Houston Journal of Mathematics. c 2008 University of Houston Volume 34, No. 4, 2008

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1 Houston Journal of Mathematics c 2008 University of Houston Volume 34, No. 4, 2008 SHARING SET AND NORMAL FAMILIES OF ENTIRE FUNCTIONS AND THEIR DERIVATIVES FENG LÜ AND JUNFENG XU Communicated by Min Ru Abstract. In this paper, we use the idea of sharing set to prove: Let F be a family of functions holomorphic in a domain, let a and b be two distinct complex numbers with a + b 0. If for all f F, f and f share S = {a, b} CM, then F is normal in D. As an application, we prove a uniqueness theorem.. Introduction and main results Let D be a domain in C and let F be a family of meromorphic functions defined in D. The family F is said to be normal in D, in the sense of Montel, if each sequence {f n } F contains a subsequence {f nj } that converges, spherically locally uniformly in D, to a meromorphic function or to.(see. [8]) Let f(z) and g(z) be two nonconstant meromorphic functions in the complex plane C and let a be a complex number. If g(z) = a whenever f(z) = a, we write f(z) = a g(z) = a. If f(z) = a g(z) = a and g(z) = a f(z) = a, we write f(z) = a g(z) = a and say that f and g share the value a IM (ignoring multiplicity). If f a and g a have the same zeros with the same multiplicities, we write f(z) = a g(z) = a and say that f and g share the value a CM (counting multiplicity). Let f and g be meromorphic functions, a be a finite value. If f(z) a = 0 whenever g(z) a = 0 and the multiplicity of each zero z 0 of f(z) a is greater than or equal to the multiplicity of the zero z 0 of g(z) a, we write g(z) a = 2000 Mathematics Subject Classification. 30D35, 30D45. Key words and phrases. Entire functions, Uniqueness, Nevanlinna theory, Normal family. Supported by Specialized Research Fund for the Doctoral Program of Higher Education (No ) and NSF of China(0772). 23

2 24 F. LÜ, J.F XU 0 f(z) a = 0. It is obvious that f and g share the value a CM if and only if g(z) a = 0 f(z) a = 0 and f(z) a = 0 g(z) a = 0.(see. []) Definition. Let f and g be two meromorphic function in a domain D, a, a 2 be two distinct finite complex numbers. We say that f and g share the set S={a, a 2 }, if f (S) = g (S). Schwick [0] was the first to draw a connection between values shared by functions in F (and their derivatives) and the normality of the family F. Specially, he showed that if there exist three distinct complex numbers a, a 2, a 3 such that f and f share a j (j =, 2, 3) in D for each f F, then F is normal in D. Pang and Zalcman [7] extended this result as follows. Theorem.. Let F be a family of meromorphic functions in a domain D, and let a, b, c, d be complex numbers such that c a and d b. If for each f F we have f(z) = a f (z) = b and f(z) = c f (z) = d, then F is normal in D. Fang [3] used the idea of sharing set to extend the result of Schwick and obtain: Theorem.2. Let F be a family of holomorphic functions in a domain D, and let a, a 2, a 3 be three distinct finite complex numbers. If f and f share the set S = {a, a 2, a 3 }, then F is normal in D. Naturally, we ask whether the number of elements in the set S in Theorem B can be reduced or not. We have a negative example. Example. Let S = {, }. Set F = {f n (z) : n = 2, 3, 4,...}, where f n (z) = n + 2n enz + n 2n e nz, D = {z : z < }. Then, for any f n F, we have n 2 [f 2 n(z) ] = f 2 n (z). Thus f n and f n share S CM, but F is not normal in D. From the example we see that a + a 2 = + ( ) = 0. One may ask what will happen if a + a 2 0. In fact we prove the following result. Theorem.3. Let F be a family of functions holomorphic in a domain, let a and b be two distinct finite complex numbers with a + b 0. If for all f F, f and f share S = {a, b} CM, then F is normal in D. Remark. The proof of Theorem.3 is based on some facts mostly from the ideas of Liu and Pang in a recent paper in [5].

3 SHARING SET AND NORMAL FAMILIES 25 In 999, Xu [9] proved the following normality criterion theorem. Theorem.4. Let F be a family of holomorphic functions in a domain D and let b be non-zero finite complex number. If f and f share 0, b in D, then F is normal in D. From Case in the proof of Theorem.3, we can easily get the following corollary which has greatly extended Theorem C. Corollary.5. Let F be a family of functions holomorphic in a domain, let a be a nonzero finite complex number. If for all f F, f and f share S = {0, a} IM, then F is normal in D. The following example shows that it is necessary that the complex number a is finite. Example 2. Let S = {0, }. Set F = {e nz : n =, 2,...} in the unite disc, thus f n = e nz and f n = ne nz share S, but F is not normal in. In the end, we use the theory of normal families to prove a uniqueness theorem which was discussed by Li and Yang [6] with Nevanlinna theory. Theorem.6. Let a and b be two distinct finite complex numbers with a+b 0, and let f(z) be a nonconstant entire function. If f and f share the set {a, b} CM, then one and only one of the following conclusions holds: (i) f = Ae z or (ii) f = Ae z + a + b, where A is a nonzero constant. 2. Some Lemmas Lemma 2.. [7] Let F be a family of functions holomorphic on the unit disc, suppose that there exists A such that f (z) A whenever f F and f(z) = 0, if F is not normal, then there exist, for each 0 α, (a) a number 0 < r < ; (b) points z n, z n < ; (c) functions f n F, and (d) positive number a n 0 such that a α n f n (z n + a n ξ) = g n (ξ) g(ξ) locally uniformly, where g is a nonconstant holomorphic function on C with order at most, such that g (ξ) g (0) = A +. Here, as usual, g (ξ) = g (ξ) + g(ξ) 2 is the spherical derivative. Lemma 2.2. [8] Let f be an entire function, let M be a positive number, if f (z) M for any z C, then f is of exponential type.

4 26 F. LÜ, J.F XU Lemma 2.3. [8] Let ζ be a family of meromorphic functions in a domain D, then ζ is normal in D if and only if the spherical derivatives of functions f ζ are uniformly bounded on compact subsets of D. Lemma 2.4. [] Let f be a nonconstant meromorphic functions, P k (f) denote a polynomial in f of degree k, and a i, i =,..., n denote finite constants. Let g = P k (f)f (f a ) (f a n ). If k < n, then m(r, g) = S(r, f), where and in the sequel S(r, f) will be used to denote any quantity o(t (r, f)), r, except a set of finite measure of r (0, ). In particular, if ρ(g) <, then m(r, g) = O(log r). 3. Proof of Theorem.3 Since normality is a local property, we may assume that the domain D is the unit disc. Suppose, to the contrary, that F is not normal in D. By Lemma 2., we can find z n <, ρ n 0 and f n F such that (3.) g n (ζ) = f n (z n + ρ n ζ) g(ζ) locally uniformly on C, where g is a nonconstant entire function such that g (ζ) g (0) = M = max{ a, b } +. In particular ρ(g). We know g is a nonconstant entire function. Without loss of generality, we can assume that g a has zeros in C. Let ζ 0 is a zero of g a. Consider the family G = {G n (ζ) : G n (ζ) = g n(ζ) a ρ n }. We claim G is not normal at ζ 0. In fact, g(ζ 0 ) = a and g(ζ) a. From (3.) and Hurwitz s Theorem, there exist ζ n, ζ n ζ 0 and g n (ζ n ) = a. Then G n (ζ n ) = 0. However, there exists a positive number δ such that δ = {z D : 0 < ζ ζ 0 < δ} D and g(ζ) a in δ. Thus for each ζ δ, g n (ζ) a (for n sufficiently large ). Therefore for each ζ δ, we have G(ζ) =. Thus we have proved that G is not normal at ζ 0. Noting that G n (ζ) = 0 G n(ζ) = a or b, and using the Lemma 2. again, we can find ζ n ζ 0, η n 0 and G n G such that F n (ξ) = G n(ζ n + η n ξ) = f n(z n + ρ n (ζ n + η n ξ)) a F (ξ) η n ρ n η n locally uniformly on C, where F is a nonconstant entire function such that F (ξ) F (0) = M. In particular ρ(f ).

5 SHARING SET AND NORMAL FAMILIES 27 We claim that () F only has finitely many zeros. (2) F (ξ) = 0 F (ξ) = a or b. We first prove Claim (). Suppose ζ 0 is a zero of g(ζ) a with multiplicity k. If F (ξ) has infinitely many zeros, then there exist k + distinct points ξ j (j =,, k + ) satisfying F (ξ j ) = 0 (j =,, k + ). Noting that F (ξ) 0, by Hurwitz s Theorem, there exists N, if n > N, we have F n (ξ jn ) = 0 (j =,, k + ) and g n (ζ n + η n ξ jn ) a = 0. Note that lim ζ n + η n ξ jn = ζ 0, (j =,, k + ) n then ζ 0 is a zero of g(ζ) a with multiplicity at least k+, which is a contradiction. Thus we have proved Claim (). Next we prove Claim (2). Suppose that F (ξ 0 ) = 0, then by Hurwitz s Theorem, there exist ξ n, ξ n ξ 0, such that (for n sufficiently large) F n (ξ n ) = f n(z n + ρ n (ζ n + η n ξ n )) a ρ n η n = 0. Thus f n (z n + ρ n (ζ n + η n ξ n )) = a. By the assumption, we have f n(z n + ρ n (ζ n + η n ξ n )) = a or b, hence F (ζ 0 ) = lim f n(z n + ρ n (ζ n + η n ξ n )) = a or b. n This proves F (ξ) = 0 F (ξ) = a or b. In the following, we will prove F (ξ) = a or b F (ξ) = 0. Suppose that F (ξ 0 ) = a. Obviously F a, for otherwise F (0) F (0) = a < M, which is a contradiction. Then by Hurwitz s Theorem, there exist ξ n, ξ n ξ 0, such that (for n sufficiently large) F n(ξ n ) = f n(z n + ρ n (ζ n + η n ξ n )) = a. It follows that F n (ξ n ) = f n (z n + ρ n (ζ n + η n ξ n )) = a or b. If there exists a positive integer N, for each n > N, we have f n (z n + ρ n (ζ n + η n ξ n )) = b. Then f n (z n + ρ n (ζ n + η n ξ n )) a F (ξ 0 ) = lim =, n ρ n η n it contradicts with F (ξ 0 ) = a. Hence there exists a subsequence of {f n }(which, renumbering, we continue to denote by {f n }) satisfying that f n (z n + ρ n (ζ n + η n ξ n )) = a.

6 28 F. LÜ, J.F XU Thus we derive f n (z n + ρ n (ζ n + η n ξ n )) a F (ξ 0 ) = lim = 0, n ρ n η n which implies F = a F = 0. Similarly, we can get F = b F = 0. Hence we have proved Claim (2). Since ρ(f ) = ρ(f ), then by the Nevanlinna s second fundamental theorem, (3.2) T (r, F ) N(r, F a ) + N(r, F b ) + S(r, F ) N(r, F a ) + N(r, F ) + O(log r) b N(r, ) + O(log r). F From Claim (), we get N(r, F ) = O(log r). Thus T (r, F ) = O(log r), it is clear that F is a polynomial. In the following, we consider two cases: Case : ab = 0. Without loss of generality we assume a = 0. We know that F has zeros, then F has multiple zeros. We assume deg(f ) = n, then T (r, F ) = (n ) log r and S(r, F ) = O(). By (3.2) we get T (r, F ) = (n ) log r N(r, ) + O() (n ) log r. F Thus we derive that F only has one multiple zeros with multiplicity 2 and F only has one zero with multiplicity, which yields that n = 2. Set F = B(ξ ξ 0 ), then F = (B/2)(ξ ξ 0 ) 2, which contradicts with F = b F = 0. This completes the proof of Case. Case 2: ab 0. We first prove F = 0 F = a or b. From ab 0, we get F = 0 F = a or b. Thus we only need to prove F = a or b F = 0. Suppose ξ 0 is a zero of F a with multiplicity m. By Rouché theorem, there exist m sequences {ξ in }(i =, 2, m) on D δ/2 = {ξ : ξ ξ 0 < δ/2} such that F n(ξ in ) = a. Then f n(z n + ρ n (ζ n + η n ξ in )) = F n(ξ in ) = a (i =, 2,, m). By f and f share {a, b} CM, we get f a only has simple zeros. Thus ξ in ξ jn ( i j m). We obtain f n (z n + ρ n (ζ n + η n ξ in )) = a or b (i =, 2,, m).

7 SHARING SET AND NORMAL FAMILIES 29 We claim that there exist infinitely many n satisfying (3.3) f n (z n + ρ n (ζ n + η n ξ in )) = a (i =, 2,, m). Otherwise we assume that for all n, there exist j (,..., m) satisfying f n (z n + ρ n (ζ n + η n ξ jn )) = b. We take a fixed number l (,..., m) satisfying (for infinitely many n) Hence f n (z n + ρ n (ζ n + η n ξ ln )) = b. f n (z n + ρ n (ζ n + η n ξ ln )) a b a F (ξ 0 ) = lim = lim =, n ρ n η n n ρ n η n which contradicts with F (ξ 0 ) = a. This proves (3.3). Therefore, F n (ξ in ) = 0, (i =, 2,, m) and ξ in ξ jn ( i j m). As n, we get ξ 0 is a zero of F with multiplicity at least m. This proves F = a F = 0. Similarly we can get F = b F = 0. Thus we have proved F = 0 F = a or b. From this, we know F a and F b only have simple zeros. Suppose that deg(f ) = n, then n = 2(n ) and n = 2. Set F = A(ξ ξ )(ξ ξ 2 ), then F = A(2ξ ξ ξ 2 ). Without loss of generality, we assume that F (ξ ) = a and F (ξ 2 ) = b, we get a + b = 0. It is a contradiction. Thus we complete the proof of Theorem Proof of Theorem.6 We first prove that f is of exponential type. Set F = {f(z + w) : w C}, then F is a family of holomorphic functions on the unit disc. By the assumption, for any function g(z) = f(z + w), we have g and g share S = {a, b} CM for all z. Hence by Theorem.3, F is normal in. Thus by Lemma 2.3, there exists M > 0 satisfying f (z) M for all z C. By Lemma 2.2, we get f is of exponential type. By the assumption, we know that f is a transcendental entire function. Now, we consider two cases: Case : ab = 0. Without loss of generality, we may assume a = 0. Then from the assumption f and f share S = {0, b} CM, we get f 0. By ρ(f ) = ρ(f)

8 220 F. LÜ, J.F XU, we set f = ABe Bz, where A, B are two nonzero constants. Then we get f = Ae Bz + c, where c is a constant. If 0 is a picard value of f, we get c = 0. From f = b f = b, we get B =, then f = f = Ae z. If b is a picard value of f, we get c = b. From f = 0 f = b, we get B =, then f = Ae z + b and f + f = b = a + b. If neither 0 nor b is a picard value of f, then from f = b f = b and f = 0 f = b, we get a contradiction. Case 2: ab 0. Set (4.) h = (f a)(f b) (f a)(f b). By the assumption, f and f share S = {a, b} CM, we get h is an entire function without zeros and poles. From (4.) and Lemma 2.4, we get T (r, h) = m(r, h) m(r, f 2 (a + b)f (f a)(f b) )+m(r, ab ) 2T (r, f)+s(r, f). (f a)(f b) Then ρ(h) ρ(f), hence we can assume that h = Ae Cz, where A 0 and C are two constants. From (4.), we get (4.2) C = h h = f (2f a b) (f a)(f b) f (2f a b) (f a)(f b). If z n is a zero of f, by (4.2) we get C = f (z n)(a+b) ab. Then for all n, there exists a positive number M satisfying (4.3) f (z n ) M. Set (4.4) ϕ = (f h f )(f h + f ) (f a)(f b) If ϕ 0. Then (f h f )(f h + f ) = 0, without loss of generality, we assume f h = f. Thus h = f f. We know h has no poles, then f 0. Hence we can set f = ABe Bz. Similarly as Case, we obtain the conclusion that f = Ae z or f = Ae z + a + b, where A is a nonzero constant. Now suppose that ϕ 0. We will prove ϕ is a polynomial. Rewriting (4.) in the form (4.5) (f a)(f b) = (f a)(f b)h,

9 SHARING SET AND NORMAL FAMILIES 22 and differentiating the both sides of (4.5), we have (4.6) (2f a b)f = (2f a b)f h + (f a)(f b)h. Suppose that z = a 0 is the zero of (f (z) a)(f (z) b), then (f(a 0 ) a)(f(a 0 ) b) = 0, we have 2f (a 0 ) a b 2f(a 0 ) a b = ±. From this and (4.6), we obtain (4.7) (f (a 0 )h(a 0 ) f (a 0 ))(f (a 0 )h(a 0 ) + f (a 0 )) = 0. We know that f a and f b only have simple zeros, thus by (4.7) we get ϕ is a entire function. Furthermore, since (4.8) by Lemma 2.4, we get f h f f a = f 2 bf (f a)(f b) f f a, (4.9) m(r, f h f f ) = O(log r). a Similarly, we have (4.0) m(r, f h + f f ) = O(log r). b Thus we get T (r, ϕ) = m(r, ϕ) = O(log r), which implies that ϕ is a polynomial. Noting that f (z n ) = 0 and (4.3), substituting z n into (4.4), we get (4.) ϕ(z n ) = f (z n ) 2 ab M 2 ab. Suppose first that f has infinitely many zeros, since ϕ is a polynomial, we get ϕ is a nonzero constant by (4.). Rewrite (4.4) as (4.2) f 2 A 2 e 2Cz = ϕ(f a)(f b) + f 2. Differentiating the both sides of (4.2), we get (4.3) f e 2Cz (2f A 2 C + 2f ) = ϕf (2f a b) + 2f f. Eliminating e Cz by (4.2) and (4.3), we obtain (4.4) A 2 ϕf (2f a b)f + 2A 2 f f f = 2A 2 Cf f 2 + 2f 3 + 2A 2 Cϕf (f a)(f b) + 2ϕ(f a)(f b)f.

10 222 F. LÜ, J.F XU Suppose that f has a zero b 0. Note that any a-points and b-points of f are simple, by (4.4) and ϕ 0, we get f (b 0 ) = 0 or C = 0. If f (b 0 ) = 0. Substituting b 0 into (4.4), we get ϕ = 0. It is a contradiction. If C = 0. Note that f (z n ) = 0, by (4.2) we get f (z n ) = 0. Substituting z n into (4.4), we also get ϕ = 0. It is also a contradiction. Hence 0 is a picard value of f. Set Then we get f = A B 2 e Bz. f = A e Bz + a z + a 2 and f = A B e Bz + a, where A, B and a, a 2 are constants. From the assumption f and f share S CM, we can easily get the conclusion of Theorem.6. We next suppose f only has finitely many zeros. Set f = P (z)e B2z, where B 2 is a nonzero constant and P (z) is a polynomial. We get f = Q(z)e B2z + c, where c is a constant and Q(z) is a polynomial. If c a, b. By (4.8), we have ϕ f c = [ f 2 bf (f c)(f a)(f b) f f c Thus m(r, f c ) = S(r, f). Note that Hence T (r, f) = T (r, N(r, f f (f a) f c ) = N(r, ) = S(r, f). Q(z) ]f h + f f b. f c ) + O() = m(r, f c ) + N(r, ) + O() = S(r, f), f c which is a contradiction. If c = a or b. Without loss of generality, suppose that c = a. Then (4.5) f = Q(z)e B2z + a and f = P (z)e B2z. From (4.5), we get N(r, f a ) = S(r, eb2z ), N(r, f b ) = T (r, eb2z ) + S(r, e B2z ), N(r, f a ) = T (r, eb2z ) + S(r, e B2z ) and N(r, f b ) = T (r, eb2z ) + S(r, e B2z ).

11 SHARING SET AND NORMAL FAMILIES 223 Next, by the assumption f and f share S = {a, b} CM, we obtain T (r, e B2z ) = N(r, f a ) + N(r, f b ) + S(r, eb2z ) = N(r, f a ) + N(r, f b ) + S(r, eb2z ) = 2T (r, e B2z ) + S(r, e B2z ). Thus we get T (r, e B2z ) = S(r, e B2z ), which is a contradiction. Hence we complete the proof of Theorem.6. Acknowledgement. The authors would like to thank Professor Hong-Xun Yi and the referee for their valuable suggestions concerning this paper. References [] C. A. Berenstein and R. Gay, Complex Variables. New York, 99. [2] L. A. Rubel and C. C. Yang, Values shared by an entire function and its derivative, Complex analysis, Lecture notes in Math., 599, Berlin: Springer, 977, [3] M. L. Fang, A note on sharing values and normality, J. of Math. Study, 29(996), [4] G. Gundersen, Meromorphic functions that share finite values with their derivative, J. of Math. Anal and Appl. 75(980), [5] X. J. Liu and X. C. Pang, Shared values and normal function, Acta Mathematica Sinca, Chinese Series, 50(2007), [6] P. Li and C. C. Yang, Value sharing of an entire functions and their derivatives, J. Math. Soc. Japan, 5(999), [7] X. C. Pang and L. Zalcman, Normal families and shared values, Bull. London Math. Soc. 32(2000), [8] J. Schiff, Normal families, Springer (993). [9] Y. Xu, Normality criteria concerning sharing values, Indian J. Pure Appl Math., 30(999), [0] W. Schwick, Sharing values and normality, Arch Math.(Basel) 59(992), [] H. X. Yi and C. C. Yang, The Uniqueness Theory of Meromorphic Functions, Beijing/New York: Science Press/Kluwer Academic Publishers, Received May 2, 2007 School of Mathematics & System Sciences Shandong University, Jinan, Shandong, 25000, P.R. China. address: lvfeng@mail.sdu.edu.cn School of Mathematics & System Sciences Shandong University, Jinan, Shandong, 25000, P.R. China. address: xujunf@gmail.com