Normal Criteria of Function Families Concerning Shared Values

Theorem 1.1. Let ℱ be a family of holomorphic (meromorphic) functions defined in a domain D, n ∈ ℕ, a ≠ 0, b ∈ ℂ. If f^′(z) + afⁿ(z) − b ≠ 0 for each function f(z) ∈ ℱ and n ≥ 2  (n ≥ 3), then ℱ is normal in D.

Example 1.2. The family of meromorphic functions $$ is not normal in D = {z:|z | < 1}. This is deduced by $$, as j → ∞ and Marty′s criterion [2], although, for any $$.

Theorem 1.3. Let ℱ be a family of meromorphic functions in a domain D, and a ≠ 0, b ∈ ℂ. If f^′(z) + a(f(z)) ² − b ≠ 0 and the poles of f(z) are of multiplicity ≥3 for each f(z) ∈ ℱ, then ℱ is normal in D.

Theorem 1.4. Let ℱ be a family of meromorphic (holomorphic) functions in D, n a positive integer, and a, b two finite complex numbers such that a ≠ 0. If n ≥ 4  (n ≥ 2) and, for every pair of functions f and g in ℱ, f^′ − afⁿ and g^′ − agⁿ share the value b, then ℱ is normal in D.

Example 1.5 (see [11].)The family of meromorphic functions $$ is not normal in D = {z:|z | < 1}. Obviously $$. So for each pair $$ and $$ share the value 0 in D, but ℱ is not normal at the point z = 0, since $$, as j → ∞.

Remark 1.6. Example 1.5 shows that Theorem 1.4 is not valid when n = 3, and the condition n = 4 is best possible for meromorphic case.

Theorem 1.7. Let ℱ be a family of meromorphic functions in D, n a positive integer, and a, b two finite complex numbers such that a ≠ 0. If n ≥ 2 and, for every pair of functions f and g in ℱ, f^′ − af⁻ⁿ and g^′ − ag⁻ⁿ share the value b, then ℱ is normal in D.

Example 1.8. The family of holomorphic functions $$ is not normal in D = {z:|z | < 1}. This is deduced by $$, as j → ∞ and Marty′s criterion [2], although, for any $$.

Remark 1.9. Example 1.8 shows that the condition that added n ≥ 2 in Theorem 1.7 is best possible. In Theorem 1.7 taking b = 0 we get Corollary 1.10 obtained by Zhang [22].

Corollary 1.10. Let ℱ be a family of meromorphic functions in D, n ≥ 2, and let a be a nonzero finite complex number. If, for every pair of functions f and g in ℱ, fⁿf^′ and gⁿg^′ share the value a, then ℱ is normal in D.

Theorem 1.11. Let ℱ be a family of meromorphic functions in D, and let a and b be two finite complex numbers such that a ≠ 0. Suppose that all of zeros are multiple for each f(z) ∈ ℱ. If, for every pair of functions f and g in ℱ, f^′ − af⁻¹ and g^′ − ag⁻¹ share the value b, then ℱ is normal in D.

Remark 1.12. Example 1.8 shows that the condition that all of zeros are multiple for each f(z) ∈ ℱ added in Theorem 1.7 is best possible. In Theorem 1.11 taking b = 0 we get Corollary 1.13.

Corollary 1.13. Let ℱ be a family of meromorphic functions in D, and let a be a nonzero finite complex number. Suppose that all of zeros are multiple for each f(z) ∈ ℱ. If, for every pair of functions f and g in ℱ, ff^′ and gg^′ share the value a, then ℱ is normal in D.

Corollary 1.14. Let ℱ be a family of meromorphic functions in D, n be a positive integer and a, b be two finite complex numbers such that a ≠ 0. If for each function f in ℱ, f^′ − af⁻ⁿ ≠ b, then ℱ is normal in D.

Lemma 2.1 (see [24].)Let ℱ be a family of meromorphic functions on the unit disc satisfying all zeros of functions in ℱ that have multiplicity ≥p and all poles of functions in ℱ that have multiplicity ≥q. Let α be a real number satisfying −q < α < p. Then ℱ is not normal at 0 if and only if there exist

• (a)

  a  number  0 < r < 1;

• (b)

  points    z_n  with  |z_n| < r;

• (c)

  functions    f_n ∈ ℱ;

• (d)

  positive  numbers    ρ_n → 0

such that g_n(ζ): = ρ^−αf_n(z_n + ρ_nζ) converges spherically uniformly on each compact subset of ℂ to a nonconstant meromorphic function g(ζ), whose all zeros have multiplicity ≥p and all poles have multiplicity ≥q and order is at most 2.

Remark 2.2. If ℱ is a family of holomorphic functions on the unit disc in Lemma 2.1, then g(ζ) is a nonconstant entire function whose order is at most 1.

Lemma 2.3 (see [25] or [26].)Let f(z) be a meromorphic function and c ∈ ℂ∖{0}. If f(z) has neither simple zero nor simple pole, and f′(z) ≠ c, then f(z) is constant.

Lemma 2.4 (see [27].)Let f(z) be a transcendental meromorphic function of finite order in ℂ and have no simple zero, then f′(z) assumes every nonzero finite value infinitely often.

Proof of Theorem 1.7. Suppose that ℱ is not normal in D. Then there exists at least one point z₀ such that ℱ is not normal at the point z₀. Without loss of generality we assume that z₀ = 0. By Lemma 2.1, there exist points z_j → 0, positive numbers ρ_j → 0, and functions f_j ∈ ℱ such that

locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function in ℂ. Moreover, the order of g is ≤2.

From (3.1) we know

in ℂ∖S locally uniformly with respect to the spherical metric, where S is the set of all poles of g(ξ).

If g′gⁿ − a ≡ 0, then −1/(n + 1)gⁿ⁺¹ ≡ aξ + c, where c is a constant. This contradicts with g being a meromorphic function. So g′gⁿ − a≢0.

If g′gⁿ − a ≠ 0, by Lemma 2.3, then g is also a constant which is a contradiction with g being a nonconstant. Hence, g′gⁿ − a is a nonconstant meromorphic function and has at least one zero.

Next we prove that g′gⁿ − a has just a unique zero. On the contrary, let ξ₀ and $$ be two distinct zeros of g′gⁿ − a, and choose δ(>0) small enough such that $$, where D(ξ₀, δ) = {ξ:|ξ − ξ₀ | < δ} and $$. From (3.2), by Hurwitz′s theorem, there exist points ξ_j ∈ D(ξ₀, δ), $$ such that for sufficiently large j

By the hypothesis that, for each pair of functions f and g in ℱ, f′ − af⁻ⁿ and g′ − ag⁻ⁿ share b in D, we know that for any positive integer  m

Fix m, take j → ∞, and note z_j + ρ_jξ_j → 0, $$, then $$. Since the zeros of $$ have no accumulation point, so

Hence, $$. This contradicts with $$, and $$. So g′gⁿ − a has just a unique zero, which can be denoted by ξ₀. By Lemma 2.4, g is not any transcendental function.

If g is a nonconstant polynomial, then g′gⁿ − a = A(ξ − ξ₀) ^l, where A is a nonzero constant, l is a positive integer, because l ≥ n ≥ 3. Set ϕ = (1/(n + 1))gⁿ⁺¹, then ϕ^′ = A(ξ − ξ₀) ^l + a and ϕ^′′ = Al(ξ − ξ₀) ^l−1. Note that the zeros of ϕ are of multiplicity ≥4. But ϕ^′′ has only one zero ξ₀, so ϕ has only the same zero ξ₀ too. Hence, ϕ^′(ξ₀) = 0 which contradicts with ϕ^′(ξ₀) = a ≠ 0. Therefore, g and ϕ are rational functions which are not polynomials, and ϕ^′ − a has just a unique zero ξ₀.

Next we prove that there exists no rational function such as ϕ. Noting that ϕ = (1/(n + 1))gⁿ⁺¹, we can set

where A is a nonzero constant, s ≥ 1,   t ≥ 1,   m_i ≥ n + 1 ≥ 3  (i = 1,2, …, s),   n_j ≥ n + 1 ≥ 3  (j = 1,2, …, t). For stating briefly, denote From (3.6), where are polynomials. Since ϕ^′(ξ) + a has only a unique zero ξ₀, set where B is a nonzero constant, so where p₂(ξ) = B(l − N − 2t)ξ^t + b_t−1ξ^t−1 + ⋯+b₀ is a polynomial. From (3.8) we also have where p₃(ξ) is also a polynomial.

Let deg (p) denote the degree of a polynomial p(ξ).

From (3.8) and (3.9),

Similarly from (3.11), (3.12) and noting (3.13),

Note that m_i ≥ 3  (i = 1,2, …, s), it follows from (3.8) and (3.10) that ϕ^′(ξ_i) = 0  (i = 1,2, …, s) and ϕ^′(ξ₀) = a ≠ 0. Thus, ξ₀ ≠ ξ_i  (i = 1,2, …, s), and then (ξ − ξ₀) ^l−1 is a factor of p₃(ξ). Hence, we get that l − 1 ≤ deg (p₃). Combining (3.11) and (3.12) we also have m − 2s = deg (p₂) + l − 1 − deg (p₃) ≤ deg (p₂). By (3.14) we obtain

Since m ≥ 3s, we know by (3.16) that

If l ≥ N + t, by (3.15), then

Noting (3.17), we obtain 1 ≤ 0; a contradiction.

If l < N + t, from (3.8) and (3.10), then deg (p₁) = deg (q₁). Noting that deg (h) ≤ s + t − 1, deg (p₁) ≤ m + t − 1, and deg (q₁) = N + t, hence m ≥ N + 1 ≥ 3t + 1. By (3.16), 2t + 1 ≤ 2s. From (3.17), we obtain 1 ≤ 0; a contradiction.

The proof of Theorem 1.7 is complete.

Proof of Theorem 1.11. The proof of this theorem is the same as the proof of Theorem 1.7, some different places are stated as follows.

The zeros of g are multiple;

The zeros of ϕ are of multiplicity ≥4:

Noting m ≥ 4s, by (3.16) we have

If l ≥ N + t, by (3.15), then

Noting (3.17), we obtain 1 ≤ 0; a contradiction.

If l < N + t, from (3.8) and (3.10), then deg (p₁) = deg (q₁). Noting that deg (h) ≤ s + t − 1, deg (p₁) ≤ m + t − 1, and deg (q₁) = N + t, hence m ≥ N + 1 ≥ 2t + 1. By (3.16), 2t + 1 ≤ 2s + t. From ((3.17)′), we obtain 1 ≤ 0; a contradiction.

The proof of Theorem 1.11 is complete.