof of Theorem1.The difference analogue of the logarithmic derivative lemma was given by Halburd-Korhonen [8, Corollary 2.2] and Chiang-Feng [7, Corollary 2.6], independently. The following Lemma 8 is a variant of [8, Corollary 2.2].Lemma 8.
Letf(z) be a nonconstant meromorphic function of finite order, and let η1,η2 be two arbitrary complex numbers. Then,
(23)m(r,f(z+η1)f(z+η2))=S(r,f).In the remark of [15, page 15], it is pointed out that the following lemma holds.Lemma 9.
Letf(z) be a nonconstant finite order meromorphic function and let c≠0 be an arbitrary complex number. Then,
(24)T(r+|c|,f)=T(r,f)+S(r,f),N(r+|c|,f)=N(r,f)+S(r,f).Letf(z) be a meromorphic function. It is shown in [16, page 66] that for an arbitrary c≠0, the following inequalities:
(25)(1+o(1))T(r-|c|,f(z))≤T(r,f(z+c))≤(1+o(1))T(r+|c|,f(z))
hold as r→∞. From its proof we see that the above relations are also true for counting functions. So by these relations and Lemma 9, we get the following lemma.Lemma 10.
Letf(z) be a nonconstant finite order meromorphic function and let c≠0 be an arbitrary complex number. Then,
(26)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).Remark 11.
In [7], Chiang and Feng proved a similar result. Let f(z) be a meromorphic function with σ(f)<∞, and let η≠0 be fixed; then for each ε>0, we have
(27)T(r,f(z+η))=T(r,f)+O(rσ(f)-1+ε)+O(logr).Proof of Theorem1.
Let(28)P(z,f)=∑λ∈I‍aλ(z)∏j=1σλ‍f(z+αλ,j)lλ,j,
and degfP=p.
Rearranging the expression ofP(z,f) by collecting together all terms having the same total degree, we get
(29)P(z,f)=∑i=0p‍hi(z)f(z)i,
where, for i=0,…,p,
(30)hi(z)=∑λ∈Ii‍aλ(z)∏j=1σλ‍(f(z+αλ,j)f(z))lλ,j,Ii={λ∈I∣∑j=1σλ‍lλ,j=i}.
Since the coefficientsaλ(z) of P(z,f) are small functions of f(z), we have
(31)m(r,aλ)≤T(r,aλ)=S(r,f).
So by Lemma 8, we have, for all i=0,1,…,p the estimates
(32)m(r,hi)=S(r,f).
Without loss of generality, we may assumec=0 in (6). Otherwise, substituting z-c for z, we get
(33)R1(z-c,f)=P(z-c,f)d1(z-c)f(z)+d0(z-c).
By Lemma 10, we see that
(34)T(r,R1(z-c,f))=T(r,R1(z,f))+S(r,f).
So, in the following discussion, we only discuss the form
(35)R1(z,f)=P(z,f)d1(z)f(z)+d0(z).
Assume first thatd1(z)=0. Clearly, we may assume that d0(z)=1. By (29), we get
(36)R1(z,f)=P(z,f)=hp(z)f(z)p+hp-1(z)f(z)p-1+⋯+h1(z)f(z)+h0(z).
If p=1, then R1(z,f)=h1(z)f(z)+h0(z). So by (32), we get
(37)m(r,R1)≤m(r,f)+S(r,f).
If p>1, then rewrite R1(z,f) in the form
(38)R1(z,f)=f(z)(hp(z)f(z)p-1+⋯+h1(z))+h0(z).
So we have
(39)m(r,R1)≤m(r,f)+m(r,hp(z)f(z)p-1+⋯+h1(z))+S(r,f).
By (39) and the inductive argument, we have
(40)m(r,R1)≤pm(r,f)+S(r,f).
To estimateN(r,R1), we use the form
(41)R1(z,f)=P(z,f)=∑λ∈I‍aλ(z)∏j=1σλ‍f(z+αλ,j)lλ,j.
Clearly,
(42)N(r,R1)≤∑λ∈I‍(N(r,aλ)+∑j=1σλ‍lλ,jN(r,f(z+αλ,j)))+O(1).
So by (31), N(r,f)=S(r,f), and Lemma 10, we get
(43)N(r,R1)=S(r,f).
Combining this equality with (40), we get
(44)T(r,R1)≤pT(r,f)+S(r,f),
and we have completed the case d1(z)=0.
We now proceed to the cased1(z)≠0. Clearly, in this case we may assume that d1(z)=1. By (29), we see that (6) becomes
(45)R1(z,f)=(hp(z)f(z)p+hp-1(z)f(z)p-1+⋯+h1(z)f(z)+h0(z)(hp(z)f(z)p+hp-1(z)f(z)p-1)×(f(z)+d0(z))-1.
By (45), we get
(46)R1(z,f)=hp(z)f(z)p-1+(hp-1*(z)f(z)p-1+hp-2(z)f(z)p-2+⋯+h1(z)f(z)+h0(z)hp-1*(z)f(z)p-1+hp-2(z)f(z)p-2)×(f(z)+d0(z))-1=hp(z)f(z)p-1+hp-1*(z)f(z)p-2+hp-2*(z)f(z)p-2+⋯+h1(z)f(z)+h0(z)f(z)+d0(z)=⋯=hp(z)f(z)p-1+hp-1*(z)f(z)p-2+⋯+h2*(z)f(z)+h1*(z)+h0*(z)f(z)+d0(z),
where
(47)hp-1*(z)=hp-1(z)-hp(z)d0(z),hp-2*(z)=hp-2(z)-hp-1*(z)d0(z),⋮h1*(z)=h1(z)-h2*(z)d0(z),h0*(z)=h0(z)-h1*(z)d0(z).
By (32), we get, for j=0,1,…,p-1, the estimates
(48)m(r,hj*)=S(r,f).
By (46), using the same method as in (36)–(40), we get
(49)m(r,R1)≤m(r,hp(z)f(z)p-1+hp-1*(z)f(z)p-2+⋯+h1*(z))+m(r,h0*(z)f(z)+d0(z))≤(p-1)m(r,f)+m(r,1f(z)+d0(z))+S(r,f).
To estimateN(r,R1), we use the form
(50)R1(z,f)=P(z,f)f(z)+d0(z)=∑λ∈Iaλ(z)∏j=1σλf(z+αλ,j)lλ,jf(z)+d0(z).
By (31), N(r,f)=S(r,f), and Lemma 10, we get
(51)N(r,R1)=N(r,1f(z)+d0(z))+S(r,f).
Combining this equality with (49), we get
(52)T(r,R1)≤(p-1)m(r,f)+T(r,1f(z)+d0(z))+S(r,f)≤pT(r,f)+S(r,f).
Theorem 1 is proved.


## 3. Proof of Theorem3

Proof.
LetP(z,f) be of the form (28) and degfP=p. Rearranging the expression of P(z,f), we get (29) and (30). We only discuss the case p≥n since the case p<n is easier.
RewriteR2(z,f) in the form
(53)R2(z,f)=P(z,f)s(z)f(z)n,
where
(54)s(z)=f(z+c1)⋯f(z+cn)f(z)n.
By Lemma 8, we get
(55)m(r,1s)=S(r,f).
By (29) and (53), we get
(56)R2(z,f)=∑i=0phi(z)f(z)is(z)f(z)n=∑i=np‍hi(z)s(z)f(z)i-n+hn-1(z)f(z)n-1+⋯+h0(z)s(z)f(z)n=∑i=np‍hi(z)s(z)f(z)i-n+∑j=1n‍hn-j(z)s(z)(1f(z))j.
By (32) and (55), we have, for all i=0,…,p, the estimates
(57)m(r,hi(z)s(z))=S(r,f).
By (57), using the same method as in (36)–(40), we get
(58)m(r,∑i=np‍hi(z)s(z)f(z)i-n)≤(p-n)m(r,f)+S(r,f),m(r,∑j=1n‍hn-j(z)s(z)(1f(z))j)≤nm(r,1f)+S(r,f).
Combining the above two inequalities with (56), we get
(59)m(r,R2)≤(p-n)m(r,f)+nm(r,1f)+S(r,f).
To estimateN(r,R2), we use the form
(60)R2(z,f)=∑λ∈Iaλ(z)∏j=1σλf(z+αλ,j)lλ,jf(z+c1)⋯f(z+cn).
By (31), N(r,f)=S(r,f), and Lemma 10, we get
(61)N(r,R2)=N(r,1f(z+c1)⋯f(z+cn))+S(r,f)≤nN(r,1f)+S(r,f).
Combining this inequality with (59), we get
(62)T(r,R2)≤(p-n)m(r,f)+nT(r,1f)+S(r,f)≤pT(r,f)+S(r,f).
Theorem 3 is proved.


## 4. Proof of Theorem4

We need the following lemma for the proof of Theorem4.Lemma 12 (see [11]).
Letf(z) be a meromorphic function of finite order such that N(r,f)=S(r,f). Suppose that H(z,f) is a difference polynomial in f(z) and H(z,f) contains just one term of maximal total degree. Then,
(63)T(r,H)=(degfH)T(r,f)+S(r,f).Proof of Theorem4.
We have the following.
Case  1. Suppose that degfP≥degfQ and P(z,f) contains just one term of maximal total degree.
LetdegfP=p and degfQ=q. By Lemma 12, we get
(64)T(r,P)=pT(r,f)+S(r,f).
By Theorem 1, we get
(65)T(r,Q)≤qT(r,f)+S(r,f).
By (11), we get
(66)P(z,f)=R3(z,f)Q(z,f).
By (64)–(66), we get
(67)pT(r,f)+S(r,f)=T(r,P(z,f))=T(r,R3(z,f)Q(z,f))≤T(r,R3(z,f))+T(r,Q(z,f))≤T(r,R3(z,f))+qT(r,f)+S(r,f).
So we have,
(68)T(r,R3)≥(p-q)T(r,f)+S(r,f).
Case  2. Suppose that degfP≤degfQ and Q(z,f) contains just one term of maximal total degree.
In this case, we consider1/R3(z,f). Using the same method as in Case 1, we can easily get
(69)T(r,R3)=T(r,1R3)≥(q-p)T(r,f)+S(r,f).
Theorem 4 is proved.


## 5. Proof of Theorem5

Proof.
LetP(z,f) be of the form (28) and degfP=p. Let
(70)Q(z,f)=∑μ∈J‍bμ(z)∏j=1τμ‍f(z+βμ,j)mμ,j,
and degfQ=q.
Rearranging the expression ofP(z,f), we get (29) and (30).
Similarly, rearranging the expression ofQ(z,f), we get
(71)Q(z,f)=∑k=0q‍tk(z)f(z)k,
where, for k=0,…,q,
(72)tk(z)=∑μ∈Jk‍bμ(z)∏j=1τμ‍(f(z+βμ,j)f(z))mμ,j,Jk={λ∈J∣∑j=1τμ‍mμ,j=k}.
By (29) and (71), we get
(73)R3(z,f)=∑i=0phi(z)f(z)i∑k=0qtk(z)f(z)k.
Since N(r,f)+N(r,1/f)=S(r,f), by Lemma 10, we have, for an arbitrary η,
(74)N(r,f(z+η)f(z))≤N(r,1f)+N(r,f(z+η))=N(r,1f)+N(r,f)+S(r,f)=S(r,f).
By (74) and Lemma 8, we have, for an arbitrary η,
(75)T(r,f(z+η)f(z))=S(r,f).
Since the coefficients aλ(z) and bμ(z) of P(z,f) and Q(z,f) are small functions of f(z), by (30), (72), and (75), we get
(76)T(r,hi)=S(r,f),i=0,…,pT(r,tk)=S(r,f),k=0,…,q.
By (73), we are not clear whether R3(z,f) is an irreducible rational function in f(z). So by Theorem A, we get
(77)T(r,R3)≤max{p,q}T(r,f)+S(r,f).
Theorem 5 is proved.



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