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Differential polynomials sharing one value
Acta Mathematica Scientia 2014,34B(6):1865–1874 http://actams.wipm.ac.cn
DIFFERENTIAL POLYNOMIALS SHARING ONE VALUE∗
ÜU9)
†
Jilong ZHANG (
School of Mathematics & Systems Science, Beihang University, Beijing 100191, China E-mail : [email protected]
ë¥)
Lianzhong YANG (
School of Mathematics, Shandong University, Jinan 250100, China E-mail : [email protected] Abstract In this paper, we investigate uniqueness problems of differential polynomials of meromorphic functions. Let a, b be non-zero constants and let n, k be positive integers satisfying n ≥ 3k + 12. If f n + af (k) and g n + ag (k) share b CM and the b-points of f n + af (k) are not the zeros of f and g, then f and g are either equal or closely related. Key words
meromorphic function; differential polynomial; share value
2010 MR Subject Classification
1
30D35
Introduction and Results
In this paper a meromorphic function will mean being meromorphic in the whole complex plane. We say that two meromorphic functions f and g share a finite value a CM (counting multiplicities) if f − a and g − a have the same zeros with the same multiplicities. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory (see, e.g., [7, 10, 16]). For any nonconstant meromorphic function f , we denote by S(r, f ) any quantity satisfying S(r, f ) = o (T (r, f )) as r → ∞ possibly outside a set r of finite linear measure. Let f be a transcendental meromorphic function and let n be a positive integer. Hayman [8] conjectured that f n f ′ assumes every non-zero value a ∈ C infinitely often. For the results concerning this subject, the reader is invited to see [1, 6, 9, 11, 13, 17, 19, 20]. Yang-Hua [15] got the following uniqueness theorem corresponding to the above conjecture. Theorem A Let f and g be two nonconstant entire functions, and let n ≥ 6 be an integer. If f n f ′ and g n g ′ share 1 CM, then either f (z) = c1 ecz , g(z) = c2 e−cz , where c1 , c2 and c are constants satisfying (c1 c2 )n+1 c2 = −1 or f = tg for a constant t such that tn+1 = 1. ∗ Received
May 15, 2013; revised October 10, 2013. This research was supported by the NNSF (11201014, 11171013, 11126036, 11371225) and the YWF-14-SXXY-008, YWF-ZY-302854 of Beihang University. This research was also supported by the youth talent program of Beijing (29201443). † Corresponding author: Jilong ZHANG.
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Subsequently, several papers appeared studying two differential polynomials that share one or more values. Some of these papers are due to Wang [14], Fang [3], Fang–Qiu [4] and Lin–Yi [12]. Typically, these papers are treating either (1) The number of zeros of (Pn (f ))(k) − 1 or (2) (Pn (f ))(k) and (Pn (g))(k) sharing 1 CM (or z CM), where Pn (f ) is a polynomial in f of degree n, and k is a positive integer. Another Picard type result in [7] says that each meromorphic function f satisfying f n (z) + af ′ (z) 6= b for all z ∈ C (where n ≥ 5 and a, b ∈ C with a 6= 0) is constant; if f is entire this holds for n ≥ 3. D¨ oringer [2] has shown, this remains valid for f n (z) + af (k) (z) instead of f n (z) + af ′ (z) provided n ≥ k + 4; if f is entire, it suffices to assume n ≥ 3 independently of k. Considering this, Grahl–Nevo [5] got the following three uniqueness results. Theorem B Let f and g be two nonconstant meromorphic functions, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 5k + 17. Assume that the functions φf := f n + af (k) share the value b CM. Then
or
and φg := g n + ag (k)
(1.1)
φf − b fn af (k) − b = n = (k) φg − b g ag − b
(1.2)
fn af (k) − b φf − b = (k) = φg − b gn ag − b
(1.3)
or f = g,
f (k) = g (k) = b/a.
(1.4)
Theorem C Let f and g be two nonconstant entire functions, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 11 and n ≥ k + 2. Assume that the functions φf and φg defined as in (1.1) share the value b CM. Then (1.2) or (1.4) holds. Theorem D Let f be a nonconstant meromorphic function, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 5k + 17. If the functions φf and φf ′ defined as in (1.1) share the value b CM, then f ≡ f ′ . If f is entire, this holds also for n ≥ max{11; k + 2}. In [5], the authors asked: Are the restrictions on n and k in the above three results best possible? In the present paper, we give an answer to this question and get the following results improving Theorems B–D. Theorem 1.1 Let φf and φg be given as in Theorem B. Assume that φf and φg share a non-zero constant b CM, and the b-points of φf are not the zeros of f and g. If n ≥ 3k + 12, then either (1.2) or (1.3) holds. Theorem 1.2 Let φf and φg be given as in Theorem B, where f and g are entire functions. Assume that φf and φg share a non-zero constant b CM, and the b-points of φf are not the zeros of f and g. If n ≥ 8, then (1.2) holds. Theorem 1.3 Let f be a nonconstant meromorphic function, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ k + 9. Assume that the functions φf and φf ′ defined as in (1.1) share the value b CM, and the b-points of φf are not the zeros of f and f ′ . Then f ≡ f ′ . If f is entire, this holds also for n ≥ 8.
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Remark 1 In our results, n is independent of k for the entire case. We cannot remove the condition “the b-points of φf are not the zeros of f and g” in this paper. We always assume that b 6= 0 in this paper, because the Hayman’s conjecture mentioned before considers “f n f ′ assumes every non-zero value a ∈ C infinitely often”. In the conclusions of Theorem 1.2, if (1.2) holds, it shows that entire functions f and g satisfy: (1) f and g share 0 CM; (2) f (k) and g (k) share b/a(6= 0) CM. Now, we recall Theorem E ([18]) Let f and g be entire functions satisfying (1) f and g share 0 CM; (2) f (k) and g (k) share 1 CM, where k is a positive integer. If δ(0, f ) > 1/2, then f = g or f (k) g (k) = 1. Noting Theorem E, a natural question is: If we add a deficient condition in Theorem 1.2, can we get a better relation of f and g? Considering this, we have the following result, in which we suffice to assume δ(0, f ) > 4/(n + 3). Theorem 1.4 Let f and g be two nonconstant entire functions, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 8. Assume that the functions φf and φg defined as in (1.1) share the value b CM and the b-points of φf are not the zeros of f and g. If δ(0, f ) > 4/(n + 3), then f = g.
2
Proofs of Theorems 1.1 and 1.2
We need the following lemmas to prove our results. The first lemma plays an important role when considering two meromorphic functions that share one finite value CM. Lemma 2.1 ([15, Lemma 3]) Let F and G be two nonconstant meromorphic functions. If F and G share 1 CM, then one of the following three cases holds: (1) max{T (r, F ), T (r, G)} 1 1 ≤ N2 r, + N2 r, + N2 (r, F ) + N2 (r, G) + S(r, F ) + S(r, G); F G
(2.1)
(2) F G = 1; (3) F = G, where N2 r, F1 denotes the counting function of zeros of F such that simple zeros are counted once and multiple zeros twice. The next result comes from the proof of Theorem B. Lemma 2.2 Assume that the functions φf and φg defined as in (1.1) share the value b(6= 0) CM. If af (k) = b or ag (k) = b, and one of the following holds (1) f and g are meromorphic, n ≥ k + 5; (2) f and g are entire, n ≥ 4, then f = g and f (k) = g (k) = b/a.
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Proof of Theorem 1.1 Noting that n ≥ 3k + 12, we deduce from Lemma 2.2 that f = g and f (k) = g (k) = b/a if af (k) = b or ag (k) = b. Then φf − b = f n have the same zeros as f , which contradicts with the conditions of Theorem 1.1. In the following, we assume af (k) 6= b and ag (k) 6= b. Let −af (k) + b −ag (k) + b F = , G = . (2.2) fn gn The first main theorem gives nT (r, f ) = T (r, 1/f n) + O(1) 1 −af (k) + b + T r, ≤ T r, + O(1) fn −af (k) + b ≤ T (r, F ) + T (r, f ) + kN (r, f ) + S(r, f ),
(2.3)
(n − k − 1)T (r, f ) ≤ T (r, F ) + S(r, f ).
(2.4)
which is
Then F is nonconstant, so is G. If z0 is a zero of both f and −af (k) + b, then z0 is also a zero of φf − b, a contradiction. Similarly, g and −ag (k) + b donot have common zeros. Therefore, F and G share 1 CM. By (2.2), we obtain 1 N2 (r, 1/F ) ≤ N2 r, + 2N (r, f ) , N2 (r, F ) ≤ 2N (r, 1/f ) −af (k) + b and
N2 (r, 1/G) ≤ N2 r,
1 −ag (k) + b
+ 2N (r, g) ,
N2 (r, G) ≤ 2N (r, 1/g) .
Suppose that (2.1) in Lemma 2.1 holds. The above four inequalities yield 1 1 T (r, F ) ≤ N2 r, + N r, + 2N (r, 1/f ) + 2N (r, 1/g) 2 −af (k) + b −ag (k) + b +2N (r, f ) + 2N (r, g) + S(r, f ) + S(r, g) ≤ T r, f (k) + T r, g (k) + 8T (r) + S(r)
≤ (2k + 10)T (r) + S(r),
(2.5)
where T (r) = max{T (r, f ), T (r, g)} and S(r) = S(r, f ) + S(r, g). Combining (2.4) with (2.5), we get (n − k − 1)T (r, f ) ≤ (2k + 10)T (r) + S(r). Similarly, (n − k − 1)T (r, g) ≤ (2k + 10)T (r) + S(r). Then (n − k − 1)T (r) ≤ (2k + 10)T (r) + S(r), which is (n − 3k − 11)T (r) ≤ S(r), a contradiction since n ≥ 3k + 12. Lemma 2.1 gives F = G or F G = 1. Therefore, Theorem 1.1 is thus proved. Proof of Theorem 1.2 If af (k) = b or ag (k) = b, a contradiction follows by the same reasoning as the beginning of the proof of Theorem 1.1. We now assume af (k) 6= b and ag (k) 6= b.
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Let F and G be defined as in (2.2). Suppose that (2.1) holds. By the same arguments as in the proof of Theorem 1.1, noting that f and g are entire, (2.5) becomes T (r, F ) ≤ 6T (r) + S(r). We insert the above inequality into (2.3), resulting in nT (r, f ) ≤ 7T (r) + S(r). The same inequality holds for T (r, g), then nT (r) ≤ 7T (r) + S(r), which contradicts with n ≥ 8. Then from Lemma 2.2, we have F = G or F G = 1, which means that (1.2) or (1.3) is valid. In the following, we prove that (1.3) cannot occur. Some ideas here come from [5]. φ −b
Since φf , φg share the value b CM and f , g are entire, φfg −b has no zero or pole at all. We get from (1.3) that 1 1 1 1 N (r, 1/f ) = N r, (k) , N (r, 1/g) = N r, (k) n n g − b/a f − b/a and
N r,
1 (k) f − b/a
= N (r, 1/g) ,
N r,
1 (k) g − b/a
= N (r, 1/f ) .
By the above inequalities and using Milloux inequality (see, e.g., [7, Theorem 3.2]), we have 1 T (r, f ) + T (r, g) ≤ N (r, 1/f ) + N (r, 1/g) + N r, (k) f − b/a 1 +N r, (k) + S(r) g − b/a ≤ 2N (r, 1/f ) + 2N (r, 1/g) + S(r) 1 1 2 2 ≤ N r, (k) + N r, (k) + S(r) n n f − b/a g − b/a 2 2 ≤ T r, f (k) + T r, g (k) + S(r) n n 2 2 ≤ T (r, f ) + T (r, g) + S(r), n n a contradiction since n ≥ 8.
3
Proof of Theorem 1.3 We need the following lemmas for the proof of Theorem 1.3.
Lemma 3.1 Suppose that the functions φf and φf ′ defined as in (1.1) share the value b CM. If n ≥ 2 and ′ n φf ′ − b f af (k+1) − b = = , (3.1) φf − b f af (k) − b then f = f ′ .
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−b and a n-fold pole of Proof Let z0 be a pole of f . Then z0 is a simple pole of afaf (k) −b ′ n f , which is a contradiction since n ≥ 2. Therefore, f has no pole. Noting that φf and φf ′ f share the value b CM, f has no zero either. Assume that
f = eα ,
(3.2)
where α is a nonconstant entire function satisfying T (r, α) = S(r, f ). By (3.2), we have f ′ = α′ eα = α′ f =: α1 f and ′′
f =
α′1 f
′
+ α1 f =
(3.3)
α′1 + α1 f ′ =: α2 f ′ . α1
By induction, f (k+1) = αk+1 f (k) , where αk+1 =
α′k αk
(3.4)
+ αk . Inserting (3.3) and (3.4) into (3.1), we have αk+1 f (k) − f (k) −
b a
b a
= αn1 ,
i.e., (αk+1 − αn1 ) f (k) =
b (1 − αn1 ) . a
(3.5)
If αk+1 − αn1 6= 0, then T (r, f (k) ) = S(r, f ) by the last equation. On the other hand, by (3.2), we have f (k) = P (α′ )eα = P (α′ )f , where P (α′ )(6= 0) is a differential polynomial of α′ . Thus 1 = S(r, f ), T (r, f ) ≤ T (r, f (k) ) + T r, P (α′ ) a contradiction. Therefore, αk+1 − αn1 = 0, and thus αk+1 = αn1 = 1 by (3.5). Since α1 is a constant, we deduce that αk+1 = α1 , and then α1 = 1. The assertion follows from (3.3). Lemma 3.2 Suppose that F and G are given by (2.2) with g = f ′ and n ≥ k + 2. Denote U=
F′ G′ − . F −1 G−1
(3.6)
If F and G share 1 CM and U 6= 0, then (n − k − 2)N (r, f ) ≤ N (r, 1/f ) + N (r, 1/f ′ ) + S(r, f ).
(3.7)
Proof Suppose that z0 is a pole of f with the multiplicity p. Then z0 is a zero of F with the multiplicity np − k − p and a zero of G with the multiplicity n(p + 1) − k − p − 1. So z0 is zero of U with the multiplicity at least n − k − 2. Noting that m(r, U ) = S(r, f ) and F and G share 1 CM, we have by (2.2) that 1 (n − k − 2)N (r, f ) ≤ N r, + S(r, f ) ≤ T (r, U ) + S(r, f ) U ≤ N (r, U ) + S(r, f ) ≤ N (r, F ) + N (r, G) + S(r, f ) ≤ N (r, 1/f ) + N (r, 1/f ′ ) + S(r, f ).
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Lemma 3.3 Suppose that F and G are nonconstant meromorphic functions given by (2.2) with g = f ′ and n ≥ k + 2. Let U be given by (3.6). If U = 0, then F = G. Proof By integration, we get from U = 0 that F = DG + 1 − D, where D is a non-zero constant. Then −af (k) + b −af (k+1) + b = D + 1 − D. (3.8) fn (f ′ )n Assume to the contrary that F 6= G. Then D 6= 1. It is easy to find that f has no pole by (3.8) and f is an entire function. Rewrite (3.8) as ′ n ′ n f (k+1) f f ′ n (k) D (k) − a (1 − D)(f ) = af +b − Db. f f f From the logarithmic derivative lemma, we deduce from the above equation that nT (r, f ′ ) = m (r, (f ′ )n ) ≤ m r, f (k) + S(r, f ) ≤ m (r, f ′ ) + m r, f (k) /f ′ + S(r, f ) ≤ T (r, f ′ ) + S(r, f ),
which gives T (r, f ′ ) = S(r, f ) since n ≥ k + 2. Then T (r, f (k) ) = m(r, f (k) ) ≤ m r, f (k) /f ′ + m (r, f ′ ) = S(r, f ).
Similarly, T (r, f (k+1) ) = S(r, f ). Inserting these into (3.8), it is not difficult to find that T (r, f ) = S(r, f ), a contradiction once more. Then D = 1 and F = G. Proof of Theorem 1.3 Suppose that f is a meromorphic function. Denote g = f ′ . If af = b or ag (k) = b, we will get a contradiction as before. So af (k) 6= b and ag (k) 6= b, suppose that F and G are defined as in (2.2). a) Suppose that (2.1) holds. By the same arguments as in the proof of Theorem 1.1, we get from (2.3) and (2.5) that (k)
(n − 1)T (r, f ) ≤ T (r, F ) + kN (r, f ) + S(r, f ) and
T (r, F ) ≤ T r, f (k) + T r, g (k) + 2N (r, 1/f ) + 2N (r, 1/g) + 4N (r, f ) + S(r, f )
≤ T (r, f ) + T (r, g) + 2N (r, 1/f ) + 2N (r, 1/g) + (2k + 4)N (r, f ) + S(r, f ).
Combining with the above two inequalities, results in (n − 2)T (r, f ) ≤ T (r, g) + 2N (r, 1/f ) + 2N (r, 1/g) + (3k + 4)N (r, f ) + S(r, f ). Similarly, (n − 2)T (r, g) ≤ T (r, f ) + 2N (r, 1/f ) + 2N (r, 1/g) + (3k + 4)N (r, f ) + S(r, f ). Then (n − 3){T (r, f ) + T (r, g)} ≤ 4{N (r, 1/f ) + N (r, 1/g)} + (6k + 8)N (r, f ) + S(r, f ). Let U be given by (3.6). If U 6= 0, then (3.7) holds. Noting the last inequality, we have 6k + 8 (n − 3){T (r, f ) + T (r, g)} ≤ 4 + {N (r, 1/f ) + N (r, 1/g)} + S(r, f ) n−k−2
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{T (r, f ) + T (r, g)} + S(r, f ),
which contradicts with n ≥ k + 9. Therefore, U = 0, and Lemma 3.3 means F = G. b) Suppose that (2.1) does not hold. Then F = G or F G = 1 by Lemma 2.1. If F G = 1, then (1.3) holds. It is easy to prove that f is an entire function since g = f ′ . By the same reasoning discussed in the proof of Theorem 1.2, this case cannot occur, thus F = G. Now we have F = G. Then f = f ′ follows from Lemma 3.1. If f is entire, Theorem 1.2 tells (1.2) holds, i.e., F = G. We get f = f ′ from Lemma 3.1. The proof of Theorem 1.3 is complete.
4
Proof of Theorem 1.4
Proof We have (1.2) from Theorem 1.2. It shows that f and g share 0 CM, f (k) and g (k) share b/a CM. By Milloux inequality, we have 1 T (r, f ) ≤ N (r, 1/f ) + N r, (k) + S(r, f ) f − b/a 1 ≤ N (r, 1/g) + N r, (k) + S(r, f ) g − b/a ≤ 2T (r, g) + S(r, f ). Similarly, T (r, g) ≤ 2T (r, f ) + S(r, g). Denote f = eh g,
(4.1)
where h is an entire function. Combining the last three inequalities, gives T (r, eh ) = O (T (r, f )) ,
T (r, eh ) = O (T (r, g)) .
Then T (r, h) = S(r, f ),
T (r, h) = S(r, g).
(4.2)
Now, taking derivatives of both sides of (4.1), yields f ′ = h ′ eh g + eh g ′ = h ′ f + eh g ′ , f ′′ = h′′ f + h′ f ′ + h′ eh g ′ + eh g ′′ = h′′ f + h′ f ′ + h′ (f ′ − h′ f ) + eh g ′′ = h′′ − h′2 f + 2h′ f ′ + eh g ′′ , .. . f (k) = a0 f + a1 f ′ + · · · + ak−1 f (k−1) + eh g (k) , where aj (j = 0, . . . , k − 1) are differential polynomials of h′ satisfying T (r, aj ) = S(r, f ) by (4.2). Denote the last equation as Lk (f ) := f (k) − a0 f − a1 f ′ + · · · − ak−1 f (k−1) = eh g (k) .
(4.3)
(k)
−b On the other hand, we get from (4.1) and (1.2) that af = enh , i.e., ag(k) −b b b g (k) = e−nh f (k) − + . a a
(4.4)
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Inserting (4.3) into the last equation, we obtain b b Lk (f )e−h = e−nh f (k) − + . a a Then e−nh =
b Lk (f )e−h 1 − . f (k) − b/a a f (k) − b/a
The logarithmic derivative lemma means Lk (f ) 1 f e−h nT r, eh ≤ m r, + m r, (k) + O(1) + m r, (k) f f − b/a f − b/a f (k+1) e−h f 1 ≤ m r, (k) + m r, (k+1) + m r, (k) + S(r, f ) f − b/a f f − b/a 1 f + S(r, f ) ≤ m r, (k+1) + T r, eh + m r, (k) f f − b/a 1 1 ≤ N r, + S(r, f ) + T r, eh + m r, (k) f f − b/a 1 1 ≤ N r, + T r, eh + T r, f (k) − N r, (k) + S(r, f ) f f − b/a 1 1 ≤ N r, + T r, eh + T (r, f ) − N r, (k) + S(r, f ). f f − b/a Using Milloux inequality again to the above inequality, (n − 1)T r, eh ≤ 2N (r, 1/f ) + S(r, f ). Rewrite (4.4) as
f (k) − b/a = enh g (k) − b/a .
Taking derivative of the last equation, we get f (k+1) = nh′ f (k) − b/a + enh g (k+1) .
Eliminating enh from the above two equations, results in f (k+1) g (k) − b/a = nh′ f (k) − b/a g (k) − b/a + f (k) − b/a g (k+1) .
Inserting (4.3) into the above equation, gives −nh′
f (k) Lk+1 (f ) − f (k+1) Lk (f ) + nh′ f (k) Lk (f ) b (k+1) b2 −h = + f − e L (f ) k+1 a2 eh a b ′ (k) −h −nh f + e Lk (f ) . a
If h′ 6= 0, both sides of the last equation are divided by f 2 , we have 1 1 2m r, ≤ 2T r, eh + m r, + S(r, f ). f f
By (4.5), we have
m (r, 1/f ) ≤
4 N (r, 1/f ) + S(r, f ), n−1
i.e., m (r, 1/f ) ≤
4 T (r, f ) + S(r, f ), n+3
(4.5)
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which is δ(0, f ) ≤ 4/(n + 3), a contradiction. Therefore, h′ = 0 and h is a constant. Noting (4.1), assume that f = cg, (4.6) where c is non-zero constant. Then f (k) = cg (k) . Inserting this into (1.2), gives b c cn−1 − 1 g (k) = (cn − 1) . a
(4.7)
If cn−1 − 1 6= 0, then g (k) = d by (4.7), where d is a constant. Thus g is a polynomial, so is f from (4.6), which is impossible. Therefore, cn−1 − 1 = 0. Then cn − 1 = 0 from (4.7), i.e., c = 1. The assertion follows by (4.6). References [1] Bergweiler W, Eremenko A. On the singularities of the inverse to a meromorphic function of finite order. Revista Matem´ atica Iberoamerivana, 1995, 11: 355–373 [2] D¨ oringer W. Exceptional values of differential polynomials. Pacific J Math, 1982, 98: 55–62 [3] Fang M L. Uniqueness and value-sharing of entire functions. Comput Math Appl, 2002, 44: 823–831 [4] Fang M L, Qiu H L. Meromorphic functions that share fixed-points. J Math Anal Appl, 2000, 268: 426–439 [5] Grahl J, Nevo S. Differential polynomials and shared values. Ann Acad Sci Fenn Math, 2011, 36: 47–70 [6] Hayman W K. Picard values of meromorphic functions and their derivatives. Ann Math, 1959, 70: 9–42 [7] Hayman W K. Meromorphic Functions. Oxford: Clarendon Press, 1964 [8] Hayman W K. Research Problems in Function Theory. The Athlone Press, 1967 [9] Jiang Yunbo, Gao Zongsheng. Normal families of meromorphic functions sharing a holomorphic function and the converse of the Bloch principle. Acta Math Sci, 2012, 32B(4): 1503–1512 [10] Laine I. Nevanlinna Theory and Complex Differential Equations. Berlin: Walter de Gruyter, 1993 [11] Li Xiaomin, Yi Hongxun. Uniqueness of meromorphic functions whose nonlinear differential polynomials share one value or have the same fixed points in an angular domain. Acta Math Sci, 2014, 34B(3): 593–609 [12] Lin W C, Yi H X. Uniqueness theorems for meromorphic function concering fixed-points. Complex Var Theory Appl, 2004, 49: 793–806 ¨ [13] Mues E. Uber ein problem von Hayman. Math Z, 1979, 164: 239–259 [14] Wang Y F. On Mues conjecture and Picard values. Science in China, 1993, 36: 28–35 [15] Yang C C, Hua X H. Uniqueness and value-sharing of meromorphic functions. Ann Acad Sci Fenn Math, 1997, 22: 395–406 [16] Yang C C, Yi H X. Uniqueness Theory of Meromorphic Functions. Dordrecht: Kluwer Academic Publishers, 2003 (Chinese original: Beijing: Science Press, 1995) [17] Yang Lianzhong, ZHang Jilong. On some results of Yi and their applications in the functional equations. Acta Math Sci, 2010, 30B(5): 1649–1660 [18] Yi H X. A question of C. C. Yang on the uniqueness of entire functions. Kodai Math J, 1990, 13: 39–46 [19] Zhang J L, Yang L Z. A power of an entire function sharing one value with its derivative. Computers and Mathematics with Applications, 2010, 60: 2153–2160 [20] Zhang J L, Gao Z S, Li S. Distribution of zeros and shared values of difference operators. Ann Polo Math, 2011, 102(3): 213–221
DIFFERENTIAL POLYNOMIALS SHARING ONE VALUE∗
ÜU9)
†
Jilong ZHANG (
School of Mathematics & Systems Science, Beihang University, Beijing 100191, China E-mail : [email protected]
ë¥)
Lianzhong YANG (
School of Mathematics, Shandong University, Jinan 250100, China E-mail : [email protected] Abstract In this paper, we investigate uniqueness problems of differential polynomials of meromorphic functions. Let a, b be non-zero constants and let n, k be positive integers satisfying n ≥ 3k + 12. If f n + af (k) and g n + ag (k) share b CM and the b-points of f n + af (k) are not the zeros of f and g, then f and g are either equal or closely related. Key words
meromorphic function; differential polynomial; share value
2010 MR Subject Classification
1
30D35
Introduction and Results
In this paper a meromorphic function will mean being meromorphic in the whole complex plane. We say that two meromorphic functions f and g share a finite value a CM (counting multiplicities) if f − a and g − a have the same zeros with the same multiplicities. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory (see, e.g., [7, 10, 16]). For any nonconstant meromorphic function f , we denote by S(r, f ) any quantity satisfying S(r, f ) = o (T (r, f )) as r → ∞ possibly outside a set r of finite linear measure. Let f be a transcendental meromorphic function and let n be a positive integer. Hayman [8] conjectured that f n f ′ assumes every non-zero value a ∈ C infinitely often. For the results concerning this subject, the reader is invited to see [1, 6, 9, 11, 13, 17, 19, 20]. Yang-Hua [15] got the following uniqueness theorem corresponding to the above conjecture. Theorem A Let f and g be two nonconstant entire functions, and let n ≥ 6 be an integer. If f n f ′ and g n g ′ share 1 CM, then either f (z) = c1 ecz , g(z) = c2 e−cz , where c1 , c2 and c are constants satisfying (c1 c2 )n+1 c2 = −1 or f = tg for a constant t such that tn+1 = 1. ∗ Received
May 15, 2013; revised October 10, 2013. This research was supported by the NNSF (11201014, 11171013, 11126036, 11371225) and the YWF-14-SXXY-008, YWF-ZY-302854 of Beihang University. This research was also supported by the youth talent program of Beijing (29201443). † Corresponding author: Jilong ZHANG.
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Subsequently, several papers appeared studying two differential polynomials that share one or more values. Some of these papers are due to Wang [14], Fang [3], Fang–Qiu [4] and Lin–Yi [12]. Typically, these papers are treating either (1) The number of zeros of (Pn (f ))(k) − 1 or (2) (Pn (f ))(k) and (Pn (g))(k) sharing 1 CM (or z CM), where Pn (f ) is a polynomial in f of degree n, and k is a positive integer. Another Picard type result in [7] says that each meromorphic function f satisfying f n (z) + af ′ (z) 6= b for all z ∈ C (where n ≥ 5 and a, b ∈ C with a 6= 0) is constant; if f is entire this holds for n ≥ 3. D¨ oringer [2] has shown, this remains valid for f n (z) + af (k) (z) instead of f n (z) + af ′ (z) provided n ≥ k + 4; if f is entire, it suffices to assume n ≥ 3 independently of k. Considering this, Grahl–Nevo [5] got the following three uniqueness results. Theorem B Let f and g be two nonconstant meromorphic functions, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 5k + 17. Assume that the functions φf := f n + af (k) share the value b CM. Then
or
and φg := g n + ag (k)
(1.1)
φf − b fn af (k) − b = n = (k) φg − b g ag − b
(1.2)
fn af (k) − b φf − b = (k) = φg − b gn ag − b
(1.3)
or f = g,
f (k) = g (k) = b/a.
(1.4)
Theorem C Let f and g be two nonconstant entire functions, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 11 and n ≥ k + 2. Assume that the functions φf and φg defined as in (1.1) share the value b CM. Then (1.2) or (1.4) holds. Theorem D Let f be a nonconstant meromorphic function, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 5k + 17. If the functions φf and φf ′ defined as in (1.1) share the value b CM, then f ≡ f ′ . If f is entire, this holds also for n ≥ max{11; k + 2}. In [5], the authors asked: Are the restrictions on n and k in the above three results best possible? In the present paper, we give an answer to this question and get the following results improving Theorems B–D. Theorem 1.1 Let φf and φg be given as in Theorem B. Assume that φf and φg share a non-zero constant b CM, and the b-points of φf are not the zeros of f and g. If n ≥ 3k + 12, then either (1.2) or (1.3) holds. Theorem 1.2 Let φf and φg be given as in Theorem B, where f and g are entire functions. Assume that φf and φg share a non-zero constant b CM, and the b-points of φf are not the zeros of f and g. If n ≥ 8, then (1.2) holds. Theorem 1.3 Let f be a nonconstant meromorphic function, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ k + 9. Assume that the functions φf and φf ′ defined as in (1.1) share the value b CM, and the b-points of φf are not the zeros of f and f ′ . Then f ≡ f ′ . If f is entire, this holds also for n ≥ 8.
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Remark 1 In our results, n is independent of k for the entire case. We cannot remove the condition “the b-points of φf are not the zeros of f and g” in this paper. We always assume that b 6= 0 in this paper, because the Hayman’s conjecture mentioned before considers “f n f ′ assumes every non-zero value a ∈ C infinitely often”. In the conclusions of Theorem 1.2, if (1.2) holds, it shows that entire functions f and g satisfy: (1) f and g share 0 CM; (2) f (k) and g (k) share b/a(6= 0) CM. Now, we recall Theorem E ([18]) Let f and g be entire functions satisfying (1) f and g share 0 CM; (2) f (k) and g (k) share 1 CM, where k is a positive integer. If δ(0, f ) > 1/2, then f = g or f (k) g (k) = 1. Noting Theorem E, a natural question is: If we add a deficient condition in Theorem 1.2, can we get a better relation of f and g? Considering this, we have the following result, in which we suffice to assume δ(0, f ) > 4/(n + 3). Theorem 1.4 Let f and g be two nonconstant entire functions, a, b ∈ C \ {0}, and let n, k be positive integers satisfying n ≥ 8. Assume that the functions φf and φg defined as in (1.1) share the value b CM and the b-points of φf are not the zeros of f and g. If δ(0, f ) > 4/(n + 3), then f = g.
2
Proofs of Theorems 1.1 and 1.2
We need the following lemmas to prove our results. The first lemma plays an important role when considering two meromorphic functions that share one finite value CM. Lemma 2.1 ([15, Lemma 3]) Let F and G be two nonconstant meromorphic functions. If F and G share 1 CM, then one of the following three cases holds: (1) max{T (r, F ), T (r, G)} 1 1 ≤ N2 r, + N2 r, + N2 (r, F ) + N2 (r, G) + S(r, F ) + S(r, G); F G
(2.1)
(2) F G = 1; (3) F = G, where N2 r, F1 denotes the counting function of zeros of F such that simple zeros are counted once and multiple zeros twice. The next result comes from the proof of Theorem B. Lemma 2.2 Assume that the functions φf and φg defined as in (1.1) share the value b(6= 0) CM. If af (k) = b or ag (k) = b, and one of the following holds (1) f and g are meromorphic, n ≥ k + 5; (2) f and g are entire, n ≥ 4, then f = g and f (k) = g (k) = b/a.
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Proof of Theorem 1.1 Noting that n ≥ 3k + 12, we deduce from Lemma 2.2 that f = g and f (k) = g (k) = b/a if af (k) = b or ag (k) = b. Then φf − b = f n have the same zeros as f , which contradicts with the conditions of Theorem 1.1. In the following, we assume af (k) 6= b and ag (k) 6= b. Let −af (k) + b −ag (k) + b F = , G = . (2.2) fn gn The first main theorem gives nT (r, f ) = T (r, 1/f n) + O(1) 1 −af (k) + b + T r, ≤ T r, + O(1) fn −af (k) + b ≤ T (r, F ) + T (r, f ) + kN (r, f ) + S(r, f ),
(2.3)
(n − k − 1)T (r, f ) ≤ T (r, F ) + S(r, f ).
(2.4)
which is
Then F is nonconstant, so is G. If z0 is a zero of both f and −af (k) + b, then z0 is also a zero of φf − b, a contradiction. Similarly, g and −ag (k) + b donot have common zeros. Therefore, F and G share 1 CM. By (2.2), we obtain 1 N2 (r, 1/F ) ≤ N2 r, + 2N (r, f ) , N2 (r, F ) ≤ 2N (r, 1/f ) −af (k) + b and
N2 (r, 1/G) ≤ N2 r,
1 −ag (k) + b
+ 2N (r, g) ,
N2 (r, G) ≤ 2N (r, 1/g) .
Suppose that (2.1) in Lemma 2.1 holds. The above four inequalities yield 1 1 T (r, F ) ≤ N2 r, + N r, + 2N (r, 1/f ) + 2N (r, 1/g) 2 −af (k) + b −ag (k) + b +2N (r, f ) + 2N (r, g) + S(r, f ) + S(r, g) ≤ T r, f (k) + T r, g (k) + 8T (r) + S(r)
≤ (2k + 10)T (r) + S(r),
(2.5)
where T (r) = max{T (r, f ), T (r, g)} and S(r) = S(r, f ) + S(r, g). Combining (2.4) with (2.5), we get (n − k − 1)T (r, f ) ≤ (2k + 10)T (r) + S(r). Similarly, (n − k − 1)T (r, g) ≤ (2k + 10)T (r) + S(r). Then (n − k − 1)T (r) ≤ (2k + 10)T (r) + S(r), which is (n − 3k − 11)T (r) ≤ S(r), a contradiction since n ≥ 3k + 12. Lemma 2.1 gives F = G or F G = 1. Therefore, Theorem 1.1 is thus proved. Proof of Theorem 1.2 If af (k) = b or ag (k) = b, a contradiction follows by the same reasoning as the beginning of the proof of Theorem 1.1. We now assume af (k) 6= b and ag (k) 6= b.
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Let F and G be defined as in (2.2). Suppose that (2.1) holds. By the same arguments as in the proof of Theorem 1.1, noting that f and g are entire, (2.5) becomes T (r, F ) ≤ 6T (r) + S(r). We insert the above inequality into (2.3), resulting in nT (r, f ) ≤ 7T (r) + S(r). The same inequality holds for T (r, g), then nT (r) ≤ 7T (r) + S(r), which contradicts with n ≥ 8. Then from Lemma 2.2, we have F = G or F G = 1, which means that (1.2) or (1.3) is valid. In the following, we prove that (1.3) cannot occur. Some ideas here come from [5]. φ −b
Since φf , φg share the value b CM and f , g are entire, φfg −b has no zero or pole at all. We get from (1.3) that 1 1 1 1 N (r, 1/f ) = N r, (k) , N (r, 1/g) = N r, (k) n n g − b/a f − b/a and
N r,
1 (k) f − b/a
= N (r, 1/g) ,
N r,
1 (k) g − b/a
= N (r, 1/f ) .
By the above inequalities and using Milloux inequality (see, e.g., [7, Theorem 3.2]), we have 1 T (r, f ) + T (r, g) ≤ N (r, 1/f ) + N (r, 1/g) + N r, (k) f − b/a 1 +N r, (k) + S(r) g − b/a ≤ 2N (r, 1/f ) + 2N (r, 1/g) + S(r) 1 1 2 2 ≤ N r, (k) + N r, (k) + S(r) n n f − b/a g − b/a 2 2 ≤ T r, f (k) + T r, g (k) + S(r) n n 2 2 ≤ T (r, f ) + T (r, g) + S(r), n n a contradiction since n ≥ 8.
3
Proof of Theorem 1.3 We need the following lemmas for the proof of Theorem 1.3.
Lemma 3.1 Suppose that the functions φf and φf ′ defined as in (1.1) share the value b CM. If n ≥ 2 and ′ n φf ′ − b f af (k+1) − b = = , (3.1) φf − b f af (k) − b then f = f ′ .
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−b and a n-fold pole of Proof Let z0 be a pole of f . Then z0 is a simple pole of afaf (k) −b ′ n f , which is a contradiction since n ≥ 2. Therefore, f has no pole. Noting that φf and φf ′ f share the value b CM, f has no zero either. Assume that
f = eα ,
(3.2)
where α is a nonconstant entire function satisfying T (r, α) = S(r, f ). By (3.2), we have f ′ = α′ eα = α′ f =: α1 f and ′′
f =
α′1 f
′
+ α1 f =
(3.3)
α′1 + α1 f ′ =: α2 f ′ . α1
By induction, f (k+1) = αk+1 f (k) , where αk+1 =
α′k αk
(3.4)
+ αk . Inserting (3.3) and (3.4) into (3.1), we have αk+1 f (k) − f (k) −
b a
b a
= αn1 ,
i.e., (αk+1 − αn1 ) f (k) =
b (1 − αn1 ) . a
(3.5)
If αk+1 − αn1 6= 0, then T (r, f (k) ) = S(r, f ) by the last equation. On the other hand, by (3.2), we have f (k) = P (α′ )eα = P (α′ )f , where P (α′ )(6= 0) is a differential polynomial of α′ . Thus 1 = S(r, f ), T (r, f ) ≤ T (r, f (k) ) + T r, P (α′ ) a contradiction. Therefore, αk+1 − αn1 = 0, and thus αk+1 = αn1 = 1 by (3.5). Since α1 is a constant, we deduce that αk+1 = α1 , and then α1 = 1. The assertion follows from (3.3). Lemma 3.2 Suppose that F and G are given by (2.2) with g = f ′ and n ≥ k + 2. Denote U=
F′ G′ − . F −1 G−1
(3.6)
If F and G share 1 CM and U 6= 0, then (n − k − 2)N (r, f ) ≤ N (r, 1/f ) + N (r, 1/f ′ ) + S(r, f ).
(3.7)
Proof Suppose that z0 is a pole of f with the multiplicity p. Then z0 is a zero of F with the multiplicity np − k − p and a zero of G with the multiplicity n(p + 1) − k − p − 1. So z0 is zero of U with the multiplicity at least n − k − 2. Noting that m(r, U ) = S(r, f ) and F and G share 1 CM, we have by (2.2) that 1 (n − k − 2)N (r, f ) ≤ N r, + S(r, f ) ≤ T (r, U ) + S(r, f ) U ≤ N (r, U ) + S(r, f ) ≤ N (r, F ) + N (r, G) + S(r, f ) ≤ N (r, 1/f ) + N (r, 1/f ′ ) + S(r, f ).
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Lemma 3.3 Suppose that F and G are nonconstant meromorphic functions given by (2.2) with g = f ′ and n ≥ k + 2. Let U be given by (3.6). If U = 0, then F = G. Proof By integration, we get from U = 0 that F = DG + 1 − D, where D is a non-zero constant. Then −af (k) + b −af (k+1) + b = D + 1 − D. (3.8) fn (f ′ )n Assume to the contrary that F 6= G. Then D 6= 1. It is easy to find that f has no pole by (3.8) and f is an entire function. Rewrite (3.8) as ′ n ′ n f (k+1) f f ′ n (k) D (k) − a (1 − D)(f ) = af +b − Db. f f f From the logarithmic derivative lemma, we deduce from the above equation that nT (r, f ′ ) = m (r, (f ′ )n ) ≤ m r, f (k) + S(r, f ) ≤ m (r, f ′ ) + m r, f (k) /f ′ + S(r, f ) ≤ T (r, f ′ ) + S(r, f ),
which gives T (r, f ′ ) = S(r, f ) since n ≥ k + 2. Then T (r, f (k) ) = m(r, f (k) ) ≤ m r, f (k) /f ′ + m (r, f ′ ) = S(r, f ).
Similarly, T (r, f (k+1) ) = S(r, f ). Inserting these into (3.8), it is not difficult to find that T (r, f ) = S(r, f ), a contradiction once more. Then D = 1 and F = G. Proof of Theorem 1.3 Suppose that f is a meromorphic function. Denote g = f ′ . If af = b or ag (k) = b, we will get a contradiction as before. So af (k) 6= b and ag (k) 6= b, suppose that F and G are defined as in (2.2). a) Suppose that (2.1) holds. By the same arguments as in the proof of Theorem 1.1, we get from (2.3) and (2.5) that (k)
(n − 1)T (r, f ) ≤ T (r, F ) + kN (r, f ) + S(r, f ) and
T (r, F ) ≤ T r, f (k) + T r, g (k) + 2N (r, 1/f ) + 2N (r, 1/g) + 4N (r, f ) + S(r, f )
≤ T (r, f ) + T (r, g) + 2N (r, 1/f ) + 2N (r, 1/g) + (2k + 4)N (r, f ) + S(r, f ).
Combining with the above two inequalities, results in (n − 2)T (r, f ) ≤ T (r, g) + 2N (r, 1/f ) + 2N (r, 1/g) + (3k + 4)N (r, f ) + S(r, f ). Similarly, (n − 2)T (r, g) ≤ T (r, f ) + 2N (r, 1/f ) + 2N (r, 1/g) + (3k + 4)N (r, f ) + S(r, f ). Then (n − 3){T (r, f ) + T (r, g)} ≤ 4{N (r, 1/f ) + N (r, 1/g)} + (6k + 8)N (r, f ) + S(r, f ). Let U be given by (3.6). If U 6= 0, then (3.7) holds. Noting the last inequality, we have 6k + 8 (n − 3){T (r, f ) + T (r, g)} ≤ 4 + {N (r, 1/f ) + N (r, 1/g)} + S(r, f ) n−k−2
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{T (r, f ) + T (r, g)} + S(r, f ),
which contradicts with n ≥ k + 9. Therefore, U = 0, and Lemma 3.3 means F = G. b) Suppose that (2.1) does not hold. Then F = G or F G = 1 by Lemma 2.1. If F G = 1, then (1.3) holds. It is easy to prove that f is an entire function since g = f ′ . By the same reasoning discussed in the proof of Theorem 1.2, this case cannot occur, thus F = G. Now we have F = G. Then f = f ′ follows from Lemma 3.1. If f is entire, Theorem 1.2 tells (1.2) holds, i.e., F = G. We get f = f ′ from Lemma 3.1. The proof of Theorem 1.3 is complete.
4
Proof of Theorem 1.4
Proof We have (1.2) from Theorem 1.2. It shows that f and g share 0 CM, f (k) and g (k) share b/a CM. By Milloux inequality, we have 1 T (r, f ) ≤ N (r, 1/f ) + N r, (k) + S(r, f ) f − b/a 1 ≤ N (r, 1/g) + N r, (k) + S(r, f ) g − b/a ≤ 2T (r, g) + S(r, f ). Similarly, T (r, g) ≤ 2T (r, f ) + S(r, g). Denote f = eh g,
(4.1)
where h is an entire function. Combining the last three inequalities, gives T (r, eh ) = O (T (r, f )) ,
T (r, eh ) = O (T (r, g)) .
Then T (r, h) = S(r, f ),
T (r, h) = S(r, g).
(4.2)
Now, taking derivatives of both sides of (4.1), yields f ′ = h ′ eh g + eh g ′ = h ′ f + eh g ′ , f ′′ = h′′ f + h′ f ′ + h′ eh g ′ + eh g ′′ = h′′ f + h′ f ′ + h′ (f ′ − h′ f ) + eh g ′′ = h′′ − h′2 f + 2h′ f ′ + eh g ′′ , .. . f (k) = a0 f + a1 f ′ + · · · + ak−1 f (k−1) + eh g (k) , where aj (j = 0, . . . , k − 1) are differential polynomials of h′ satisfying T (r, aj ) = S(r, f ) by (4.2). Denote the last equation as Lk (f ) := f (k) − a0 f − a1 f ′ + · · · − ak−1 f (k−1) = eh g (k) .
(4.3)
(k)
−b On the other hand, we get from (4.1) and (1.2) that af = enh , i.e., ag(k) −b b b g (k) = e−nh f (k) − + . a a
(4.4)
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Inserting (4.3) into the last equation, we obtain b b Lk (f )e−h = e−nh f (k) − + . a a Then e−nh =
b Lk (f )e−h 1 − . f (k) − b/a a f (k) − b/a
The logarithmic derivative lemma means Lk (f ) 1 f e−h nT r, eh ≤ m r, + m r, (k) + O(1) + m r, (k) f f − b/a f − b/a f (k+1) e−h f 1 ≤ m r, (k) + m r, (k+1) + m r, (k) + S(r, f ) f − b/a f f − b/a 1 f + S(r, f ) ≤ m r, (k+1) + T r, eh + m r, (k) f f − b/a 1 1 ≤ N r, + S(r, f ) + T r, eh + m r, (k) f f − b/a 1 1 ≤ N r, + T r, eh + T r, f (k) − N r, (k) + S(r, f ) f f − b/a 1 1 ≤ N r, + T r, eh + T (r, f ) − N r, (k) + S(r, f ). f f − b/a Using Milloux inequality again to the above inequality, (n − 1)T r, eh ≤ 2N (r, 1/f ) + S(r, f ). Rewrite (4.4) as
f (k) − b/a = enh g (k) − b/a .
Taking derivative of the last equation, we get f (k+1) = nh′ f (k) − b/a + enh g (k+1) .
Eliminating enh from the above two equations, results in f (k+1) g (k) − b/a = nh′ f (k) − b/a g (k) − b/a + f (k) − b/a g (k+1) .
Inserting (4.3) into the above equation, gives −nh′
f (k) Lk+1 (f ) − f (k+1) Lk (f ) + nh′ f (k) Lk (f ) b (k+1) b2 −h = + f − e L (f ) k+1 a2 eh a b ′ (k) −h −nh f + e Lk (f ) . a
If h′ 6= 0, both sides of the last equation are divided by f 2 , we have 1 1 2m r, ≤ 2T r, eh + m r, + S(r, f ). f f
By (4.5), we have
m (r, 1/f ) ≤
4 N (r, 1/f ) + S(r, f ), n−1
i.e., m (r, 1/f ) ≤
4 T (r, f ) + S(r, f ), n+3
(4.5)
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Vol.34 Ser.B
which is δ(0, f ) ≤ 4/(n + 3), a contradiction. Therefore, h′ = 0 and h is a constant. Noting (4.1), assume that f = cg, (4.6) where c is non-zero constant. Then f (k) = cg (k) . Inserting this into (1.2), gives b c cn−1 − 1 g (k) = (cn − 1) . a
(4.7)
If cn−1 − 1 6= 0, then g (k) = d by (4.7), where d is a constant. Thus g is a polynomial, so is f from (4.6), which is impossible. Therefore, cn−1 − 1 = 0. Then cn − 1 = 0 from (4.7), i.e., c = 1. The assertion follows by (4.6). References [1] Bergweiler W, Eremenko A. On the singularities of the inverse to a meromorphic function of finite order. Revista Matem´ atica Iberoamerivana, 1995, 11: 355–373 [2] D¨ oringer W. Exceptional values of differential polynomials. Pacific J Math, 1982, 98: 55–62 [3] Fang M L. Uniqueness and value-sharing of entire functions. Comput Math Appl, 2002, 44: 823–831 [4] Fang M L, Qiu H L. Meromorphic functions that share fixed-points. J Math Anal Appl, 2000, 268: 426–439 [5] Grahl J, Nevo S. Differential polynomials and shared values. Ann Acad Sci Fenn Math, 2011, 36: 47–70 [6] Hayman W K. Picard values of meromorphic functions and their derivatives. Ann Math, 1959, 70: 9–42 [7] Hayman W K. Meromorphic Functions. Oxford: Clarendon Press, 1964 [8] Hayman W K. Research Problems in Function Theory. The Athlone Press, 1967 [9] Jiang Yunbo, Gao Zongsheng. Normal families of meromorphic functions sharing a holomorphic function and the converse of the Bloch principle. Acta Math Sci, 2012, 32B(4): 1503–1512 [10] Laine I. Nevanlinna Theory and Complex Differential Equations. Berlin: Walter de Gruyter, 1993 [11] Li Xiaomin, Yi Hongxun. Uniqueness of meromorphic functions whose nonlinear differential polynomials share one value or have the same fixed points in an angular domain. Acta Math Sci, 2014, 34B(3): 593–609 [12] Lin W C, Yi H X. Uniqueness theorems for meromorphic function concering fixed-points. Complex Var Theory Appl, 2004, 49: 793–806 ¨ [13] Mues E. Uber ein problem von Hayman. Math Z, 1979, 164: 239–259 [14] Wang Y F. On Mues conjecture and Picard values. Science in China, 1993, 36: 28–35 [15] Yang C C, Hua X H. Uniqueness and value-sharing of meromorphic functions. Ann Acad Sci Fenn Math, 1997, 22: 395–406 [16] Yang C C, Yi H X. Uniqueness Theory of Meromorphic Functions. Dordrecht: Kluwer Academic Publishers, 2003 (Chinese original: Beijing: Science Press, 1995) [17] Yang Lianzhong, ZHang Jilong. On some results of Yi and their applications in the functional equations. Acta Math Sci, 2010, 30B(5): 1649–1660 [18] Yi H X. A question of C. C. Yang on the uniqueness of entire functions. Kodai Math J, 1990, 13: 39–46 [19] Zhang J L, Yang L Z. A power of an entire function sharing one value with its derivative. Computers and Mathematics with Applications, 2010, 60: 2153–2160 [20] Zhang J L, Gao Z S, Li S. Distribution of zeros and shared values of difference operators. Ann Polo Math, 2011, 102(3): 213–221