J. Math. Anal. Appl. 377 20 44 449 Contents lists available at ScienceDirect Journal o Mathematical Analysis and Applications www.elsevier.com/locate/jmaa Value sharing results or shits o meromorphic unctions Xudan Luo, Wei-Chuan Lin Department o Mathematics, Fujian Normal University, Fuzhou 350007, Fujian Province, PR China article ino abstract Article history: Received 7 May 200 Available online 5 November 200 Submitted by Steven G. Krantz Keywords: Uniqueness Entire unction Dierence product In this paper, we deal with the value distribution o dierence products o entire unctions, and present some result on two dierence products o entire unctions sharing one value with the same multiplicities. The research indings also include an analogue or shit o a well-known conjecture by Brück. Our theorems improve the results o I. Laine and C.C. Yang [I. Laine, C.C. Yang, Value distribution o dierence polynomials, Proc. Japan Acad. Ser. A Math. Sci. 83 2007 48 5], K. Liu and L.Z. Yang [K. Liu, L.Z. Yang, Value distribution o the dierence operator, Arch. Math. 92 2009 270 278], and J. Heittokangas et al. [J. Heittokangas, R. Korhonen, I. Laine, J. Rieppo, J.L. Zhang, Value sharing results or shits o meromorphic unction, and suicient conditions or periodicity, J. Math. Anal. Appl. 355 2009 352 363]. Moreover, we show by illustrating a number o examples that our results are best possible in certain senses. Published by Elsevier Inc.. Introduction and main results In the paper, we assume all the unctions are nonconstant meromorphic unctions in the complex plane C. We shall use the ollowing standard notations o value distribution theory: T r,, mr,, Nr,, Nr,, Sr,,... See, e.g. [7,3]. We denote by Sr, any unction satisying Sr, = o { T r, }, as r +, possibly outside o a set with inite measure. We speciy the notion o small unctions as ollows: Given a meromorphic unction, the amily o all meromorphic unctions az such that T r, a = Sr, is denoted by S. For convenience, we also include all constant unctions in S. Moreover,Ŝ = S { }. I or some a C { },thezeroso a and g a coincide in locations and multiplicity, we say that and g share the value a CM. There has been an increasing interest in studying dierence equations and dierence product in the complex plane. Halburd and Korhonen [5] established a version o Nevanlinna theory based on dierence operators. Bergweiler and Langley [2] considered the value distribution o zeros o dierence operators that can be viewed as discrete analogues o zeros o z. Ishizaki and Yanagihara [9] developed a version o Wiman Valion theory or dierence equations o entire unctions o small growth. Growth estimates or the dierence analogue o the logarithmic derivative z+c z were given by Halburd and Korhonen [4] and Chiang and Feng [3] independently. Recently, Laine and Yang [0] investigated the value distribution o dierence products o entire unctions, and obtained the ollowing result. The project sponsored by SRF or ROCS, SEM, and the National Natural Science Foundation o China Grant No. 06709. * Corresponding author. E-mail address: sxlwc936@jnu.edu.cn W.-C. Lin. 0022-247X/$ see ront matter Published by Elsevier Inc. doi:0.06/j.jmaa.200.0.055
442 X. Luo, W.-C. Lin / J. Math. Anal. Appl. 377 20 44 449 Theorem A. Let z be a transcendental entire unction o inite order, and c be a nonzero complex constant. Then or n 2, z n z + c assumes every nonzero value a C ininitely oten. The restriction in Theorem A to the inite order case is essential. As an example, take z = exp e z and e c = n, then is o ininite order such that z n z + c. Theorem A does not remain valid i n =. Indeed, take z = + e z,then z z + π i = e 2z. Aterwards, Liu and Yang [] improved Theorem A, and proved the ollowing result. Theorem B. Let z be a transcendental entire unction o inite order, and c be a nonzero complex constant. Then or n 2, z n z + c pz has ininitely many zeros, where Pz 0 is a polynomial in z. In this paper, we will establish an improvement o Theorem A and Theorem B, which is stated as ollows. Theorem. Let be a transcendental entire unction o inite order σ and c be a ixed nonzero complex constant, let Pz = a n z n + a n z n + +a z + a 0 be a nonzero polynomial, where a 0, a,...,a n 0 are complex constants, and m is the number o the distinct zeros o Pz. Then or n > m, P z + c = az has ininitely many solutions, where az S \{0}. We shall give a much simple proo o Theorem in Section 3, which is dierent rom Res. [0,]. Remark. The ollowing examples show that Theorem A, Theorem B and Theorem may ail to occur or meromorphic unctionsoiniteorder. Example. Let Pz = z 3 z 2 + 2, = tan z, c = π 2.WegetP z + π 2 = or this example. Example. Let Pz = z 2, z = 3tanz +, c = 2π 3.WegetP z + 2π 3 = 6 and Theorem ail to occur or this example. +. Clearly, Theorem ails to occur cos 4 z 8 8. Clearly, Theorem A, Theorem B cos 2 z Corresponding to the above result, we investigate the uniqueness o dierence products o entire unctions, and obtain the next result. For the sake o simplicity, we use the deinition as ollows. Deinition. Let Pz = a n z n + a n z n + +a z + a 0 a n 0, wedenoteγ 0 = m + 2m 2, where m is the number o the simple zero o Pz, and m 2 is the number o multiple zeros o Pz. Wedenoted = GCD{λ 0,λ,...,λ n }, where { i +, ai 0, λ i = n +, a i = 0, i = 0,, 2,...,n. Theorem 2. Let and g be transcendental entire unctions o inite order, c be a nonzero complex constant, Pz = a n z n +a n z n + +a z + a 0 be a nonzero polynomial, where a 0, a,...,a n 0 are complex constants, and let n > 2Γ 0 + be an integer. I P z + c and Pggz + c share CM, then one o the ollowing results holds: tg or a constant t such that t d =,wheredisdeinedabove; 2 and g satisy the algebraic equation R, g 0,whereRw, w 2 = Pw w z + c Pw 2 w 2 z + c; 3 z = e αz,gz = e βz,whereαz and βz are two polynomials, b is a constant satisying α + β banda 2 n en+b =. Remark 2. The ollowing example shows that the second case o Theorem 2 may occur. Let Pz = z 6 z + 6 z, z = sin z, gz = cos z and c = 2π. It is easy to see that n > 2Γ 0 + and P z + c Pggz + c, sop z + c and Pggz + c share CM. Clearly, we get tg or a constant t such that t m =, where m Z +, but and g satisy the algebraic equation R, g 0, where Rw, w 2 = Pw w z + c Pw 2 w 2 z + c. However, when Pz is a nonzero monomial, the second case o Theorem 2 may be deleted. Indeed, or instance, let Pz = z n, n > 2, then rom the algebraic equation R, g 0, we have n = g gz + c z + c.
X. Luo, W.-C. Lin / J. Math. Anal. Appl. 377 20 44 449 443 This relation and Lemma 3 yields that n T r, = S r,. g g It ollows that g is a constant, that is, tg or a constant t such that tn+ =. Thereore, we obtain the ollowing result. Corollary. Let and g be transcendental entire unctions o inite order, c be a nonzero complex constant. Suppose that n z + c and g n gz + c share CM, and n > 5 is an integer, then one o the ollowing results holds: tg or a constant t such that t n+ = ; 2 g t,wheret n+ =. We continue to our study in this paper by establishing shared value problems related to a meromorphic unction z and its shit z + c, where c C. Currently, J. Heittokangas, R. Korhonen, I. Laine, J. Rieppo and J. Zhang [8] obtained that i z is o inite order and shares two values CM and one value IM Ignoring multiplicities with its shit z + c, then is a periodic unction with period c. It is natural to investigate whether there exist uniqueness theorems under reducing the number o the shared small periodic unctions. However, the ollowing two counterexamples show that may not be a periodic unction i z and z + c only have two shared values. Example. Let = e z sin z +, c = 2π. Clearly, we get and z + c share CM, but is not periodic unctions. Example. Let = ez sin z +, c = 2π. sin z Clearly, we get and z + c share, CM, but is not periodic unctions. z However, rom the above examples, we ind that the two unctions satisy = τ or some constant τ.thesuch z+c result can be named as a shit analogue o Brück s conjecture []. The ollowing result is due to Heittokangas et al., see [8, Theorem ]. Theorem C. Let be a meromorphic unction o order o growth ρ := lim sup r log T r, < 2, log r and let c C.I z and z + C shared the values a C and CM, then z a z + c a = τ or some constant τ. Here, we also study the shit analogue o Brück s conjecture by relaxing the growth condition, and obtain the result as ollows. Theorem 3. Let be a nonconstant meromorphic unction o inite order, n 6 be an integer. I n and n z+c share az S \{0} and CM, then wz + c, or a constant w satisying w n =. 2. Some lemmas Next, or the proo o our theorems, we still need the ollowing lemmas. Lemma. See [2]. Let and g be two nonconstant meromorphic unction. I and g share CM, one o the ollowing three cases holds:
444 X. Luo, W.-C. Lin / J. Math. Anal. Appl. 377 20 44 449 T r, N 2 r, + N 2 r, g + N 2 r, + N 2r, + Sr, + Sr, g, the same inequality holding or T r, g; g 2 g; 3 g. Lemma 2. See [4,5]. Let z be a meromorphic unction o inite order σ, and let c be a ixed nonzero complex constant. Then or each ε > 0,wehave m r, z + c z + m r, z = O r σ +ε. z + c Lemma 3. See [3]. Let be a meromorphic unction o inite order σ,c 0 be ixed. Then or each ε > 0, wehave T r, z + c = T r, + O r σ +ε + Olog r. Lemma 4. See [6]. Let T : 0, + 0, + be a non-decreasing continuous unction, s > 0, α <,andletf R + be the set o all r such that T r αt r + s. I the logarithmic o F is ininite, then lim sup r log + T r, =. log r Lemma 5. Let be an entire unction o inite order σ, c be a ixed nonzero complex constant, and let Pz = a n z n + a n z n + + a z + a 0 be a nonzero polynomial, where a 0, a,...,a n 0 are complex constants. Then or each ε > 0, wehave T r, P z + c = T r, P z + O r σ +ε. Proo. Since has inite order σ, rom Lemma 2, we deduce that T r, P z + c = m r, P z + c m r, P z + c + m r, m r, P + O r σ +ε = T r, P + O r σ +ε. Similarly, we deduce T r, P = m r, P m r, P z + c + m r, z + c m r, P z + c + O r σ +ε = T r, P z + c + O r σ +ε. Thereore, T r, P z + c = T r, P z + Or σ +ε. Remark 3. Under the condition o Lemma 5, we have Sr, P z + c = Sr, and Sr, = Sr, P z + c. Remark 4. The ollowing example shows Lemma 5 may ail to occur or meromorphic unctions o inite order. Example. Let Pz = z n, n Z +, = z tan z, c = π 2. We get P z + π 2 = zn z + π 2 tann z and P z = z n+ tan n+ z. Clearly, Lemma 5 ails to occur or this example. Lemma 6. See [3]. Let and g be two nonconstant meromorphic unctions such that and g share, CM. I N 2 r, + N 2 r, + 2Nr, <λtr + Sr, g where λ<, Tr = max{t r,, T r, g}, Sr = 0{T r} r,r / E, and E has inite linear measure. Then gor g.
X. Luo, W.-C. Lin / J. Math. Anal. Appl. 377 20 44 449 445 Lemma 7. See [3]. Let be a nonconstant meromorphic unction, and let a 0, a,...,a n be inite complex numbers such that a n 0. Then T r, a n n + a n n + +a 0 = ntr, + Sr,. Lemma 8. Let be an meromorphic unction o inite order, c 0 be ixed. Then N r, N r, + Sr,, z + c N r, z + c Nr, + Sr,, N r, N r, + Sr,, z + c N r, z + c Nr, + Sr,, outside o a possible exceptional set with inite logarithmic measure. Proo. We will use the method o proo o Re. [8, Theorems 6, 7] to prove this lemma. By a simple geometric observation, we have N r, N r + c,. z + c Since the order o is inite, by Lemma 4, we obtain N r + c, N r, + Sr,, outside o a possible exceptional set with inite logarithmic measure. On the other hand, we have Nr, Nr + c,. Thereore, Nr + c, = Nr, + Sr,. From above, we get N r, z + c N Similarly, we obtain that r, + Sr,. N r, z + c Nr, + Sr,, N r, N r, + Sr,, z + c N r, z + c Nr, + Sr,, outside o a possible exceptional set with inite logarithmic measure. 3. Proo o Theorem Contrary to the assertion, suppose that P z + c = az has initely solutions, then by the Second Fundamental Theorem, Lemma 5 and Lemma 8, we have T r, P z + c N r, + N r, + Sr, P z + c P z + c az N r, + N r, + Sr, P z + c N r, + N r, + Sr, P m + T r, + Sr,. From Lemma 5, we get T r, P m + T r, + Sr,,
446 X. Luo, W.-C. Lin / J. Math. Anal. Appl. 377 20 44 449 i.e. n + T r, m + T r, + Sr,, which contradicts with n > m. Thus, we have completed the proo o Theorem. 4. Proo o Theorem 2 Let F = p z + c, G = pggz + c, thenf and G share CM. Applying Lemma to F and G, wegetthatoneo the ollowing three cases holds. Case. I T r, F N 2 r, F + N 2r, + Sr, F + Sr, G, by Lemma 5 and Lemma 8, we have G T r, F N 2 r, + N 2 r, + Sr, + Sr, g F G = N 2 r, + N 2 r, p z + c pggz + c N 2 r, + N 2 r, + N 2 r, p z + c pg Γ 0 T r, + Γ 0 T r, g + N r, + N r, Γ 0 T r, + Γ 0 T r, g + N r, z + c + N r, g Γ 0 + T r, + Γ 0 + T r, g + Sr, + Sr, g. From Lemma 5, we have + Sr, + Sr, g + N 2 r, gz + c + Sr, + Sr, g T r, P Γ 0 + T r, + Γ 0 + T r, g + Sr, + Sr, g. By Lemma 7, we deduce + Sr, + Sr, g gz + c + Sr, + Sr, g n + T r, Γ 0 + T r, + Γ 0 + T r, g + Sr, + Sr, g. 4. Similarly, we obtain n + T r, g Γ 0 + T r, + Γ 0 + T r, g + Sr, + Sr, g. 4.2 Combining 4. and 4.2, we have n + [ T r, + T r, g ] 2Γ 0 + 2 [ T r, + T r, g ] + Sr, + Sr, g, which contradicts with n > 2Γ 0 +. Case 2. I F G, that is P z + c Pggz + c. 4.3 Set h =,ih is a constant, then substituting = gh into 4.3, we deduce that g gz + c [ a n g n h n+ + a n g n h n + +a 0 h ] 0, where a 0, a,...,a n 0 are complex constants. Since g is transcendental entire unction, hence gz + c 0. From above, we get a n g n h n+ + a n g n h n + +a 0 h 0. 4.4 We claim that h d =, where d is deined as in Deinition. Thus, tg or a constant t such that t d =. In act, we discuss the ollowing subcases. Subcase. Suppose that a n is the only nonzero coeicient. Since g is transcendental entire unction, we have h n+ =.
X. Luo, W.-C. Lin / J. Math. Anal. Appl. 377 20 44 449 447 Subcase 2. Suppose that a n is not the only nonzero coeicient. I h n+, by Lemma 7 and 4.4, wededucet r, g = Sr, g, which is a contradiction. Hence, h n+ =. According to the similar discussion, we obtain that h k+ = when a k 0 or some k = 0,...,n. Thereore, we get tg or a constant t such that t d =, where d = GCDλ 0,λ,...,λn. I h is not a constant,then we know by 4.3 that and g satisy the algebraic equation R, g 0, where Rw, w 2 = Pw w z + c Pw 2 w 2 z + c. Case 3. I FG, that is P z + cpggz + c. From the assumption that and g are two nonconstant entire unctions, we deduce by 4.5 that P 0, Pg 0. By Picard s theorem, we claim that P = a n a n, Pg = a n g a n, where a is a complex constant. Otherwise, the Picard s exceptional values are at least three, which is a contradiction. Hence, rom the assumption that and g be transcendental entire unctions o inite order, we obtain that z = e αz + a, gz = e βz + a, where αz and βz are two nonconstant polynomials. By 4.5, we also get z + c 0, gz + c 0. So a = 0, i.e. z = e αz, gz = e βz, Pz = a n z n, and a 2 n enαz+βz+αz+c+βz+c. Dierentiating this yields n α z + β z + α z + c + β z + c 0. 4.6 Let ωz = α z + β z, we deduce by 4.6 that nωz + ωz + c 0. Since ωz is a polynomial, we suppose degωz = m, and z,...,z m are the zeros o ωz. Thus,z + c,...,z m + c are also the zeros o ωz. Thereore, ω 0, α + β b, where b is a constant. From this we can easily obtain that z = e αz, gz = e βz, where αz and βz are two polynomials, b is a constant satisying α + β b and a 2 n en+b =. This completes the proo o Theorem 2. 5. Proo o Theorem 3 4.5 4.7 Let F = n az, G = n z+c az,thenf and G share, CM. By Lemma 3, we have Sr, G = Sr,. Set F H = F F G GG. We distinguish two cases as ollows. 5. Case. I H 0, we get F B B G, where B is a nonzero constant. We claim that F G, that is, n n z + c, which implies that wz + c, or a constant w satisying w n =. In act, we discuss the ollowing subcases. Subcase. Suppose that Nr, Sr,, then there exists z which is not a zero or pole o az such that F and F z + c share CM, so F z = Gz = 0. We get rom 5. that B =, so F G. Subcase 2. Suppose that Nr, = Sr,. IB, then we have N r, = Nr, G = Sr, F. B Hence, T r, F Nr, F + N r, + N r, F F B which contradicts with n 6. Thereore, B =. Thus, F G. + Sr, F N r, + Sr,, z = 0. Since
448 X. Luo, W.-C. Lin / J. Math. Anal. Appl. 377 20 44 449 Case 2. I H 0, by Lemma 8 and 5., we deduce that Nr, H N r, + N r, + Sr, z + c 2N r, + Sr, 2T r, + Sr,. 5.2 Thereore, by a logarithmic derivative theorem and 5.2, we get T r, H 2T r, + Sr,. 5.3 Suppose that z 0 is a pole o with multiplicity p, then an elementary calculation gives that z 0 is the zero o H with multiplicity at least np. From this and 5.3, we have n Nr, N Hence, r, H Nr, 2 T r, + Sr,. n By 5.4 and Lemma 8, we obtain that N 2 r, + N 2 r, F G 2T r, + Sr,. + 2Nr, F 2N 4N r, + 2N r, z + c r, + 2Nr, + Sr, Set T r = max{t r, F, T r, G}=nTr, + Sr,. From 5.5, we have N 2 r, + N 2 r, + 2Nr, F 4 + 4 n T r + Sr,, F G n + 2Nr, + Sr, 5.4 4 + 4 T r, + Sr,. 5.5 n which contradicts with n 6. Using Lemma 6, we get F G or FG. I F G, that is, n n z + c, which implies that wz + c, or a constant w satisying w n =. I FG, that is F zf z + c a 2 z, which implies Nr, F = Nr, = Sr,. Since F 2 z = a2 zf z F z+c, mr, F = Sr,, F z F z+c = N r, a2 z F 2 zf z+c,wehave F z F z + c = Sr,. Hence, T r. = Sr,, which is a contradiction. This completes the proo o Theorem 3. 6. Discussion We irstly denote M by the set o meromorphic unctions in complex plane such that Nr, = Sr,. In irst section, we have discussed the value distribution o dierence products o entire unctions, and presented the examples to show that Theorem is not valid or meromorphic unctions. It remains an open question under what conditions Theorem holds or meromorphic unctions o inite order. Here, we show that this problem is valid or the meromorphic unction M with inite order. Theorem 4. Let M be a meromorphic unction o inite order σ and c be a ixed nonzero complex constant, let Pz = a n z n + a n z n + +a z + a 0 be a nonzero polynomial, where a 0, a,...,a n 0 are complex constants, and m is the number o the distinct zeros o Pz. Then or n > m, P z + c = az has ininitely many solutions, where az S \{0}. F z+c
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