aa r X i v : . [ m a t h . C V ] M a r A NOTE ON HAYMAN’S CONJECTURE
TA THI HOAI AN AND NGUYEN VIET PHUONG
Abstract.
In this paper, we will give suitable conditions on differential poly-nomials Q ( f ) such that they take every finite non-zero value infinitely often,where f is a meromorphic function in complex plane. These results are relatedto Problem 1.19 and Problem 1.20 in a book of Hayman and Lingham [9]. Asconsequences, we give a new proof of the Hayman conjecture. Moreover, ourresults allow differential polynomials Q ( f ) to have some terms of any degreeof f and also the hypothesis n > k in [1, Theorem 2] is replaced by n ≥ Introduction and main results
In 1940, Milloux obtained the first result involving the value distribution ofmeromorphic functions together with their derivatives and called usually Milloux’sinequality. This result has led to a series of important research questions, forexample Hayman [7] (or Problem 1.19 in [9]) proved that if f is a meromorphicfunction in the plane omitting a finite value a , and if its k -th derivative f ( k ) ,for some k ≥
1, omits a finite nonzero value b , then f is constant. This resultis known as Hayman’s alternative . Closely linked to Milloux’s inequality andHayman’s alternative is Hayman’s conjecture on the value distribution of certaindifferential monomials, if f is a transcendental meromorphic function and n ∈ N ,then Hayman conjectured that ( f n ) ′ takes every finite nonzero value infinitelyoften for all n ≥
2. Hayman [7] himself proved this conjecture for n ≥ , Mues[13] proved the case of n = 3. In 1995, Bergweiler and Eremenko [1], Chen andFang [3], and Zalcman [15] independently proved the conjecture for n = 2 . In [2,5, 10, 13, 15], the authors generalized Hayman’s conjecture to f ( k ) . The effectiveresult in this direction was given in [1, Theorem 2]: if f is a transcendentalmeromorphic function in the plane and m > k ≥ f m ) ( k ) assumes everyfinite non-zero value infinitely often. A closed connection was given in [9, Problem1.20], which was proven for n ≥ n = 4 in [13] and n = 3 in [2]: If f is Key words: Meromorphic Functions, Entire Functions, Nevanlinna theory, Value Distribution,Differential Polynomial.The authors are supported by Vietnam’s National Foundation for Science and TechnologyDevelopment (NAFOSTED), under grant number 101.04-2017.320. non-constant and meromorphic in the plane and n ≥
3, then f ′ − f n assumes allfinite complex values.A question arising in connection with Hayman’s alternative was given by Ere-menko and Langley [6]: whether ( f m ) ( k ) can be replaced by a more general term,such as a linear differential polynomial F = L [ f ] = f ( k ) + a k − f ( k − + · · · + a f, where the coefficients a j are small functions of f. The following counterexamplegives a negative answer to above question. Let f ( z ) = 1 + e − z and F = L [ f ] = f ′′ + 2 zf ′ + 3 f = 3 + e − z . Then f ( z ) = 1 and F ( z ) = 3, but f is not constant.Until now, most results related to Hayman’s conjecture and Hayman’s inequalitywere considered for non-constant differential polynomials Q in f , where all ofwhose terms have degree at least 2 in f and its derivatives (see [4], [11]).In this paper, we will generalize Hayman’s conjecture to differential polynomi-als. Here, terms of the polynomials are not necessary all to have degree at least2. Let P ( z ) and Q ( z ) be polynomials in C [ z ] of degree p and q respectively. Wewrite Q ( z ) = b ( z − α ) m ( z − α ) m . . . ( z − α l ) m l , and P ( z ) = c ( z − β ) n ( z − β ) n . . . ( z − β h ) n h . where b, c ∈ C ∗ . Our results are as follows.
Theorem 1.
Let k be an positive integer. If f is a transcendental meromorphicfunction and q ≥ l + 1 , then [ Q ( f )] ( k ) takes every finite non-zero value infinitelyoften. As a consequence, when we consider k = 1 and Q ( z ) = z n , Hayman conjectureis obtained: Corollary 1 (Hayman conjecture [1, 7, 8, 13] ) . If f is a transcendental mero-morphic function, then f n f ′ takes every finite non-zero complex value infinitelyoften, for any integer n ≥ . More generally, for any n ≥ Corollary 2. [1, Theorem 2] If f is a transcendental meromorphic function,and if n ≥ and k are positive integers, then ( f n ) ( k ) takes every finite non-zerocomplex value infinitely often. Note that in [1, Theorem 2], the authors need n > k . However, in Corollary 2,we only need n ≥ Corollary 3.
Let k, l be positive integers and a , . . . , a l be complex numbers. If f is a transcendental meromorphic function then [( f − a ) ( f − a ) . . . ( f − a l )] ( k ) takes every finite non-zero value infinitely often. Theorem 2.
Let P and Q be polynomials of degree p and q respectively, and k ≥ be a positive integer. Let f be a transcendental meromorphic function. If q ≥ ( k + 1) p + l + 2 then Q ( f ) + P ( f ( k ) ) has infinitely many zeros. In the special case that P ( z ) = z and Q ( z ) = − az n + b where a = 0 , b areconstants, we recover many known results, for example the results in [7], as specialcases of our result. Corollary 4. [7, Theorem 9] If f is a transcendental meromorphic function, n ≥ and a = 0 , then f ′ ( z ) − af ( z ) n takes every finite complex value infinitelyoften. Theorem 3.
Let P and Q be polynomials of degree p and q respectively. Let α = 0 be a small function with respect to f . If f is a transcendental meromorphicfunction of finite order and p − q − h − l − > , then P ( f ) Q ( f ( z + c )+ c f ( z )) − α has infinitely many zeros, for any c , c ∈ C .In particular, if f is a transcendental entire function of finite order and p − h − l > , then P ( f ) Q ( f ( z + c ) + c f ( z )) − α has infinitely many zeros, for any c , c ∈ C . Corollary 5. [12, Theorem 1.2]
Suppose that f is a transcendental meromorphicfunction, α is a small function with respect to f and n, s are integers. If n ≥ s +6 ,then f ( z ) n ( f ( z + c ) − f ( z )) s − α takes every finite non-zero complex value infinitelyoften.Acknowledgments. A part of this article was written while the first name authorwas visiting Vietnam Institute for Advanced Study in Mathematics (VIASM). Shewould like to thank the institute for warm hospitality and partial support. Thefirst name author also would like to thank the International Centre for Researchand Postgraduate Training in Mathematics (ICRTM, grant number ICRTM01-2020.05) for a part of support.2.
Proof of theorems
Let f be a meromorphic function on the complex plane C . We use standardnotations, definitions and results of Nevanlinna theory in [8, 14]. We denote by S ( r, f ) any function satisfying S ( r, f ) = o ( T ( r, f )) as r → + ∞ outside of apossible exceptional set with finite measure. We first recall the following lemmas Lemma 1. [3, Theorem 2.1]
Let f ( z ) be a transcendental meromorphic functionof finite order. Then, for any c ∈ C , (i) m (cid:16) r, f ( z + c ) f ( z ) (cid:17) = S ( r, f ) for all r outside of a set of finite logarithmic mea-sure. (ii) T(r,f(z+c))=T(r,f(z))+S(r,f ).
Lemma 2. (Milloux, see Hayman [8, Theorem 3.2] ) For k ≥ ,T ( r, f ) ≤ N ( r, f ) + N (cid:16) r, f (cid:17) + N (cid:16) r, f ( k ) − (cid:17) − N (cid:16) r, f ( k +1) (cid:17) + S ( r, f ) . Recently, Yamanoi [16] proved the following result.
Lemma 3. [16]
Let f be a transcendental meromorphic function in the complexplane, k ≥ be an integer, and ǫ > let A ⊂ C be a finite set of complexnumbers. Then we have kN ( r, f ) + X a ∈ A N (cid:16) r, f − a (cid:17) ≤ N (cid:16) r, f ( k +1) (cid:17) + ǫT ( r, f ) , for all r > e outside a set E ⊂ ( e, ∞ ) of logarithmic density . Here, E dependson f, k, ǫ and A, and N (cid:16) r, f − a (cid:17) = N (cid:16) r, f − a (cid:17) − N (cid:16) r, f − a (cid:17) . Proof of Theorem 1.
We will prove [ Q ( f )] ( k ) takes a non-zero constant a infinitelyoften. Without loss of generality, we may assume a = 1. Applying Lemma 3 tothe transcendental meromorphic function Q ( f ) and k ≥ , set A = { } ⊂ C and ǫ = 12 q , we have kN ( r, f ) + N (cid:16) r, Q ( f ) (cid:17) − N (cid:16) r, Q ( f ) (cid:17) ≤ N (cid:16) r, Q ( f ) ( k +1) (cid:17) + 12 q T ( r, Q ( f )) ≤ N (cid:16) r, Q ( f ) ( k +1) (cid:17) + 12 T ( r, f ) + O (1) Together with Lemma 2, we get T ( r, Q ( f )) ≤ kN ( r, f ) + N (cid:16) r, Q ( f ) (cid:17) − N (cid:16) r, Q ( f ) (cid:17) − ( k − N ( r, f ) + N (cid:16) r, Q ( f ) (cid:17) + N (cid:16) r, Q ( f )) ( k ) − (cid:17) − N (cid:16) r, Q ( f )) ( k +1) (cid:17) + S ( r, f ) ≤ N (cid:16) r, Q ( f ) (cid:17) + N (cid:16) r, Q ( f )) ( k ) − (cid:17) + 12 T ( r, f ) + S ( r, f ) ≤ l X i =1 N (cid:16) r, f − α i (cid:17) + N (cid:16) r, Q ( f )) ( k ) − (cid:17) + 12 T ( r, f ) + S ( r, f ) ≤ (cid:16) l + 12 (cid:17) T ( r, f ) + N (cid:16) r, Q ( f )) ( k ) − (cid:17) + S ( r, f ) . Hence, (cid:16) q − l − (cid:17) T ( r, f ) ≤ N (cid:16) r, Q ( f )) ( k ) − (cid:17) + S ( r, f ) . Thus ( Q ( f )) ( k ) = 1 has infinitely many roots when q ≥ l + 1 . It follows that( Q ( f )) ( k ) assumes every finite non-zero value infinitely often. The proof is com-plete. (cid:3) Proof of Theorem 2.
Set F = Q ( f ) + P ( f ( k ) ) , (2.1) R ( f ) = [ Q ( f )] ′ Q ( f ) − F ′ F , H ( f ) = P ( f ( k ) ) (cid:16) F ′ F − [ P ( f ( k ) )] ′ P ( f ( k ) ) (cid:17) . (2.2)We have m ( r, R ( f )) = S ( r, f ) , (2.3) Q ( f ) R ( f ) = H ( f ) , (2.4)and T ( r, F ) ≤ T ( r, Q ( f )) + T ( r, P ( f ( k ) ) + S ( r, f ) = qT ( r, f ) + pT ( r, f ( k ) ) + S ( r, f ) ≤ ( q + p ( k + 1)) T ( r, f ) + S ( r, f ) . Hence S ( r, F ) = S ( r, f ).If R ( f ) ≡ , then H ( f ) ≡
0. This means Q ( f ) = ( c − P ( f ( k ) ) , (2.5) where c = 0 , q ≥ ( k + 1) p + l + 2, Equation (2.5) impliesthat f cannot have poles. On the other hand, we have qm ( r, f ) = m ( r, Q ( f )) ≤ m ( r, P ( f ( k ) ) + O (1) ≤ pm ( r, f ( k ) ) + O (1) ≤ pm ( r, f ) + pm (cid:0) r, f ( k ) f (cid:1) + O (1) = pm ( r, f ) + S ( r, f ) . Therefore m ( r, f ) = S ( r, f ) , (2.6)and T ( r, f ) = N ( r, f ) + m ( r, f ) = S ( r, f ) , which is a contradiction and then R ( f ) . It is easy to see that m ( r, H ( f )) ≤ m ( r, P ( f ( k ) )) + m (cid:16) r, F ′ F − [ P ( f ( k ) )] ′ P ( f ( k ) ) (cid:17) + O (1) ≤ pm ( r, f ( k ) ) + S ( r, f ) ≤ p (cid:16) m ( r, f ) + m ( r, f ( k ) f ) (cid:17) + S ( r, f )= pm ( r, f ) + S ( r, f ) . (2.7)We have qm ( r, f ) = m ( r, Q ( f )) + S ( r, f ) = m (cid:16) r, Q ( f ) R ( f ) 1 R ( f ) (cid:17) + S ( r, f ) ≤ m ( r, Q ( f ) R ( f )) + m (cid:16) r, R ( f ) (cid:17) + S ( r, f )= m ( r, H ( f )) + m (cid:16) r, R ( f ) (cid:17) + S ( r, f ) ≤ pm ( r, f ) + m (cid:16) r, R ( f ) (cid:17) + S ( r, f ) , hence ( q − p ) m ( r, f ) ≤ m (cid:16) r, R ( f ) (cid:17) + S ( r, f ) . (2.8)By the first main theorem and (2.3), we have m ( r, R ( f ) ) = T ( r, R ( f )) − N ( r, R ( f ) )+ O (1) = N ( r, R ( f )) − N ( r, R ( f ) )+ S ( r, f ) . (2.9)From the definition of R ( f ), we see immediately that the possible poles of R ( f )occur only at the poles of f and the zeros of Q ( f ) and F. Note that R ( f ) canhave only simple poles. Now, suppose that z is a pole of f of order s. Then z is a pole of Q ( f ) of order qs and H ( f ) of order at most ( s + k ) p + 1 . Since q ≥ ( k + 1) p + l + 2 , we have qs − ( s + k ) p − q − p ) s − kp − > . Thus, by (2.4), we deduce z must be a zero of R ( f ) of order at least qs − (( s + k ) p + 1) = ( q − p ) s − kp − . Hence, we get N ( r, R ( f )) ≤ N (cid:0) r, F (cid:1) + N (cid:0) r, Q ( f ) (cid:1) ≤ N (cid:0) r, F (cid:1) + l X i =1 N (cid:0) r, f − α i (cid:1) ≤ N (cid:0) r, F (cid:1) + lT ( r, f ) + O (1) , (2.10)and N (cid:0) r, R ( f ) (cid:1) ≥ ( q − p ) N ( r, f ) − ( kp + 1) N ( r, f ) . (2.11)Combining (2.8), (2.9), (2.10), (2.11) and by the first main theorem, we get( q − p ) T ( r, f ) = ( q − p ) N ( r, f ) + ( q − p ) m ( r, f ) ≤ N (cid:0) r, F (cid:1) + lT ( r, f ) + ( kp + 1) N ( r, f ) + S ( r, f ) ≤ N (cid:0) r, F (cid:1) + ( l + kp + 1) T ( r, f ) + S ( r, f ) . Hence, (cid:0) q − ( k + 1) p − l − (cid:1) T ( r, f ) ≤ N (cid:0) r, F (cid:1) + S ( r, f ) . Thus F = Q ( f ) + P ( f ( k ) ) assumes every finite value infinitely often when q ≥ ( k + 1) p + l + 2 . (cid:3) Proof of Theorem 3.
Denote by G ( z ) = P ( f ( z )) Q ( f ( z + c ) + c f ( z )) . Applyingthe Second Main Theorem for the meromorphic function G and 0 , ∞ , α , we have T ( r, G ) ≤ N ( r, G ) + N ( r, G ) + N ( r, G − α ) + S ( r, f ) ≤ T ( r, f ) + h X i =1 N ( r, f − β i ) + l X i =1 N ( r, f ( z + c ) + c f ( z )) − α i )+ N ( r, G − α ) + S ( r, f ) ≤ ( h + 2 l + 2) T ( r, f ) + N ( r, G − α ) + S ( r, f ) . (2.12)On the other hand, we have1 f q P ( f ) = 1 G Q ( f ( z + c ) + c f ( z )) f q = 1 G l Y i =1 h f ( z + c ) + c f ( z ) − α i f ( z ) i m i . (2.13) Therefore,( p + q ) T ( r, f ) = T ( r, f q P ( f )) + O (1) ≤ T ( G ) + l X i =1 m i T (cid:16) r, f ( z + c ) + c f ( z ) − α i f ( z ) (cid:17) + O (1) ≤ T ( G ) + l X i =1 m i T (cid:16) r, f ( z + c ) − α i f ( z ) (cid:17) + O (1) ≤ T ( G ) + 2 l X i =1 m i T ( r, f ) + O (1) ≤ T ( G ) + 2 qT ( r, f ) + O (1) , which implies T ( r, G ) ≥ ( p − q ) T ( r, f ) + O (1) . (2.14)Combining (2.12) and (2.14), we have( p − q − h − l − T ( r, f ) ≤ N ( r, G − α ) + S ( r, f ) . Thus, P ( f ) Q ( f ( z + c ) + c f ( z )) − α has infinitely many zeros if f is a transcen-dental meromorphic function of finite order and p − q − h − l − > f is an entire function then Lemma 1 (i) implies T ( r, f ( z + c ) + c f ( z )) − α i ) = T ( r, f ( z + c ) + c f ( z )) + O (1)= m ( r, f ( z + c ) + c f ( z )) + O (1)= m (cid:16) r, f ( z ) (cid:0) f ( z + c ) f ( z ) + c (cid:1)(cid:17) + O (1) ≤ m ( r, f ) + m (cid:16) r, f ( z + c ) f ( z ) + c (cid:17) + O (1) ≤ m ( r, f ) + m (cid:16) r, f ( z + c ) f ( z ) (cid:17) + O (1) ≤ T ( r, f ) + S ( r, f ) . Together with (2.12), we have T ( r, G ) ≤ N ( r, G ) + N ( r, G − α ) + S ( r, f ) ≤ h X i =1 N ( r, f − β i ) + l X i =1 N ( r, f ( z + c ) + c f ( z ) − α i )+ N ( r, G − α ) + S ( r, f ) ≤ h X i =1 N ( r, f − β i ) + l X i =1 T ( r, f ( z + c ) + c f ( z ) − α i )+ N ( r, G − α ) + S ( r, f ) ≤ ( h + l ) T ( r, f ) + N ( r, G − α ) + S ( r, f ) . (2.15)On the other hand, by (2.13), we obtain( p + q ) T ( r, f ) = T ( r, f q P ( f )) + O (1) ≤ T ( G ) + l X i =1 m i T (cid:16) r, f ( z + c ) + c f ( z ) − α i f ( z ) (cid:17) + O (1) ≤ T ( G ) + l X i =1 m i T (cid:16) r, f ( z + c ) − α i f ( z ) (cid:17) + O (1) ≤ T ( G ) + l X i =1 m i T ( r, f ) + O (1) ≤ T ( G ) + qT ( r, f ) + O (1) , where the fourth inequality follows from T (cid:16) r, f ( z + c ) − α i f ( z ) (cid:17) = m (cid:16) r, f ( z + c ) − α i f ( z ) (cid:17) + N (cid:16) r, f ( z + c ) − α i f ( z ) (cid:17) ≤ m (cid:16) r, f ( z + c ) f ( z ) (cid:17) + m (cid:16) r, f ( z ) (cid:17) + N ( r, f ( z + c ) − α i )+ N (cid:16) r, f ( z ) (cid:17) + O (1) ≤ T ( r, f ) + S ( r, f ) . Hence, we get T ( r, G ) ≥ pT ( r, f ) + S ( r, f ) . (2.16)Combining (2.15) and (2.16), we have( p − h − l ) T ( r, G ) ≤ ¯ N ( r, G − α ) + S ( r, f ) . Hence, P ( f ) Q ( f ( z + c )+ c f ( z )) − α has infinitely many zeros when p − h − l > f is a transcendental entire function of finite order. (cid:3) References [1] W. Bergweiler and A. Eremenko,
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Institute of Mathematics, Vietnam Academy of Science and Technology, 18Hoang Quoc Viet Road, Cau Giay District, 10307 Hanoi, Vietnamand: Institute of Mathematics and Applied Sciences (TIMAS), Thang Long Uni-versity, Hanoi, Vietnam
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