Proceedings of the Japan Academy, Series A, Mathematical Sciences

Meromorphic functions sharing four values

Xiao-Min Li and Hong-Xun Yi

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Abstract

In this paper, we deal with the problem of uniqueness of meromorphic functions that share three values IM and a fourth value CM, and prove some results which answer a open question on uniqueness of meromorphic functions. Examples show that the conditions of theorems in this paper are necessary.

Article information

Source
Proc. Japan Acad. Ser. A Math. Sci., Volume 83, Number 7 (2007), 123-128.

Dates
First available in Project Euclid: 18 January 2008

Permanent link to this document
https://projecteuclid.org/euclid.pja/1200672013

Digital Object Identifier
doi:10.3792/pjaa.83.123

Mathematical Reviews number (MathSciNet)
MR2361424

Zentralblatt MATH identifier
1154.30026

Subjects
Primary: 30D35: Distribution of values, Nevanlinna theory
Secondary: 30D30: Meromorphic functions, general theory

Keywords
Nevanlinna theory uniqueness of meromorphic functions shared value normal growth

Citation

Yi, Hong-Xun; Li, Xiao-Min. Meromorphic functions sharing four values. Proc. Japan Acad. Ser. A Math. Sci. 83 (2007), no. 7, 123--128. doi:10.3792/pjaa.83.123. https://projecteuclid.org/euclid.pja/1200672013


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  • \item[(i)]$T(r,f)=T(r,g)+S(r,f)$;
  • \item[(ii)]$\sum\limits_{j=1}^{4}\overline{N}(r,\frac{1}{f-a_{j}})=2T(r,f)+S(r,f)$. \endenumerate \endlem \beginlem[see [10, Lemma 3] or [2, Theorem 4.4]] Let $f$ and $g$ be distinct nonconstant meromorphic functions that share four values $a_{1}$, $a_{2}$, $a_{3}$ and $a_{4}$ IM, where $a_{4}=\infty$. Then the following statements hold: \beginenumerate
  • \item[(i)]$N_{1}(r,0,f')=S(r,f)$ and $N_{1}(r,0,g')=S(r,f)$, where $N_{1}(r,0,f')$ and $N_{1}(r,0,g')$ “count” respectively only those points in $N(r,0,f')$ and $N(r,0,g')$ which do not occur when $f(z)=g(z)=a_{i}$ for some $i=1,2,3$.
  • \item[(ii)]For $i=1,2,3,4$, let $N_{2}(r,a_{i})$ refer only to those $a_{i}$-points that are multiple for both $f$ and $g$ and “count” each such point the number of times of the smaller of the two multiplicities. Then \beginequation* \sum_i=1^4N_2(r,a_i)=S(r,f). \endequation* \endenumerate \endlem \beginlem[see [2, Lemma 4.3]] Suppose that $f(z)$ is a nonconstant meromorphic function, and $P(f)=a_{0}f^{p}+a_{1}f^{p-1}+\cdots+a_{p}\, (a_{0}\neq 0)$ is a polynomial in $f$ with degree $p$ and coefficients $a_{j}$ $(j=0, 1, 2, \cdots, p)$ are constants, suppose furthermore that $b_{j}$ $(j=1, 2,\cdots, q)\, (q>p)$ are distinct finite values. Then \beginequation* \fracP(f)(f-b_1)(f-b_2)\cdots (f-b_q)=\sum_j=1^q\fracA_jf-b_j \endequation* where $A_{j}\, (j=1, 2, \cdots, q)$ are nonzero constants, and \beginequation* m\left(r,\fracP(f)f'(f-b_1)(f-b_2)\cdots (f-b_q)\right)=S(r,f). \endequation* \endlem Let $f$ and $g$ be two nonconstant meromorphic functions sharing the value $a$ IM. We denote by $\overline{N}_{(p,q)}(r,a)$ the reduced counting function of those common zeros of $f-a$ and $g-a$ such that $a$ is taken by $f$ with multiplicity $p$, and by $g$ with multiplicity $q$. \beginlem Let $f$ and $g$ be distinct nonconstant meromorphic functions that share four values $a_{1}$, $a_{2}$, $a_{3}$ and $a_{4}$ IM, where $a_{4}=\infty$. Then \beginalign \frac17 T(r,f) & \leq \sum_j=1^3\sum_k=1^6\overlineN_(1,k)(r,a_j) \label_2_1 & +\sum_j=1^3\sum_k=2^6\overlineN_(k,1)(r,a_j) + S(r,f). \notag \endalign \endlem \beginproof From Lemma 6 (i) we have \beginequation T(r,f)=T(r,g)+S(r,f). \label_2_2 \endequation By Lemma 7 (ii) we obtain \beginequation \sum_j=1^3 N_2 (r,a_j)=S(r,f). \label_2_3 \endequation From (\ref_2_2) and (\ref_2_3) we get for $j=1,2,3$ \beginalign & \label_2_4 & \overlineN(r,a_j)=\sum_k=1^\infty\overlineN_(1,k)(r,a_j) +\sum_k=2^\infty\overlineN_(k,1)(r,a_j)+S(r,f)\notag & =\sum_k=1^6\overlineN_(1,k)(r,a_j) +\sum_k=7^\infty\overlineN_(1,k)(r,a_j)\notag & \qquad +\sum_k=2^6\overlineN_(k,1)(r,a_j) +\sum_k=7^\infty\overlineN_(k,1)(r,a_j) +S(r,f)\notag & \leq \sum_k=1^6\overlineN_(1,k)(r,a_j) +\frac17 N\left(r,\frac1g-a_j\right)\notag & \qquad +\sum_k=2^6\overlineN_(k,1)(r,a_j) +\frac17 N\left(r,\frac1f-a_j\right)+S(r,f)\notag & \leq \sum_k=1^6\overlineN_(1,k)(r,a_j) + \frac17 T(r,g)\notag & \qquad+\sum_k=2^6\overlineN_(k,1)(r,a_j) +\frac17 T(r,f)+S(r,f)\notag & \leq \sum_k=1^6\overlineN_(1,k)(r,a_j) +\sum_k=2^6\overlineN_(k,1)(r,a_j)\notag & \qquad +\frac27 T(r,f)+S(r,f).\notag \endalign By the second fundamental theorem, we have \beginequation T(r,f)\leq \sum_j=1^3 \overlineN(r,a_j) +S(r,f). \label_2_5 \endequation From (\ref_2_4) and (\ref_2_5) we get (\ref_2_1). \endproof \sectionProof of Theorem 1. Since $f\not\equiv g$, by Lemma 1 we know that $f$ and $g$ are transcendental meromorphic functions. From (\ref_2_2) and Lemma 3, we obtain \beginequation \mu (f)=\mu (g), \sigma (f)=\sigma(g). \label_3_1 \endequation Suppose that $\mu (f)=\infty$. Noting that $\mu (f)\leq{\sigma}(f)$, we have $\mu(f)={\sigma}(f)=\infty$. Thus, the conclusions of Theorem 1 hold. In the following we suppose $\mu (f)<\infty$. Without loss of generality, we assume that $a_{1}=0$, $a_{2}=1$, $a_{3}=c$ and $a_{4}=\infty$. Put \beginalign \Phi & :=\fracf”f'-\fracf'f-\fracf'f-1-\fracf'f-c\label_3_2 & -\fracg”g'+\fracg'g+\fracg'g-1+\fracg'g-c. \notag \endalign By Lemma 5 and (\ref_2_2), we have \beginequation m(r,\Phi)=S(r,f). \label_3_3 \endequation Since $f$ and $g$ share $a_{1}$, $a_{2}$, $a_{3}$ IM and $a_{4}$ CM, by simply calculating we can see that $\Phi$ is analytic at any point $z$ such that $f(z)=g(z)=a_{i}$ for some $i=1, 2, 3, 4$. Thus, from (\ref_2_2) and Lemma 7 (i) we get \beginalign N(r,\Phi) & \leq N_1(r,0,f')+N_1(r,0,g')\label_3_4 & =S(r,f). \notag \endalign From (\ref_3_3) and (\ref_3_4) we can easily deduce \beginequation T(r,\Phi)=S(r,f). \label_3_5 \endequation Noting that $\mu (f)<\infty$, from (\ref_3_5) we can get \beginequation \liminf_r\to\infty\fracT(r,\Phi)\log r<\infty. \label_3_6 \endequation From (\ref_3_6) and Lemma 2 we can see that $\Phi$ is a rational function. From (\ref_3_2) we can deduce \beginequation \Phi=P_1+\sum_j=1^q\fracm_jz-z_j, \label_3_7 \endequation where $q\, (\geq 0)$, $m_{j}\, (1\leq j\leq q)$ are integers, and $z_{j}\, (1\leq j\leq q)$ are those points such that $f'(z_{j})=0$ and $g'(z_{j})(f(z_{j})-a_{i})(g(z_{j})-a_{i})\neq 0\, (i=1, 2, 3)$, or $g'(z_{j})=0$ and $f'(z_{j})(f(z_{j})-a_{i})(g(z_{j})-a_{i})\neq 0$ $(i=1, 2, 3)$, or $f'(z_{j})=0$ and $g'(z_{j})=0$ with the different multiplicities, but $(f(z_{j})-a_{i})(g(z_{j})-a_{i})\neq 0$ for $i=1, 2, 3$. By (\ref_3_7) and integrating two sides of (\ref_3_2), we can easily deduce \beginequation \fracf'g(g-1)(g-c)g'f(f-1)(f-c) = Q(z) e^P, \label_3_8 \endequation where $Q(z)=\prod\limits_{j=1}^{q}(z-z_{j})^{m_{j}}$ is a rational function, and $P=\int_{0}^{z} \, P_{1}(\eta) \, d\eta+A$ is a polynomial, $A$ is a constant. Set \beginequation H:=Q(z) e^P, \label_3_9 \endequation Then we have \beginequation H=\fracf'g(g-1)(g-c)g'f(f-1)(f-c). \label_3_10 \endequation Assume that $z_{0}$ is a point such that $f(z_{0})=a$ with multiplicity $p$ and $g(z_{0})=a$ with multiplicity $q$, where $a\in \{a_{1}, a_{2}, a_{3}\}$. From (\ref_3_10) we obtain \beginequation H(z_0)=\fracpq. \label_3_11 \endequation We discuss the following two cases. Case 1. Suppose that $H\equiv C$, where $C$ is a nonzero constant. By Lemma 9 we know that at least one of \beginalign* & \sum_k=1^6\overlineN_(1,k)(r,a_j)+ \sum_k=2^6\overlineN_(k,1)(r,a_j) & \neq S(r,f) (j=1,2,3) \endalign* must occur. Without loss of generality, we assume that \beginequation \sum_k=1^6\overlineN_(1,k)(r,0)+ \sum_k=2^6\overlineN_(k,1)(r,0)\neq S(r,f). \label_3_12 \endequation From (\ref_3_12) we know that at least one of \beginalign* & \overlineN_(1,1)(r,0)\neq S(r,f), & \overlineN_(1,k)(r,0)\neq S(r,f) (k=2,3,\cdots,6) \intertextand & \overlineN_(k,1)(r,0)\neq S(r,f) (k=2,3,\cdots,6) \endalign* must occur. We distinguish the following three subcases. Subcase 1.1. Suppose that $\overline{N}_{(1,1)}(r,0)\neq S(r,f)$. Then there exists a point $z_{0}$ such that $z_{0}$ is a simple zero of both $f$ and $g$. From (\ref_3_11) we have $C=1$. By (\ref_3_10) we obtain \beginequation \fracf'f(f-1)(f-c)\equiv\fracg'g(g-1)(g-c). \label_3_13 \endequation Noting that $f$ and $g$ share $0$, $1$, $c$ IM and $\infty$ CM, from (\ref_3_13) we can easily see that $f$ and $g$ share $0$, $1$, $c$ and $\infty$ CM. By Theorem A and Lemma 4, we can obtain the conclusions of Theorem 1. Subcase 1.2. Suppose that $\overline{N}_{(1,k)}(r,0)\neq S(r,f)$ $(k=2,3,\cdots,6)$. Then there exists a point $z_{0}$ such that $z_{0}$ is a simple zero of $f$, a zero of $g$ of order $k$, where $2\leq k\leq 6$. From (\ref_3_11) we have $C=\frac{1}{k}$. By (\ref_3_10) we obtain \beginequation \frack f'f(f-1)(f-c)\equiv\fracg'g(g-1)(g-c). \label_3_14 \endequation Noting that $f$ and $g$ share $0$, $1$, $c$ IM, from (\ref_3_14) we can easily see that for the common zeros of $f-a$ and $g-a$, where $a\in \{0,1,c\}$, if $a$ is taken by $f$ with multiplicity $p$, and by $g$ with multiplicity $q$, then $\frac{p}{q}=\frac{1}{k}$. Let \beginequation \phi_1:=\frackf'f(f-1)-\fracg'g(g-1). \label_3_15 \endequation It is easy to see that $\phi_{1}$ is a entire function. By Lemma 5 and Lemma 8, from (\ref_2_2) and (\ref_3_15) we obtain $m(r,\phi_{1})=S(r,f)$. Thus, \beginequation T(r,\phi_1)=S(r,f). \label_3_16 \endequation Noting that $\mu (f)<\infty$, from (\ref_3_16) we can get \beginequation \liminf_r\to\infty\fracT(r,\phi_1)\log r<\infty. \label_3_17 \endequation From (\ref_3_17) and Lemma 2 we can see that $\phi_{1}$ is a polynomial. From (\ref_3_15) we can deduce \beginequation \frac(f-1)^k gf^k (g-1)= e^\phi, \label_3_18 \endequation where $\phi=\int_{0}^{z} \, \phi_{1}(\eta) \, d\eta + A$ is a polynomial, $A$ is a constant. By (\ref_2_2) and (\ref_3_18) we have \beginalign T(r,e^\phi) & \leq T\left(r,\frac(f-1)^kf^k\right)+T\left(r,\fracgg-1\right)+S(r,f)\notag & =(k+1)T(r,f)+S(r,f). \label_3_19 \endalign Again by Lemma 3, we obtain \beginequation \mu (e^\phi)\leq \mu (f). \label_3_20 \endequation From (\ref_3_18) we obtain \beginequation \frac(f-1)^kf^k= e^\phi\cdot \fracg-1g. \label_3_21 \endequation From (\ref_3_21) we get \beginequation k T(r,f) \leq T(r,e^\phi)+T(r,g)+S(r,f). \label_3_22 \endequation By (\ref_2_2) and (\ref_3_22) we obtain \beginequation T(r,e^\phi)\geq (k-1)T(r,f)+S(r,f). \label_3_23 \endequation Noting $f$ is a transcendental, from (\ref_3_23) we know that $\phi$ is a nonconstant polynomial. Thus, \beginequation \mu (e^\phi)=\sigma(e^\phi)=\gamma, \label_3_24 \endequation where ${\gamma}$ is the degree of $\phi$. By Lemma 3 and (\ref_3_23), we obtain \beginequation \sigma (f) \leq\sigma (e^\phi). \label_3_25 \endequation Noting that $\mu (f)\leq{\sigma}(f)$, from (\ref_3_20), (\ref_3_24) and (\ref_3_25) we obtain the conclusions of Theorem 1. Subcase 1.3. Suppose that $\overline{N}_{(k,1)}(r,0)\neq S(r,f)$ $(k=2,3,\cdots,6)$. Then there exists a point $z_{0}$ such that $z_{0}$ is a simple zero of $g$, a zero of $f$ of order $k$, where $2\leq k\leq 6$. Similar to Subcase 1.2, we can obtain the conclusions of Theorem 1. Case 2. Suppose that $H$ is not a constant. By (\ref_2_2) and (\ref_3_10) and Lemma 6 (ii) we have \beginalign T(r,H) & \leq T\left(r,\fracf'f(f-1)(f-c)\right)\label_3_26 & +T\left(r,\fracg'g(g-1)(g-c)\right)+S(r,f)\notag & \leq 2 \sum_j=1^3 \overlineN(r,a_j) + S(r,f)\notag & \leq 4 T(r,f)+S(r,f). \notag \endalign Again by Lemma 3, we obtain \beginequation \mu (H)\leq \mu (f). \label_3_27 \endequation From (\ref_3_11) we obtain \beginalign & \sum_j=1^3 \sum_k=1^6\overlineN_(1,k)(r,a_j) + \sum_j=1^3\sum_k=2^6\overlineN_(k,1)(r,a_j)\label_3_28 & \leq \sum_k=1^6 \overlineN\left(r,\frac1H-\frac1k\right) +\sum_k=2^6 \overlineN\left(r,\frac1H-k\right)\notag & \leq 11 T(r,H)+ O(1). \notag \endalign From (\ref_2_1) and (\ref_3_28) we get \beginequation T(r,f)\leq 77 T(r,H)+S(r,f). \label_3_29 \endequation Noting $f$ is transcendental, from (\ref_3_29) we know that $H$ is transcendental. Again from (\ref_3_9) we know that $P$ is a nonconstant polynomial. Thus, \beginequation \mu (H)=\sigma(H)=\gamma, \label_3_30 \endequation where ${\gamma}$ is the degree of $P$. By Lemma 3 and (\ref_3_29), we obtain \beginequation \sigma (f) \leq\sigma (H). \label_3_31 \endequation Noting that $\mu (f)\leq{\sigma}(f)$, from (\ref_3_27), (\ref_3_30) and (\ref_3_31) we obtain the conclusions of Theorem 1. Theorem 1 is thus completely proved. \section*Acknowledgments. This work is supported by the National Natural Science Foundation of China (No. 10371065), and the Research Foundation of Doctor Points of China (No. 20060422049). The authors would like to thank the referee for valuable suggestions concerning this paper. \beginthebibliography99
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