The value distribution of solutions of certain difference equations
is investigated. As its applications, we investigate the difference analogue of the Brück
conjecture. We obtain some results on entire functions sharing a finite value with their
difference operators. Examples are provided to show that our results are the best possible.

1. Introduction and Main Results

In this paper, the term meromorphic function will mean being meromorphic in the whole complex plane ℂ. It is assumed that the reader is familiar with the standard notations and the fundamental results of the Nevanlinna theory; see, for example, [1–3]. In addition, we use notations σ(f),λ(f) to denote the order and the exponent of convergence of the sequence of zeros of a meromorphic function f, respectively. The notation S(r,f) is defined to be any quantity satisfying S(r,f)=o(T(r,f)) as r→∞, possibly outside a set E of r of finite logarithmic measure.

Let f and g be two nonconstant meromorphic functions, and let a∈ℂ. We say that f and g share a CM, provided that f-a and g-a have the same zeros with the same multiplicities. Similarly, we say that f and g share a IM, provided that f-a and g-a have the same zeros ignoring multiplicities.

The famous results in the uniqueness theory of meromorphic functions are the 5 IM and 4 CM shared values theorems due to Nevanlinna [4]. It shows that if two nonconstant meromorphic functions f and g share five different values IM or four different values CM, then f≡g or f is a linear fractional transformation of g. Condition 4 CM shared values have been improved to 2 CM + 2 IM by Gundersen [5], while the case 1 CM + 3 IM still remains an open problem. Specifically, Brück posed the following conjecture.

Conjecture 1 (see [<xref ref-type="bibr" rid="B1">6</xref>]).

Let f be a nonconstant entire function satisfying the hyperorder σ2(f)<∞, where σ2(f) is not a positive integer. If f and f′ share a finite value a CM, then f-a≡c(f′-a) for some nonzero constant c.

In [6], Brück proved that the conjecture is true provided that a=0 or N(r,1/f′)=S(r,f). He also gave counterexamples to show that the restriction on the growth of f is necessary.

In recent years, as the research on the difference analogues of Nevanlinna theory is becoming active, lots of authors [7–11] started to consider the uniqueness of meromorphic functions sharing values with their shifts or their difference operators.

Heittokangas et al. proved the following result which is a shifted analogue of Brück’s conjecture.

Theorem A (see [<xref ref-type="bibr" rid="B8">8</xref>]).

Let f be a meromorphic function of σ(f)<2 and η a nonzero complex number. If f(z) and f(z+η) share a finite value a and ∞ CM, then
(1)f(z+η)-af(z)-a=τ,
for some constant τ.

In [8], Heittokangas et al. gave the example f(z)=ez2+1 which shows that σ(f)<2 cannot be relaxed to σ(f)≤2.

For a nonzero complex number η, we define difference operators as
(2)Δηf(z)=f(z+η)-f(z),Δηnf(z)=Δηn-1(Δηf(z)),n∈ℕ,n≥2.
Regarding the difference analogue of Brück’s conjecture, we mention the following results.

Theorem B (see [<xref ref-type="bibr" rid="B4">7</xref>]).

Let f be a finite order transcendental entire function which has a finite Borel exceptional value a, and let η be a constant such that f(z+η)≢f(z). If f(z) and Δηf(z) share a CM, then
(3)a=0,f(z+η)-f(z)f(z)=c,
for some nonzero constant c.

Theorem C (see [<xref ref-type="bibr" rid="B13">11</xref>]).

Let f be a nonperiodic transcendental entire function of finite order. If f(z) and Δηnf(z) share a nonzero finite value a CM, then 1≤σ(f)≤λ(f-a)+1; that is,
(4)f(z)=A(z)eQ(z)+a,
where A(z) is an entire function with σ(A)=λ(f-a) and Q(z) is a polynomial with degQ≤σ(A)+1.

Let f be a nonperiodic transcendental entire function of finite order. Theorem B shows that if a nonzero finite value a is shared by f(z) and Δηf(z), then σ(f)=λ(f-a). It is obvious that the result in Theorem B is sharper than Theorem C for n=1. In this paper, we continue to investigate the difference analogue of Brück’s conjecture and obtain the following result.

Theorem 2.

Let f be a finite order entire function, n≥2 an integer, and η a constant such that Δηnf(z)≢0. If f(z) and Δηnf(z) share a finite value a(≠0) CM, then λ(f-a)=σ(f)≥1; that is,
(5)f(z)=A(z)eQ(z)+a,
where A(z) is an entire function with 1≤σ(A)=λ(f-a)=σ(f) and Q(z) is a polynomial with degQ≤σ(A).

Remark 3.

It is obvious that Theorem 2 is sharper than Theorem C and a supplement of Theorem B for n≥2.

The discussions in Theorems C and 2 are concerning the case that shared value a≠0. When a=0, we obtain the following result.

Theorem 4.

Let f be a finite order entire function, n a positive integer, and η a constant such that Δηnf(z)≢0. If f(z) and Δηnf(z) share 0 CM, then 1≤σ(f)≤λ(f)+1; that is,
(6)f(z)=A(z)eQ(z),
where A(z) is an entire function with σ(A)=λ(f) and Q(z) is a polynomial with degQ≤σ(A)+1.

It is well known that if a finite order entire function f(z) shares a CM with Δηnf(z), then f(z) satisfies the difference equation
(7)Δηnf(z)-a=eQ(z)(f(z)-a),
where Q(z) is a polynomial. Hence in order to prove the above results, we consider the value distribution of entire solutions of the difference equation
(8)an(z)f(z+nη)+⋯+a1(z)f(z+η)+(a0(z)-eQ(z))f(z)=B(z)
and obtain the following result.

Theorem 5.

Let a0,…,an-1,an(≢0),B(≢0) be polynomials, and let Q be a polynomial with degree m(≥1). Then every entire solution f of finite order of (8) satisfies σ(f)≥m and

if σ(f)>1, then λ(f)=σ(f);

if σ(f)=1, then λ(f)=σ(f) or f has only finitely many zeros.

2. LemmasLemma 6 (see [<xref ref-type="bibr" rid="B6">12</xref>]).

Let T:(0,+∞)→(0,+∞) be a nondecreasing continuous function, s>0,α<1, and let F⊂ℝ+ be the set of all r such that T(r)≤αT(r+s). If the logarithmic measure of F is infinite, then
(9)lim¯r→∞logT(r)logr=∞.

Lemma 7 (see [<xref ref-type="bibr" rid="B7">13</xref>]).

Let f be a nonconstant meromorphic function of finite order, η∈ℂ,δ<1. Then
(10)m(r,f(z+η)f(z))=o(T(r+|η|,f)rδ)
for all r outside a possible exceptional set E with finite logarithmic measure ∫E(dr/r)<∞.

Remark 8.

By Lemmas 6 and 7, we know that, for a nonconstant meromorphic function f of finite order,
(11)m(r,f(z+η)f(z))=S(r,f).

Lemma 9 (see [<xref ref-type="bibr" rid="B16">3</xref>]).

Let fj(j=1,…,n+1) and gj(j=1,…,n) be entire functions such that

∑j=1nfj(z)egj(z)≡fn+1(z),

the order of fj is less than the order of egk for 1≤j≤n+1,1≤k≤n; and furthermore, the order of fj is less than the order of egh-gk for n≥2 and 1≤j≤n+1,1≤h<k≤n.

Then fj(z)≡0(j=1,…,n+1).Lemma 10 (see [<xref ref-type="bibr" rid="B2">14</xref>]).

Let f be a meromorphic function with finite order σ(f)=σ<1, η∈ℂ∖{0}. Then for any given ɛ>0 and integers 0≤j<k, there exists a set E⊂(1,∞) of finite logarithmic measure, so that, for all |z|=r∉E⋃[0,1], we have
(12)|Δηkf(z)Δηjf(z)|≤|z|(k-j)(σ-1)+ɛ.

Lemma 11 (see [<xref ref-type="bibr" rid="B3">15</xref>]).

Let a0(z),…,ak(z) be entire functions with finite order. If there exists an integer l(0≤l≤k) such that
(13)σ(al)>max0≤j≤kj≠l{σ(aj)}
holds, then every meromorphic solution f(≢0) of the difference equation
(14)ak(z)f(z+k)+⋯+a1(z)f(z+1)+a0(z)f(z)=0
satisfies σ(f)≥σ(al)+1.

3. Proofs of ResultsProof of Theorem <xref ref-type="statement" rid="thm1.3">5</xref>.

Let f be an entire solution of finite order of (8). By Remark 8 and (8), we get
(15)T(r,eQ)=T(r,eQ-a0)+S(r,eQ)≤∑j=1nm(r,f(z+jη)f(z))+∑j=0nm(r,aj)+m(r,B(z)f(z))+S(r,eQ)≤T(r,f)+S(r,f)+S(r,eQ).
By (15) we get σ(f)≥m.

Case 1 (σ(f)>1). Suppose that λ(f)<σ(f), by the Weierstrass factorization; we get f(z)=h1(z)eh2(z), where h1(z)(≢0) is an entire function and h2(z) is a polynomial such that
(16)σ(h1)=λ(h1)=λ(f)<σ(f)=degh2.
Substituting f(z)=h1(z)eh2(z) into (8), we get
(17)∑j=1naj(z)h1(z+jη)eh2(z+jη)-h2(z)+(a0(z)-eQ(z))h1(z)=B(z)e-h2(z).
If degh2>m, then by (16) we know that the order of the right side of (17) is degh2, and the order of the left side of (17) is less than degh2. This is a contradiction. Hence degh2=m>1. Set
(18)Q(z)=bmzm+⋯+b0,h2(z)=cmzm+⋯+c0,
where bm(≠0),…,b0,cm(≠0),…,c0 are complex numbers. By (17) we get
(19)∑j=1naj(z)h1(z+jη)eh2(z+jη)-h2(z)+a0(z)h1(z)=h1(z)eQ(z)+B(z)e-h2(z).
Next we discuss the following two subcases.

Subcase 1 (bm+cm≠0). Then by Lemma 9, (16), and (19), we get B(z)≡0,h1(z)≡0. This is impossible.

Subcase 2 (bm+cm=0). Suppose that
(20)h1(z)eQ(z)-bmzm+B(z)e-h2(z)-bmzm≡0.
Then h1(z)=-B(z)e-h2(z)-Q(z). By σ(h1)=λ(h1), we obtain that e-h2(z)-Q(z) is a nonzero constant. Hence h1(z) is a nonzero polynomial. By (19) we get
(21)∑j=1naj(z)h1(z+jη)eh2(z+jη)-h2(z)=-a0(z)h1(z).
Since deg{h2(z+jη)-h2(z+iη)}=m-1>0 for i≠j, then by Lemma 9 and (21), we get
(22)aj(z)h1(z+jη)≡0(j=0,1,…,n).
This is impossible. Hence we have h1(z)eQ(z)-bmzm+B(z)e-h2(z)-bmzm≢0. Then from the order consideration, we know that the order of the right side of (19) is m, and the order of the left side of (19) is less than m. This is a contradiction. Hence λ(f)=σ(f).

Case 2 (σ(f)=1). Then by σ(f)≥m, we get m=1. Suppose that f(z) has infinitely many zeros and λ(f)<σ(f); by the Weierstrass factorization, we get
(23)f(z)=h3(z)eβz,
where β(≠0) is a complex number and h3(z)(≢0) is an entire function such that
(24)σ(h3)=λ(h3)=λ(f)<1.
Let Q(z)=b1z+b0, where b1(≠0),b0 are complex numbers. Substituting f(z)=h3(z)eβz into (8), we get
(25)∑j=1naj(z)h3(z+jη)eβjη+a0(z)h3(z)=h3(z)eb1z+b0+B(z)e-βz.
Note that h3(z)eb0+B(z)≢0; otherwise f has only finitely many zeros. Ifb1+β=0, then the order of the right side of (25) is 1, but the order of the left side of (25) is less than 1. This is absurd. Ifb1+β≠0, then by Lemma 9, (24), and (25), we get h3(z)≡0,B(z)≡0. This is impossible. Hence λ(f)=σ(f). Theorem 5 is thus completely proved.

Proof of Theorem <xref ref-type="statement" rid="thm1.1">2</xref>.

Since f(z) and Δηnf(z) share a CM and f is of finite order, then
(26)Δηnf(z)-af(z)-a=eQ(z),
where Q(z) is a polynomial with degQ≤σ(f). Now we will take two steps to complete the proof.

Step 1. We prove that λ(f-a)=σ(f).

Let F(z)=f(z)-a; then
(27)λ(F)=λ(f-a),σ(F)=σ(f)≥degQ,
and Δηnf(z)=ΔηnF(z)=∑j=0n(nj)(-1)n-jF(z+jη). By this and (26), we get
(28)∑j=1n(nj)(-1)n-jF(z+jη)+((-1)n-eQ(z))F(z)=a.
Next we discuss the following three cases.

Case 1 (degQ≥1 and σ(F)>degQ). Then σ(F)>1. By Theorem 5(i), (27), and (28), we get λ(f-a)=σ(f).

Case 2 (degQ≥1 andσ(F)=degQ). If σ(F)=degQ>1, then by Theorem 5(i), (27), and (28), we get λ(f-a)=σ(f). If σ(F)=degQ=1, then by Theorem 5(ii) and (27), we obtain that λ(f-a)=σ(f), or F has only finitely many zeros.

If F has only finitely many zeros, set
(29)F(z)=h1(z)ebz,
where h1(z)(≢0) is a polynomial and b(≠0) is a complex number; then substituting (29) into (28), we get
(30)∑j=1n(nj)(-1)n-jh1(z+jη)ebjη+(-1)nh1(z)=h1(z)eQ(z)+ae-bz.
By (30) and ΔηnF(z)≢0, we know that the order of the left side of (30) is 0 and the order of the right side of (30) is 1 unless h1(z)=-a and Q(z)=-bz. In this case, take it into the left side of (30); we have (-a)(ebη-1)n=0. Since all a,b, and η are not zero, it is impossible. Hence we get λ(f-a)=σ(f).

Case 3. Q is a complex constant. Then by (28) we get
(31)∑j=1n(nj)(-1)n-jF(z+jη)+((-1)n-c)F(z)=a,
where c(=eQ≠0) is a complex number. Suppose that λ(F)<σ(F). Let F(z)=h2(z)eh3(z), where h2(z)(≢0) is an entire function and h3(z) is a polynomial such that
(32)σ(h2)=λ(h2)=λ(F)<σ(F)=degh3.
Substituting F(z)=h2(z)eh3(z) into (31), we get
(33)∑j=1n(nj)(-1)n-jh2(z+jη)eh3(z+jη)-h3(z)+((-1)n-c)h2(z)=ae-h3(z).
Since deg(h3(z+jη)-h3(z))=degh3(z)-1,(j=1,…,n), by (32) we obtain that the order of the left side of (33) is less than degh3 and the order of the right side of (33) is degh3. This is absurd. Hence we get λ(f-a)=σ(f).

Step 2. We prove that σ(f)≥1.

Suppose that σ(f)<1. Since f(z) and Δηnf(z) share a CM, then
(34)Δηnf(z)-af(z)-a=c,
where c is a nonzero constant. Let F(z)=f(z)-a; then by (34) we get
(35)ΔηnF(z)=cF(z)+a.
Differentiating (35), we get
(36)(ΔηnF(z))′=cF′(z).
Note that (ΔηnF(z))′=Δηn(F′(z)) and σ(F′)=σ(F)=σ(f)<1. So by Lemma 10 and (36), we get
(37)|c|=|Δηn(F′(z))F′(z)|≤|z|n(σ(F)-1)+ɛ⟶0.
This is absurd. So σ(f)≥1. Theorem 2 is thus completely proved.

Proof of Theorem <xref ref-type="statement" rid="thm1.2">4</xref>.

Since f(z) and Δηnf(z) share 0 CM and f is of finite order, then
(38)Δηnf(z)f(z)=eQ(z),
where Q(z) is a polynomial. By Δηnf=∑j=0n(nj)(-1)n-jf(z+jη) and (38), we get
(39)f(z+nη)+∑j=1n-1(nj)(-1)n-jf(z+jη)+((-1)n-eQ(z))f(z)=0.
We discuss the following two cases.

Case 1. Q is a polynomial with degQ=m≥1. Then by Lemma 11 and (39), we get σ(f)≥m+1. Now we prove σ(f)≤λ(f)+1. Suppose that σ(f)>λ(f)+1; then by the Weierstrass factorization, we get f(z)=h1(z)eh2(z), where h1(z)(≢0) is an entire function and h2(z) is a polynomial such that
(40)σ(h1)=λ(h1)=λ(f),σ(f)=degh2>σ(h1)+1.
Substituting f(z)=h1(z)eh2(z) into (39), we get
(41)∑j=1n(nj)(-1)n-jh1(z+jη)eh2(z+jη)-h2(z)+(-1)nh1(z)=h1(z)eQ(z).

If σ(f)>m+1, then by (40), (41), and deg(h2(z+jη)-h2(z))=degh2(z)-1,(j=1,…,n), we obtain that the order of the left side of (41) is degh2-1 and the order of the right side of (41) is less than degh2-1. This is absurd.

If σ(f)=m+1, then by (41) we get
(42)∑j=1n(nj)(-1)n-jh1(z+jη)eh2(z+jη)-h2(z)-h1(z)eQ(z)=(-1)n+1h1(z).
Set
(43)h2(z)=dm+1zm+1+⋯+d0,
where dm+1(≠0),…,d0 are complex numbers. Then
(44)h2(z+jη)-h2(z)=dm+1(m+1)jηzm+⋯+d1jη,hhhhhhhhhhhhhhhhh(j=1,…,n).
Now we discuss the following two subcases.

Subcase 1. deg(Q(z)-(h2(z+jη)-h2(z)))=m holds for every j∈{1,…,n}. Then by (40), (42), deg(h2(z+jη)-h2(z+iη))=m,(j≠i), and Lemma 9, we get h1(z)≡0. This is absurd.

Subcase 2. There exist some j0∈{1,…,n} such that deg(Q(z)-(h2(z+j0η)-h2(z)))≤m-1. Then by (44) we have deg(Q(z)-(h2(z+jη)-h2(z)))=m for j≠j0. Merging the term -h1(z)eQ(z) into (nj0)(-1)n-j0h1(z+j0η)eh2(z+j0η)-h2(z), by (42) we get
(45)∑j=1j≠j0n(nj)(-1)n-jh1(z+jη)eh2(z+jη)-h2(z)+A(z)eh2(z+j0η)-h2(z)hh=(-1)n+1h1(z),(n≥2),
or
(46)A(z)eh2(z+j0η)-h2(z)=(-1)n+1h1(z),(n=1),
where A(z)=(nj0)(-1)n-j0h1(z+j0η)-h1(z)eQ(z)-(h2(z+j0η)-h2(z)) satisfying σ(A)<m. If n≥2, then by (40), (45), deg(h2(z+jη)-h2(z+iη))=m,(j≠i), and Lemma 9, we get h1(z)≡0. This is absurd. If n=1, then by (46) and h1(z)≢0, we get A(z)≢0. By this we know that the order of the left side of (46) is m and the order of the right side of (46) is less than m. This is absurd. Hence we get σ(f)≤λ(f)+1.

Case 2. Q is a complex constant. Then by Lemma 10 and (38), we get σ(f)≥1. Now we prove σ(f)≤λ(f)+1. Suppose that σ(f)>λ(f)+1. If σ(f)>1, then by the similar argument to that of case 1, we get h1(z)≡0. This is absurd. If σ(f)=1, then by (40) we get 0≤λ(f)<σ(f)-1=0. Since λ(f)=0, then σ(f)=λ(f)+1. Theorem 4 is thus completely proved.

4. Some Examples

The following examples show the existence of such entire functions which satisfy Theorems 2–5. Moreover, Example 2 shows that the result in Theorem 4 is the best possible.

Example 1.

Let η=1,n=2, and f(z)=(d+1)z+((d2-1)/d2)a, where a(≠0),d(≠0,±1) are constants. Then f(z) and Δηnf(z) share a CM and σ(f)=λ(f-a)=1.

Example 2.

Let η=1,n=2, and f(z)=ez. Then f(z) and Δηnf(z) share 0 CM and σ(f)=1=λ(f)+1.

Example 3.

Let η=1,n=2, and f(z)=H(z)ez, where H(z) is an entire function with period 1 such that σ(H)>1 and σ(H)∉ℕ. Then f(z) and Δηnf(z) share 0 CM and λ(f)=λ(H)=σ(H)=σ(f)>1. (Ozawa [16] proved that for any σ∈[1,∞) there exists a period entire function of order σ.)

Example 4.

The entire function f(z)=ze-z satisfies the difference equation
(47)f(z+2η)-4f(z+η)+(4-ez)f(z)=-z,
where η=-log2. Here σ(f)=1 and f has only finitely many zeros.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported by the National Natural Science Foundation of China (nos. 11201195, 11171119) and the Natural Science Foundation of Jiangxi, China (nos. 20122BAB201012, 20132BAB201008.).

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