On Picard value problem of some difference polynomials

Arabian Journal of Mathematics, Nov 2017

In this paper, we study the value distribution of zeros of certain nonlinear difference polynomials of entire functions of finite order.

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On Picard value problem of some difference polynomials

On Picard value problem of some difference polynomials Zinelâabidine Latreuch Benharrat Belaïdi Mathematics Subject Classification In this paper, we study the value distribution of zeros of certain nonlinear difference polynomials of entire functions of finite order. Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna's value distribution theory [10,13]. In addition, we will use ρ ( f ) to denote the order of growth of f , we say that a meromorphic function a (z) is a small function of f (z) if T (r, a) = S (r, f ) , where S (r, f ) = o (T (r, f )) , as r → +∞ outside of a possible exceptional set of finite logarithmic measure, we use S ( f ) to denote the family of all small functions with respect to f (z). For a meromorphic function f (z) , we define its shift by fc (z) = f (z + c) . In 1959, Hayman proved in [11] that if f is a transcendental entire function, then f n f assume every nonzero complex number infinitely many times, provided that n ≥ 3. Later, Hayman [12] conjectured that this result remains to be valid when n = 1 and n = 2. Then Mues [18] confirmed the case when n = 2 and Bergweiler-Eremenko [2] and Chen-Fang [3] confirmed the case when n = 1, independently. Since then, there are many research publications (see [17]) regarding this type of Picard value problem. In 1997, Bergweiler obtained the following result. Theorem 1.1 [1] If f is a transcendental meromorphic function of finite order and q is a not identically zero polynomial, then f f − q has infinitely many zeros. - In the same paper, Laine and Yang showed that Theorem 1.2 does not remain valid for the case n = 1. Indeed, take f (z) = ez + 1. Then f (z) f (z + π i ) − 1 = 1 + ez 1 − ez − 1 = −e2z . After their, a stream of studies on the value distribution of nonlinear difference polynomials in f has been launched and many related results have been obtained, see, e.g. [ 5,14–16 ] . For example, Liu and Yang improved the previous result and obtained the following. Theorem 1.3 [ 15 ] Let f (z) be a transcendental entire function of finite order, and c be a nonzero complex constant. Then for n ≥ 2, f n (z) f (z + c) − p (z) has infinitely many zeros, where p (z) ≡ 0 is a polynomial in z. Hence, it is natural to ask: What can be said about the value distribution of f (z) f (z + c) − q(z); when f is a transcendental meromorphic function and q be a not identically zero small function of f? In this paper, as an attempt in resolving this question, we obtain the following results. Theorem 1.4 Let f be a transcendental entire function of finite order, let c1, c2 be two nonzero complex numbers such that f (z + c1) ≡ f (z + c2) and q be not identically zero polynomial. Then f (z) f (z + c1) − q (z) and f (z) f (z + c2) − q (z) at least one of them has infinitely many zeros. The following corollary arises directly from Theorems 1.4 and 1.3. Corollary 1.5 Let n ≥ 1 be an integer and let c1, c2 (c1c2 = 0) be two distinct complex numbers. Let α, β, p1, p2 and q (≡ 0) be nonconstant polynomials. If f is a finite order transcendental entire solution of f n (z) f (z + c1) − q (z) = p1 (z) eα(z) f n (z) f (z + c2) − q (z) = p2 (z) eβ(z) , then n = 1 and f must be a periodic function of period c1 − c2. 2 Some lemmas The following lemma is an extension of the difference analogue of the Clunie lemma obtained by Halburd and Korhonen [8]. Lemma 2.1 [ 4 ] Let f (z) be a nonconstant, finite order meromorphic solution of f n P (z, f ) = Q (z, f ) , where P (z, f ) , Q (z, f ) are the difference polynomials in f (z) with meromorphic coefficients a j (z) ( j = 1, . . . , s) , and let δ < 1. If the degree of Q (z, f ) as a polynomial in f (z) and its shifts is at most n, then m (r, P (z, f )) = o T (r + |c| , f ) r δ + o (T (r, f )) + O ⎝ ⎛ s j=1 ⎞ m r, a j ⎠ , for all r outside an exceptional set of finite logarithmic measure. Lemma 2.2 [ 6 ] Let f (z) be a nonconstant, finite order meromorphic function and let c = 0 be an arbitrary complex number. Then T (r, f (z + c)) = T (r, f (z)) + S (r, f ) . Lemma 2.3 [ 7 ] Let f (z) be a transcendental meromorphic function of finite order ρ , and let ε > 0 be a given constant. Then, there exists a set E0 ⊂ (1, +∞) that has finite logarithmic measure, such that for all z satisfying |z| ∈/ E0 ∪ [ 0, 1 ] , and for all k, j, 0 ≤ j < k, we have f (k) (z) f ( j) (z) ≤ |z|(k− j)(ρ−1+ε) . The following lemma is the lemma of the logarithmic derivative. Lemma 2.6 Let f (z) be a transcendental meromorphic solution of the system m r, f (z + η1) f (z + η2) = O r σ −1+ε . f (z) f (z + c1) − q (z) = p1 (z) eα(z), f (z) f (z + c2) − q (z) = p2 (z) eβ(z), where α, β are polynomials and p1, p2, q are not identically zero rational functions. If N (r, f ) = S (r, f ) , then deg α = deg β = deg (α + β) = ρ ( f ) > 0. Proof First, we prove that deg α = ρ ( f ) and by the same we can deduce that deg β = ρ ( f ) . It is clear from (2.1) that deg α ≤ ρ ( f ) . Suppose that deg α < ρ ( f ) , this means that f (z) f (z + c1) := F = q (z) + p1 (z) eα(z) ∈ S ( f ) . Applying Lemmas 2.1 and 2.2 into (2.2) , we obtain T (r, fc) = T (r, f ) = S (r, f ) which is a contradiction. Assume now that deg (α + β) < ρ ( f ) , this leads to p1 p2eα+β ∈ S ( f ) . From this and (2.1), we have Lemma 2.4 [ 10 ] Let f be a meromorphic function and let k ∈ N. Then f (k) f f (k) f m r, = S (r, f ) , m r, = O (log r ) . where S (r, f ) = O (log T (r, f ) + log r ) , possibly outside a set E1 ⊂ [0, +∞) of a finite linear measure. If f is of finite order of growth, then The following lemma is a difference analogue of the lemma of the logarithmic derivative for finite order meromorphic functions. Lemma 2.5 [ 6,8,9 ] Let η1, η2 be two arbitrary complex numbers such that η1 = η2 and let f (z) be a finite order meromorphic function. Let σ be the order of f (z). Then for each ε > 0, we have (2.1) (2.2) where and which leads to It is clear that P (z, f ) ≡ 0, and using Lemma 2.1, we get f 2 P (z, f ) = p1 p2eα+β + q2, P (z, f ) = a (z) f 2 − b (z) a = fc1 fc2 f f , b = q fc1 fc2 f + f . m (r, P (z, f )) = S (r, f ) 2T (r, f ) = m r, b (z) + P (z, f ) a (z) which is a contradiction. Hence, deg (α + β) = deg α = deg β. Finally, using Lemma 2.1, it is easy to see that both of α and β are nonconstant polynomials. 3 Proof of Theorem 1.4 We shall prove this theorem by contradiction. Suppose contrary to our assertion that both of f (z) f (z + c1) − q (z) and f (z) f (z + c2) − q (z) have finitely many zeros. Then, there exist four polynomials α, β, p1 and p2 such that and By differentiating (3.1) and eliminating eα, we get f (z) f (z + c1) − q (z) = p1 (z) eα(z) f (z) f (z + c2) − q (z) = p2 (z) eβ(z). p where A1 = p11 + α , B1 = p p11 + α A1 f fc1 − f fc1 − f fc1 = B1, q − q . By Lemma 2.6, we have deg α = deg β = deg (α + β) = ρ ( f ) > 0. Now, we prove that A1 ≡ 0. To show this, we suppose the contrary. Then, there exists a constant A such that A = p1 (z) eα, which implies the contradiction deg α = ρ ( f ) = 0. By the same, we can prove that B1 ≡ 0. By the same arguments as above, (3.2) gives A2 f fc2 − f fc2 − f fc2 = B2, p where A2 = p22 + β and B2 = p p22 + β (3.3) and (3.4) by f 2, we get for each ε > 0 q − q . Obviously, A2 ≡ 0 and B2 ≡ 0. Dividing both sides of 2m r, 1 f fci f ≤ m r, + m r, + m r, = O r ρ−1+ε + O (log r ) = S (r, f ) . f fci f f fci fci fci f + O (log r ) So, by the first fundamental theorem, we deduce that (3.1) (3.2) (3.3) (3.4) (3.5) (3.6) (3.7) (3.8) (3.9) T (r, f ) = N r, It is clear from (3.3) and (3.4) that any multiple zero of f is a zero of Bi (i = 1, 2) . Hence, where N(2 r, 1f denotes the counting function of zeros of f whose multiplicities are not less than 2. It follows by this and (3.5) that T (r, f ) = N1) r, and (3.4) , for every zero z0 such that f (z0) = 0 which is not zero or pole of B1 and B2, we have where N1) r, 1f and By (3.7) and (3.8), we obtain which means that the function B2 fc1 −B1 fc2 has at most a finite number of simple poles. We consider two cases: f Case 1. B2 fc1 − B1 fc2 ≡ 0. Set h (z) = B1 A2 B2 f fc2 + BB21 f fc2 + BB21 f fc2 = −B1. A1h B2 − and ≡ 0. Then, by the definition of A1 and by simple integration, A1h h B2 − B2 Then, we obtain h p1eα = C1 B2 , where C1 is a nonzero constant. This implies that deg α = ρ ( f ) − 1, which is a contradiction. Hence, ≡ 0. Next, we shall prove AB1B21 − BB21 − BB1A22 ≡ 0. Suppose that AB1B21 − BB21 − BB1A22 ≡ 0. pp21 eα−β = C2 BB21 := γ , where C2 is a nonzero constant and γ is a small function of f. From (3.1) and (3.2), we get f fc1 − γ fc2 = (1 − γ ) q. (3.10) (3.11) (3.12) (3.13) (3.15) If γ ≡ 1, then by applying Clunie’s lemma to (3.15), we obtain By this and (3.15), we have , N = 2h B2 ( A1 − A2) BB21 − B1 B2 ϕ (z) = BB21 M + Bh2 , ψ = BB21 N . Differentiation of (3.16) gives fc2 = M f + Substituting (3.16) and (3.18) into (3.4), we get M + N f + N f . Differentiating (3.19), we get M − A2 M f 2 + N − A2 N + 2M f f + N f 2 + f f = −B2. M − A2 M + 2N − A2 N + 2M f 2 + f f + N 3 f f + f f = −B2. Suppose z0 is a simple zero of f and not a zero or pole of B2. Then from (3.19) and (3.20), we have B2 N f + f 2N − A2 N + 2M f + 3N f − B2 N f + 3B2 N f . Therefore, the function satisfies T (r, H ) = S (r, f ) and f = 3BH2 N f + Substituting (3.21) into (3.19), we get B1 B2 − (3.16) (3.17) (3.18) (3.19) (3.20) (3.21) (3.22) H q1 = M − A2 M + 3B2 , q2 = 31 N + 31 BB22 − 2 A2 N + 43 M, q3 = N . where c is a nonzero constant, this leads to the contradiction deg (α + β) < deg α = deg β. Hence, q2 ≡ 0. Differentiating (3.22), we obtain Therefore, z0 is a zero of B2 q2 + q3 − B2q3 f + 2B2q3 f . Hence, the function where We prove first q2 ≡ 0. Suppose the contrary. Then satisfies T (r, R) = S (r, f ) and Substituting (3.25) into (3.24) q2 R q1 + 2B2q3 f 2 + 2q1 + q2 + 21 BB22 q2 − 2 q2 + q3 q3 + B2 1 q2 R f f + BB2q23 f 2 = −B2. Combining (3.26) and (3.22), we obtain From (3.27), we deduce that q2 R B2 q1 f + 2q1 + q2 − 21 BB22 q2 − 21 q2 + q3 qq23 + BR2 q1 + 2B2q3 − B2 f = 0. q2 R B2 q1 = 0 q1 + 2B2q3 − B2 (3.23) (3.24) (3.25) (3.26) (3.27) and By eliminating R from the above two equations, we obtain Thus, Eq. (3.25) can be rewritten as Subcase 1. If 4q1q3 − q22 ≡ 0, then from (3.28), we have f = (3.28) (3.29) (3.30) (3.31) On the other hand, Hence, On the other hand, By the definition of Ai (i = 1, 2) and simple integration, we deduce that which is a contradiction. Subcase 2. If 4q1q3 ≡ q22, then from (3.29) and (3.21), we have deg (α + β) < deg α = deg β B2 q1 q1 H B2 q2 − q2 = 3B2 N . q1 q3 − M − A2 M N H = 3B2 N . Combining (3.30) and (3.31), we obtain function, we obtain Dividing both sides of the above equation by (A1+A2)2 and since lim R(z) = 0 if R is a nonzero rational R (z) 2 z→∞ A2 A1 ( A1 + A2) 2 On the other hand, since ρ (h) ≤ ρ ( f ) − 1 and by Lemma 2.3 (3.32) and (3.33), we deduce for all z satisfying |z| ∈/ E0 ∪ [ 0, 1 ] , where E0 ⊂ (1, ∞) is a set of finite logarithmic measure. By combining h (z) h (z) ≤ |z|ρ( f )−2+ε , lim z→∞ |z|∈/E0∪[ 0,1 ] A2 A1 ( A1 + A2) 2 = lim z→∞ |z|∈/E0∪[ 0,1 ] α β (α + β ) By setting α (z) = am zm + · · · + a0 and β (z) = bm zm + · · · + b0, we deduce which implies that am 2 bm = 2 or 1 . We consider first the case abmm = 2 1 , we get from (3.1) and (3.17) and Therefore, which implies the contradiction T r, fc2 = T r, M f + N f = S (r, f ) . Case 2. B2 fc1 − B1 fc2 ≡ 0, using the same arguments as in the proof of (3.14) , we obtain that (3.35) log+ ϕ r eiθ f r eiθ + ψ r eiθ f r eiθ dθ + S (r, f ) , log+ ϕ r eiθ f r eiθ + ψ r eiθ f r eiθ dθ E1 E1 log+ f r eiθ f r eiθ log+ f r eiθ dθ + S (r, f ) dθ + S (r, f ) = S (r, f ) . log+ ϕ r eiθ f r eiθ + ψ r eiθ f r eiθ dθ q r eiθ f r eiθ dθ M r eiθ B r eiθ N r eiθ + B r eiθ f r eiθ f r eiθ q r eiθ − f 2 r eiθ dθ + S (r, f ) = S (r, f ) . which leads to where k is a nonzero complex constant. By this (3.1) and (3.2), we have p1 eα−β B1 B2 = k fc2 , (3.36) (3.37) If k = 1, then by applying Clunie lemma to (3.37) , we deduce the contradiction T r, fci = S (r, f ) . Hence, k = 1 and from the equation (3.36) , we conclude that fc1 ≡ fc2 which exclude the hypothesis of our theorem. This shows that at least one of f (z) f (z + c1) − q (z) and f (z) f (z + c2) − q (z) has infinitely many zeros. 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Zinelâabidine Latreuch, Benharrat Belaïdi. On Picard value problem of some difference polynomials, Arabian Journal of Mathematics, 2017, 1-11, DOI: 10.1007/s40065-017-0189-x