Meromorphic functions partially share three values with their difference operators
aa r X i v : . [ m a t h . C V ] J a n Meromorphic functions partially share three values withtheir difference operators
Feng L¨u ∗ and Zhenliu Yang College of Science, China University of Petroleum, Qingdao, Shandong, 266580, P.R. China.
Abstract
In this paper, we give a simple proof and strengthening of a uniqueness theorem ofmeromorphic functions which partially share 0, ∞ CM and 1 IM with their differenceoperators. Meanwhile, we partially solve a conjecture given by Chen-Yi in [4] andgeneralize some previous theorems in [2, 3].
MSC 2010 : 30D30, 39A10.
Keywords and phrases : Uniqueness, Meromorphic, Difference operator, Shared val-ues.
In this paper, the term ”meromorphic function” always means meromorphic in the wholecomplex plane C . We say two meromorphic functions f and g share a constant a IM if f − a and g − a have the same zeros or E ( a, f ) = E ( a, g ), and the notation E ( a, f ) denotes the setof all zeros of f ( z ) − a , where a zero is counted one time. In addition, we say f and g share a CM if f − a and g − a have the same zeros with multiplicities or E ( a, f ) = E ( a, g ), andthe notation E ( a, f ) denotes the set of all zeros of f ( z ) − a , where a zero with multiplicity m is counted m times.The paper mainly concerns the uniqueness problem of meromorphic function f ( z ) par-tially sharing 2 CM + 1 IM with its difference operator ∆ c f = f ( z + c ) − f ( z ), where c is anonzero constant. This problem was inspired by a conjecture given by Chen and Yi in [4].They considered the uniqueness problem of meromorphic function f ( z ) sharing e , e and ∞ CM with ∆ c f and posed the following conjecture. Conjecture.
Let f be a transcendental meromorphic function, and let c ∈ C \{ } suchthat ∆ c f
0. If ∆ c f and f ( z ) share three distinct values e , e , ∞ CM, then △ c f = f .Since then, many scholars devoted to studying this conjecture. Zhang and Liao [10]affirmed the conjecture for the case that f is an entire function of finite order. Here and ∗ Corresponding author, Email: [email protected], [email protected]. ρ ( f ) = lim sup r →∞ log T ( r, f )log r , ρ ( f ) = lim sup r →∞ log log T ( r, f )log r . Later on, L¨u and L¨u [7] proved that the conjecture was also right when f is a meromor-phic function of finite order. Some related theorems can be found in [5]. Recently, Chenand his co-workers considered the conjecture in another direction. They asked whetherthe shared value conditions “3CM” in Conjecture can be weaken or not? Actually, theypartially answered the question by the following two theorems, which were given by Chenin [2] and Chen and Xu in [3], respectively. Theorem A.
Let f be a nonconstant meromorphic function of hyper order ρ ( f ) <
1, andlet c ∈ C \{ } such that ∆ c f
0. If ∆ c f and f ( z ) share 1 CM and satisfy E (0 , f ( z )) ⊆ E (0 , ∆ c f ( z )) and E ( ∞ , ∆ c f ( z )) ⊆ E ( ∞ , f ( z )), then ∆ c f = f . Theorem B.
Let f be a nonconstant meromorphic function of hyper order ρ ( f ) <
1, andlet c ∈ C \{ } such that ∆ c f
0. If ∆ c f and f ( z ) share 0, ∞ CM and 1 IM, then ∆ c f = f .In this present paper, we still pay attention to this conjecture and give a simple proofand strengthening of a uniqueness theorem as follows. Theorem 1.
Let f be a transcendental meromorphic function of hyper order ρ ( f ) < c ∈ C \{ } , and let L ( f ) = af ( z + c ) − bf ( z ), where a, b are two nonzero constants.If E (0 , f ( z )) ⊆ E (0 , L ( f )), E ( ∞ , L ( f )) ⊆ E ( ∞ , f ( z )) and E (1 , f ( z )) ⊆ E (1 , L ( f )), then L ( f ) = f ( z ) or f ( z + c ) = f ( z ).Obviously, Theorem 1 is an improvement of Theorems A and B. More generally, wededuce the following result and Theorem 1 is an immediate consequence of it. Theorem 2.
Let f be a nonconstant meromorphic function of hyper order ρ ( f ) <
1. If E (0 , f ( z )) ⊆ E (0 , f ( z + c )) , E ( ∞ , f ( z + c )) ⊆ E ( ∞ , f ( z )) , E (1 , f ( z )) ⊆ E ( A, f ( z + c )) , where A is a nonzero constant, then f ( z + c ) = Af ( z ) or f ( z + c ) = f ( z ). In particularly,if A = 1, then f ( z + c ) = f ( z ). Remark 1.
We give the following examples to show that either f ( z + c ) = Af ( z ) or f ( z + c ) = ± f ( z ) may occur. Example 1.
Set f ( z ) = e az , where a is a nonzero constant such that e ac = A . Then, f ( z ) and f ( z + c ) satisfies all the condition of Theorem 2. Clearly, f ( z + c ) = Af ( z ). Example 2.
Set f ( z ) = 1 + e az , where a is a nonzero constant such that e ac = 1. Then, f ( z + c ) = f ( z ) and 1 is a Picard value of f ( z ). Obviously, f ( z ) and f ( z + c ) satisfies allthe condition of Theorem 2. 2 emark 2. Below, we apply Theorem 2 to derive Theorem 1. We write f ( z + c ) as f ( z + c ) = L ( f )+ bf ( z ) a . Then, the condition E (0 , f ( z )) ⊆ E (0 , L ( f ) and E (1 , f ( z )) ⊆ E (1 , L ( f )) yields that E (0 , f ( z )) ⊆ E (0 , f ( z + c )) and E (1 , f ) ⊆ E ( ba , f ( z + c )), re-spectively. Set f ( z ) = af ( z + c ) − L ( f ) b . Suppose that z is a pole of f ( z + c ) with multiplicity m . We assume that z is a pole of f ( z ) with multiplicity n (It is pointed out that n maybezero). Below, we prove m ≤ n . On the contrary, assume n < m . Then z is a pole of L ( f ) = af ( z + c ) − bf ( z ) with multiplicity m . The condition E ( ∞ , L ( f )) ⊆ E ( ∞ , f ( z ))yields that z is a pole of f ( z ) with multiplicity at least m , a contradiction. Thus m ≤ n and E ( ∞ , f ( z + c )) ⊆ E ( ∞ , f ( z )). The above discussions yields that f ( z ) and f ( z + c )satisfies all the conditions of Theorem 2. So, f ( z + c ) = ba f ( z ) or f ( z + c ) = f ( z ). Thefirst case implies that L ( f ) = f . Therefore, the proof of Theorem 1 is finished.Before to proceed, we spare the reader for a moment and assume his/her familiarity withthe basics of Nevanlinna’s theory of meromorphic functions in C and the usual notationssuch as T ( r, f ), m ( r, f ), N ( r, f ), N ( r, f ) and S ( r, f ) (see e.g., [8, 9]). Proof of Theorem 2.
The proof is based on some idea of Chen in [1]. On the contrary, weassume that f ( z + c ) Af ( z ) and f ( z + c ) f ( z ). Below, we will derive a contradiction.We firstly introduce the auxiliary function H ( z ) = f ( z + c ) f ( z ) . (1.1)By ρ ( f ) < m ( r, H ) = S ( r, f ). The conditions E (0 , f ( z )) ⊂ E (0 , f ( z + c )) and E ( ∞ , f ( z + c )) ⊂ E ( ∞ , f ( z )) yields that H is an entire function. Thus, T ( r, H ) = S ( r, f ). We claim that N ( r, f ( z ) − ) = S ( r, f ).If 1 is a Picard value of f ( z ), then the claim is right. Below, assume that 1 is not aPicard value of f ( z ) and z is a zero of f ( z ) −
1. Then, f ( z + c ) = A and H ( z ) = A . If H ( z ) ≡ A , then f ( z + c ) = Af ( z ), a contradiction. So, H ( z ) A . Thus, N ( r, f ( z ) − ≤ N ( r, H ( z ) − A ) ≤ T ( r, H ) + O (1) = S ( r, f ) , (1.2)which implies that the claim holds. Below, we employ a result, which is Lemma 8.3 ofHalburd-Korhonen-Tohge in [6]. Lemma.
Let T : [0 , + ∞ ) → [0 , + ∞ ) be a non-decreasing continuous function and let s ∈ (0 , ∞ ) . If the hyper-order of T is strictly less than one, i.e., lim sup r →∞ log log T ( r )log r = s < and δ ∈ (0 , − s) , then T ( r + s ) = T ( r ) + o (cid:18) T ( r ) r δ (cid:19) here r runs to infinity outside of a set of finite logarithmic measure. Applying the above lemma to f ( z ), one gets N ( r, f ( z ) − N ( r, f ( z − c ) − S ( r, f ) = N ( r, f ( z + c ) − S ( r, f ) . (1.3)¡¡¡¡ Therefore, N ( r, f ( z − c ) − S ( r, f ) , N ( r, f ( z + c ) − S ( r, f ) . (1.4)We rewrite (1.1) as f ( z + c ) − H ( z )[ f ( z ) − H ( z ) ] . (1.5)Then, combining (1.4) and (1.5) yields that N ( r, f ( z ) − H ( z ) ) = N ( r, f ( z + c ) − S ( r, f ) = S ( r, f ) . (1.6)We rewrite (1.1) as f ( z ) = H ( z − c ) f ( z − c ). Then¡¡¡¡¡¡¡¡ f ( z − c ) − H ( z − c ) [ f ( z ) − H ( z − c )] . (1.7)The same argument leads to N ( r, f ( z ) − H ( z − c ) ) = N ( r, f ( z − c ) − S ( r, f ) = S ( r, f ) . (1.8)Assume that the functions H ( z ) , H ( z − c ) and 1 are distinct with each other. Applying thesecond main theorem of Nevanlinna, one gets T ( r, f ( z )) ≤ N ( r, f ( z ) − H ( z − c ) )+ N ( r, f ( z ) − H ( z ) )+ N ( r, f ( z ) − S ( r, f ) = S ( r, f ) , a contradiction. Therefore, there are at least two functions in the set { H ( z ) , H ( z − c ) , } which are equal identically. If H ( z ) = 1 or H ( z − c ) = 1, then f ( z ) ≡ f ( z + c ), a contradic-tion. Therefore, it is suffice to handle the case H ( z ) = H ( z − c ).We rewrite the above equation as H ( z ) H ( z + c ) = 1. Then, H ( z ) = H ( z ) H ( z + c ) . Supposethat H is not a constant. Then,2 T ( r, H ( z )) = 2 m ( r, H ( z )) = m ( r, H ( z )) = m ( r, H ( z ) H ( z + c ) ) = S ( r, H ) , a contradiction. Thus, H is a constant. The above equation yields that H = ±
1. So, f ( z + c ) = − f ( z ). If A = −
1, then f ( z + c ) = Af ( z ), a contradiction, which implies that A = −
1. Further, the condition E (1 , f ( z )) ⊂ E ( A, f ( z + c )) yields that 1 is a Picard value of4 ( z ). Meanwhile, − f ( z + c ), so is f ( z ). Thus, we can set f ( z ) − f ( z )+1 = e β ,where β is a nonconstant entire function with order ρ ( β ) <
1. We rewrite the above equa-tion as f ( z ) = e β − e β . Substituting the form of f ( z ) into the function f ( z + c ) = − f ( z )yields that e β ( z + c )+ β ( z ) = 1, which leads to β ( z + c ) + β ( z ) = 2 kπi , k is a fixed integer.Furthermore, one gets ρ ( β ) ≥