ON FACTORIZATIONS OF ENTIRE FUNCTIONS OF BOUNDED TYPE

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1 Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 29, 2004, ON FACTORIZATIONS OF ENTIRE FUNCTIONS OF BOUNDED TYPE Liang-Wen Liao and Chung-Chun Yang Nanjing University, Department of Mathematics Nanjing, China; The Hong Kong University of Science & Technology Department of Mathematics Kowloon, Hong Kong; Abstract. We prove that if f is a transcendental entire function and the set of all finite singularities of its inverse function f 1 is bounded, then fz)+p z) is prime for any nonconstant polynomial P z), unless fz) and P z) has a nonlinear common right factor. Particularly, it is shown that fz) + az is prime for any constant a Introduction A transcendental meromorphic function F is said to be prime pseudo-prime) if, and only if, whenever F = fg) for some meromorphic functions f and g, either f or g must be bilinear rational); F is called left-prime right-prime) if every factorization of F implies that f is bilinear whenever g is transcendental g is linear if f is transcendental). It is easily seen F is prime if and only if F is leftprime as well as right-prime. We refer the readers to [3] or [4] for an introduction to the factorization theory of entire and meromorphic functions. A point a is called a singularity of f 1 the inverse function of f ), if a is either a critical value or asymptotic value of f. We denote by singf 1 ) the set of all finite singularities of f 1, i.e. singf 1 ) = {z C : z is a singularity of f 1 }. We denote by B the class of all entire functions f such that singf 1 ) is bounded and by S the class of all entire functions f such that singf 1 ) is finite. If f B f S ), we say f is of bounded finite) type. In 1981, Noda [8] proved the following result. Theorem A. Let fz) be a transcendental entire function. Then the set is at most countable NP f) = {a a C, fz) + az is not prime} 2000 Mathematics Subject Classification: Primary 30D35.

2 346 Liang-Wen Liao and Chung-Chun Yang As a further study on the cardinality of NP f), which is denoted by NP f), Ozawa and Sawada [9] posed the following interesting question: Question. Is there any f for which the exceptional set NP f) in Theorem A is really infinitely countable? Or what is the maximal cardinal number of the exceptional set NP f)? Theorem B Ozawa and Sawada [9]). Let Gw) be an entire function satisfying M R, Gw) ) expkr) for R R 0 > 0 and for some constant K > 0. Then either Ge z ) + az or Ge z ) + bz is prime if aba b) 0. This shows that the cardinality of NP Ge z )) is at most 2 if MR, Gw)) expkr) for R R 0 > 0 and for some constant K > 0. As a study of the above question, Liao Yang [6] proved the following result. Theorem C. Let f be a transcendental entire function of finite order in S. Then for any constant a 0, fz) + az is prime, i.e. NP f) 1. Recently, Wang Yang [13] proved the following theorem. Theorem D. Let P, Q be nonconstant polynomials, α B, h a periodic entire function of order one and mean type, Gz) = P h αz). If F z) = G n z) + Qz) has a factorization F z) = f gz) ), then gz) must be a common right factor of αz) and Qz). Remar 1. The original statement of Theorem D only requires that h is of order one. Here we would lie to point out that h should be at most order one of mean type, as it is needed in the proof of Theorem D, Lemma 5 in [13]. However, f in Lemma 5 should be an entire function of exponential type, i.e. f has order less than one or order one and mean type; see p. 27 in [4]. Remar 2. Let G be defined in Theorem D. Then G n z) + az is prime for any constant a 0. As a continuation of the study of our previous wor [6], we are able to extend Theorem C to a large class of functions, namely, functions of bounded type. The following is our main result. Theorem. Let f be a transcendental entire function in B, then for any nonconstant polynomial P z), fz) + P z) is prime unless fz) and P z) has a nonlinear common right factor.

3 On factorizations of entire functions of bounded type Some lemmas Lemma 1 Rippon and Stallard [11]). Let f be a meromorphic function with a bounded set of all finite critical and asymptotic values. Then there exists K > 0 such that if z > K and fz) > K, then f z) fz) log fz). 16π z Lemma 2 [5]). Let f be a transcendental entire function, and 0 < δ < 1 4. Suppose that at the point z with z = r the inequality 1) fz) > Mr, f)νr, f) 1/4)+δ holds. Then there exists a set F in R + F and of finite logarithmic measure, i.e., dt t < + such that ) m νr, f) 2) f m) ) z) = 1 + o1) fz) z holds whenever m is a fixed nonnegative integer and r / F. Lemma 3 Baer and Singh [1], also see [2]). Let f and g be two entire functions. Then sing f g) 1) singf 1 ) f singg 1 ) ). Lemma 4 Polya [10]). Let f and g be two transcendental entire functions. Then Mr, f g) lim =. r Mr, g) Lemma 5. Let f be a transcendental entire function. Then for a sufficiently large r. Mr, f ) Mr, f) 2 Remar 3. This follows easily from a result of Valiron [12]): log Mr, f ) lim r log Mr, f) = 1.

4 348 Liang-Wen Liao and Chung-Chun Yang 3. Proof of the theorem Let F z) = fz) + P z), P z) is a nonconstant polynomial. We first prove that F is pseudo-prime. Assume that F z) = g hz) ), where g is a transcendental meromorphic function with at most one pole and h is a transcendental entire function. Thus 3) fz) = g hz) ) P z), f z) = g hz) ) h z) P z). First we consider the case that g is a transcendental entire function, and then we discuss two situations. Case 1: g has at least two zeros. Then there exists a zero c of g such that hz) = c has infinitely many roots {z } =1. Thus we have By Lemma 1, we would have which leads to a contradiction. fz ) = P z ) + gc), f z ) = P z ). P z ) P z ) gc) log P z ) gc), 16π z Case 2: g has at most one zero. Thus g w) = w w 0 ) n e αw), f z) = hz) w 0 ) ne αhz)) h z) P z), where n is a non-negative integer. Let Kz) = e αhz))/n+3), and assume that Γ is a simple curve tending to infinity such that if z Γ and z = r, then Kz) = Mr, K). By Lemmas 4 and 5, we have, if z Γ and z = r is sufficiently large, g hz) ) h z) ) = ne hz) αhz)) w0 h z) 4) ) nh hz) w 0 z) 1 = Mr, K) n+3 Mr, K) 0. Let Lz) = αhz))/n + 3) and Ar, L) = max z =r Re Lz). Thus if z Γ, Kz) = Mr, K) = e Ar,L), Re Lz) = Ar, L). By Hadamard s three-circle theorem, we have, for r 1 < r 2 < r 3, 5) Ar 2, L) log r 2 log r 1 log r 3 log r 1 Ar 3, L) + log r 3 log r 2 log r 3 log r 1 Ar 1, L).

5 For z 0 Γ, we have On factorizations of entire functions of bounded type 349 6) L Lz) Lz 0 ) z 0 ) = lim z z 0,z Γ z z 0 lim z z 0,z Γ Re Lz) Re Lz 0 ). z z 0 Let z 0 = r 0 and z = r 0 + h, h > 0, then as z z 0, h 0. Thus, by 5) and 6), we have, for sufficiently large r 0, 7) L Ar 0 + h, L) Ar 0, L) z 0 ) lim z z 0,z Γ z z 0 h Ar 0 + h, L) Ar 0, L) = lim z z 0,z Γ z z 0 h = lim h 0 Ar 0 + h, L) Ar 0, L) h lim h 0 log1 + h/r 0 ) log r 0 Ar0, L) A1, L) ) = Ar 0, L) A1, L) r 0 log r o > 1. h Let w = Gz) = e αhz))/n+3) = e Lz). Thus 0 is an asymptotic value of G and Γ is the corresponding asymptotic curve, γ = GΓ) is a simple curve connecting G0) and 0. Let B be the length of γ, which is a finite number. And dw = e Lz) L z)dz. By this, 4) and 7), if z Γ, we have ) z g hz) = g hz) ) h z) dz + g hz 0 ) ) z 0 along Γ z z 0 along Γ g hz) ) h z) dz + g hz0 ) ) w 1 w 0 along γ w L z) dw + g hz0 ) ) w 0 along γ dw + g hz0 ) ) B + g hz0 ) ). Thus we can find a sequence of {z } =1 such that z as, and fz ) P z ), f z ) P z ). A contradiction follows from this and Lemma 1.

6 350 Liang-Wen Liao and Chung-Chun Yang If g has just one pole w 1, so does g, then hz) does not assume w 1, i.e., hz) = e βz) + w 1. Moreover, if g has a zero c, then hz) = c has infinitely many roots. One can derive a contradiction by arguing similarly as in Case 1. Hence g has no zeros, i.e., g 1 w) = w w 1 ) n eαw), and g hz) ) h z) = β z) exp α e βz)+w 1) + 1 n)βz) ). By the same argument as that in Case 2 above, we can get a contradiction. Thus F z) = fz) + P z) is pseudo-prime. Now we assume that F z) has the following factorization: F z) = fz) + P z) = Q gz) ), where Q is rational, g is a transcendental meromorphic function. If Q is a polynomial, then g is entire. If Q has a pole w 1, then gz) does not assume w 1. Thus hz) = 1/gz) w 1 ) is an entire function and F z) = Q 1 hz) ), where Q1 is a rational function. Without loss of generality, we may assume that gz) is entire, and Qw) has at most one pole. Now we discuss the following two sub-cases. Subcase 1: Q has one pole, say w 0, i.e., Qw) = Q 1 w)/w w 0 ) n, where Q 1 w) is a polynomial with degree m and Q 1 w 0 ) 0. Then gz) = w 0 + e hz), where hz) is a nonconstant entire function. Thus we have fz) = Q 1 w 0 + e hz) )e nhz) P z) = a 0 e nhz) + a 1 e n 1)hz) + + a m e m n)hz) P z), where a 0, a 1,..., a m are constants and a m 0, a 0 = Q 1 w 0 ) 0. Thus f z) = na 0 e nhz) n 1)a 1 e n 1)hz) + + m n)a m e m n)hz) )h z) P z) = [ na 0 n 1)a 1 e hz) + + m n)a m e mhz)] e nhz) h z) P z) = P 1 e hz) )e nhz) h z) P z), where P 1 w) is a polynomial and P 1 0) = na 0 0. If P 1 w) is a nonconstant polynomial, then P 1 w) has a zero c 0 and e hz) = c has infinitely many roots. Let {z } + =1 be zeros of ehz) c, then f z ) = P z ) and fz ) = Q 1w 0 + c) c n P z ).

7 On factorizations of entire functions of bounded type 351 Again, by Lemma 1, we have a contradiction. If P 1 w) is a constant polynomial, then fz) = a 0 e nhz) + a m P z), f z) = na 0 e nhz) h z) P z). Let Kz) = e nhz) and z = r, Kz ) = Mr, K). Then by Lemma 2, we have, for r / F, na 0 e nhz ) h z ) = a 1 0 Kz ) a 0 e nhz ) = a 0 Mr, K). K z ) Kz ) = a 1 0 Mr, K) νr, K) 1 + o1)), r ) Noting lim r νr, K)/Mr, K) = 0 for a transcendental entire function K, we can find a sequence of {z } + =1 such that fz ) P z ), f z ) P z ). A contradiction follows from this and Lemma 1. Subcase 2: Qw) has no pole, i.e., Qw) is a polynomial with degree 2. If Q w) has at least two distinct zeros, then there exists a zero w 1 of Q w) such that gz) = w 1 has infinitely many zeros {z n } + n=1. Then f z n ) = Q gz n ) ) P z n ) = P z n ), fz n ) = Qw 1 ) + P z n ). However, by Lemma 1, f z n ) fz n) log fz n ), 16π z n which will lead to a contradiction. Therefore, we only need to treat the case that Q w) has only one zero w 0. If gz) w 0 has infinitely many zeros, again a contradiction follows from Lemma 1. Hence, we have gz) = w 0 + p 1 z)e hz) and Q z) = Aw w 0 ) n 1, where p 1 z) is a polynomial, hz) a nonconstant entire function. Thus Qw) = A n w w 0) n + B, fz) = A n p 1z) n e nhz) + B P z), f z) = A n p 1 z) + p 1 z)nh z) ) e nhz) P z).

8 352 Liang-Wen Liao and Chung-Chun Yang Set Kz) = e nhz) and let z = r, Kz ) = Mr, K). Then it follows from Lemma 2, for r / F, that A p n 1 z ) + p 1 z )nh z ) ) e nhz ) = A p 1 z ) n Kz ) p ) 1z ) K z ) Kz ) Kz ) cr t Mr, K) + drt νr, K) Mr, K), where c, d are positive constants, t = deg p 1 1. Noting r t νr, K) lim r Mr, K) = 0 for a transcendental entire function K, there exists a sequence of {z n } + n=1 such that fz n ) P z n ), f z n ) P z n ). Again by Lemma 1, we get a contradiction. Thus we have proved that F z) = fz) + P z) is left-prime. Next we show that F is right-prime. Let F z) = g qz) ), where g is a transcendental entire function and qz) a polynomial with degree 2. Thus fz) = g qz) ) P z) and hence f z) = g qz) ) q z) P z). First, we prove that g w) has infinitely many zeros. In fact, if g w) has only finitely many zeros, then g w) = sw)e hw), where sw) is a polynomial and hw) is a nonconstant entire function. Let Kz) = e hz)/3. There exists a curve Γ tending to infinity such that if z Γ, then Kz) = M z, K). Noting that K is a transcendental entire function, we have that Mr, K) r 2m+2 for r r 0, where m = deg s. Let w = Gz) = e hz)/3 and λ = GΓ). Then dw = 1 3 h z)e hz)/3. If hz) is nonconstant polynomial, then there exists a positive constant c such that h z) c for sufficiently large z = r. If hz) is transcendental, then 1 3 h z) > 1 for z Γ and sufficiently large z = r, by 7). Hence, we have, for z Γ and z r 0, g 1 z) Mr, K), 2 z w gz) = g z) dz + gz 0 ) dw A, z 0 along Γ w 0 along λ

9 On factorizations of entire functions of bounded type 353 where w 0 = Gz 0 ), w = Gz) and A is a positive constant. Let γ be a component of q 1 Γ), and denote R = qz) for z γ. Then for z γ, we have ) g qz) A, g z)q z) BRm+1 0, as z, MR, K) 2 where A and B are constants. Hence, for z γ, we have fz) P z), f z) P z). Again, by Lemma 1, the above estimates will lead to a contradiction as before. Thus g has infinitely many zeros. Now let n = deg q and m = deg P. Next we will prove that n m, i.e., there is a positive integer r such that m = nr. Let {w } =1 denote the zeros of g w) and set We consider the roots of the equation which implies qz) = a n z n + a n 1 z n a 1 z + a 0. qz) = w, 8) a n z n 1 + o1) ) = w. On the other hand, the roots of the above equation can be expressed as where Thus z j) = 1/n w a n e i2jπ+φ )/n 1 + o1) ), φ = arg w a n, j = 0, 1, 2,..., n 1. P z 0) ) A w m/n, P z 1) ) e2mπi/n A w m/n, P z 0) ) B w m 1)/n, P z 1) ) e2m 1)πi/n B w m 1)/n, where A, B are constants depending on qz) and P z) only. sequences {w } =1, with w as, {z 0) } =1 and {z1) } =1 Thus we have such that

10 354 Liang-Wen Liao and Chung-Chun Yang 9) 10) 11) qz 0) ) = qz1) ) = w, P z 0) ) P z1) ) 1 e2mπi/n )A w m/n, f z 0) ) = P z 0) ) B w m 1)/n, 12) f z 1) ) = P z 1) ) e2m 1)πi/n B w m 1)/n, 13) fz 0) ) = gw ) P z 0) ), 14) fz 1) ) = gw ) P z 1) ), 15) fz 1) ) fz0) ) = P z0) ) P z1) ). If n m, then 1 e 2mπi/n 0. Now we discuss two subcases. Subcase 1: {fz 0) )} =1 is bounded. We have, by 10) 15), 16) fz 1) ) 1 e2mπi/n )A w m/n. By this and Lemma 1, we obtain that where B w m 1)/n f z 1) fz1) ) log fz1) ) 16π z 1) C w m 1)/n log 1 e 2mπi/n )A w m/n ), which is a contradiction. Subcase 2: {fz 0) )} =1 {fz 0) )} =1 C = 1 e2mπi/n )A a n 1/n, 16π is unbounded. Then there exists a sub-sequence of tending to infinity, which we may, without confusing, denote by the original sequence: {fz 0) )} =1 B w m 1)/n f z 0). Thus by Lemma 1, we have fz0) ) log fz0) ) 16π z 0) a n 1/n fz 0) ) log fz0) ). 16π w 1/n Hence, fz 0) ) = o wm/n) ).

11 Thus On factorizations of entire functions of bounded type 355 fz 1) ) 1 e2mπi/n )A w m/n. By arguing similarly as in Subcase 1, we will arrive at a contradiction. Hence n m. Finally, we will prove that qz) is a common right factor of fz) and P z). If qz) is not a right factor of P z), then there exist polynomials Q and P 1 with 0 < deg P 1 < n = deg q such that P z) = Q qz) ) + P 1 z). Thus Gz) = fz) + P 1 z) = g qz) ) Q qz) ) = g 1 qz) ), where g 1 w) = gw) Qw) is a transcendental entire function. By arguing similarly as in the subcase above, it follows that n deg P 1, which is a contradiction. Thus, P z) = Q qz) ) and fz) = g qz) ) Q qz) ). The conclusion follows. 4. Concluding remars Corollary. Let f be a transcendental entire function in B, then for any constant a 0, fz) + az is prime. Remar 4. This corollary shows that if fz) az B for some constant a, then NP f) 1. Remar 5. If h is a periodic entire function of order one and mean type, then h B. Thus if Gz) is as stated in Theorem D, then G n B. Remar 6. The condition f B in the above theorem and corollary is not removable. For example, fz) = e z e ez + e z, then fz) = we w + w) e z, and fz) + z = e w + w) e z + z). This example shows the cardinality of NP f) may be greater than one if f / B. Remar 7. If f is an entire function such that singf 1 ) R, then, by Lemma 3, sin fz) ) B and cos fz) ) B. Thus, for any constant a 0, sin fz) ) + az and cos fz) ) + az are prime. It was mentioned in [2] that the Pólya Laguerre class LP consists of all entire functions f which have a representation fz) = exp az 2 + bz + c)z n 1 z ) ) z exp, z z where a, b, c R, a 0, n N 0, z R\{0} for all N, and =1 z 2 <. Furthermore, if f 1, f 2,..., f n LP, and f = f 1 f 2 f n, then singf 1 ) R. Thus, for example, sin fz) ) + az is prime for a 0, when f LP.

12 356 Liang-Wen Liao and Chung-Chun Yang References [1] Baer, I. N., and A. Singh: Wandering domains in the iteration of compositions of entire functions. - Ann. Acad. Sci. Fenn. Math. 20, 1995, [2] Bergweiler, W., and Y. F. Wang: On the dynamics of composite entire functions. - Ar. Mat. 36, 1998, [3] Chuang, Chi-Tai, and Chuang-Chun Yang: Fix-Points and Factorization of Meromorphic Functions. - World Scientific, Singapore New Jersey London Hong Kong, [4] Gross, F.: Factorization of Meromorphic Functions. - U. S. Government Printing Office, [5] Laine, I.: Nevanlinna Theory and Complex Differential Equations. - Walter de Gruyter, Berlin New Yor, [6] Liao, L. W., and C. C. Yang: On factorizations of entire functions of finite type. - Preprint. [7] Nevanlinna, R.: Eindeutige analytische Funtionen. - Springer-Verlag, Berlin Göttingen Heidelberg, [8] Noda, Y.: On factorization of entire functions. - Kodai Math. J. 4, 1981, [9] Ozawa, M., and K. Sawada: A theorem on factorization of certain entire functions. - Complex Variables Theory Appl. 34, 1997, [10] Polya, G.: On an integral of an integral function. - J. London Math. Soc. 1, 1926, [11] Rippon, P. J., and G. M. Stallard: Iteration of a class of hyperbolic meromorphic functions. - Proc. Amer. Math. Soc. 127, 1999, [12] Valiron, G.: Lecture on the general theory of integral functions. -Toulous, Edouard privat, [13] Wang, X. L., and C. C. Yang: On the factorization of a certain class of entire functions. - Indian J. Pure Appl. Math. 33, 2002, Received 1 October 2003