Results on Difference Analogues of Valiron-Mohon’ko Theorem

and Applied Analysis 3 2. Proof of Theorem 1 We need the following lemmas for the proof of Theorem 1. The difference analogue of the logarithmic derivative lemma was given by Halburd-Korhonen [8, Corollary 2.2] and Chiang-Feng [7, Corollary 2.6], independently. The following Lemma 8 is a variant of [8, Corollary 2.2]. Lemma 8. Let f(z) be a nonconstant meromorphic function of finite order, and let η1, η2 be two arbitrary complex numbers. Then, m(r, f (z + η1) f (z + η2) ) = S (r, f) . (23) In the remark of [15, page 15], it is pointed out that the following lemma holds. Lemma 9. Let f(z) be a nonconstant finite order meromorphic function and let c ̸ = 0 be an arbitrary complex number. Then, T (r + |c| , f) = T (r, f) + S (r, f) , N (r + |c| , f) = N (r, f) + S (r, f) . (24) Let f(z) be a meromorphic function. It is shown in [16, page 66] that for an arbitrary c ̸ = 0, the following inequalities: (1 + o (1)) T (r − |c| , f (z)) ≤ T (r, f (z + c)) ≤ (1 + o (1)) T (r + |c| , f (z)) (25) hold as r → ∞. From its proofwe see that the above relations are also true for counting functions. So by these relations and Lemma 9, we get the following lemma. Lemma 10. 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) + S (r, f) , N (r, f (z + c)) = N (r, f) + S (r, f) , N(r, 1 f (z + c) ) = N(r, 1 f ) + S (r, f) . (26) Remark 11. In [7], Chiang and Feng proved a similar result. Let f(z) be a meromorphic function with σ(f) < ∞, and let η ̸ = 0 be fixed; then for each ε > 0, we have T (r, f (z + η)) = T (r, f) + O (r) + O (log r) . (27) Proof of Theorem 1. Let P (z, f) = ∑ λ∈I aλ (z) σλ ∏ j=1 f(z + αλ,j) lλ,j , (28) and deg f P = p. Rearranging the expression of P(z, f) by collecting together all terms having the same total degree, we get P (z, f) = p ∑ i=0 hi (z) f(z) , (29) where, for i = 0, . . . , p, hi (z) = ∑ λ∈Ii aλ (z) σλ ∏ j=1 ( f(z + αλ,j) f (z) ) lλ,j , Ii = { { { λ ∈ I | σλ ∑ j=1 lλ,j = i } } } . (30) Since the coefficients aλ(z) of P(z, f) are small functions of f(z), we have m(r, aλ) ≤ T (r, aλ) = S (r, f) . (31) So by Lemma 8, we have, for all i = 0, 1, . . . , p the estimates m(r, hi) = S (r, f) . (32) Without loss of generality, we may assume c = 0 in (6). Otherwise, substituting z − c for z, we get R1 (z − c, f) = P (z − c, f) d1 (z − c) f (z) + d0 (z − c) . (33) By Lemma 10, we see that T (r, R1 (z − c, f)) = T (r, R1 (z, f)) + S (r, f) . (34) So, in the following discussion, we only discuss the form R1 (z, f) = P (z, f) d1 (z) f (z) + d0 (z) . (35) Assume first that d1(z) = 0. Clearly, we may assume that d0(z) = 1. By (29), we get R1 (z, f) = P (z, f) = hp (z) f(z) p + hp−1 (z) f(z) p−1 + ⋅ ⋅ ⋅ + h1 (z) f (z) + h0 (z) . (36) If p = 1, then R1(z, f) = h1(z)f(z) + h0(z). So by (32), we get m(r, R1) ≤ m (r, f) + S (r, f) . (37) If p > 1, then rewrite R1(z, f) in the form R1 (z, f) = f (z) (hp (z) f(z) p−1 + ⋅ ⋅ ⋅ + h1 (z)) + h0 (z) . (38) So we have m(r, R1) ≤ m (r, f) + m (r, hp (z) f(z) p−1 + ⋅ ⋅ ⋅ + h1 (z)) + S (r, f) . (39) 4 Abstract and Applied Analysis By (39) and the inductive argument, we have m(r, R1) ≤ pm (r, f) + S (r, f) . (40) To estimateN(r, R1), we use the form R1 (z, f) = P (z, f) = ∑ λ∈I aλ (z) σλ ∏ j=1 f(z + αλ,j) lλ,j . (41)