Let f be a meromorphic function in ℂ , define M f = f ′ ( f − 1 ( 0 ) ) = { f ′ ( z ) : z ∈ ℂ     and     f ( z ) = 0 } . W. Bergweiler [1] gave a conjecture in 2001 as follow: Conjecture 1: Let f be a transcendental meromorphic function in ℂ . If f ′ ( z ) ≠ 1 for all z ∈ ℂ , then M f is unbounded.

Bergweiler pointed that let g ( z ) = z − f ( z ) , Conjecture 1 is equivalent to the following one.

Conjecture 2: Let g be a transcendental meromorphic function in ℂ . Suppose that g ′ does not have zeros. Then there exist a sequence { z n } n = 1 ∞ of fixed points of g such that | g ′ ( z n ) | → ∞ .

Bergweiler [1] has separately proved Conjecture 1 is affirmative for finite order meromorphic functions and entire functions; Jianming Chang [2] has confirmed the conjecture for infinite order meromorphic functions for the first time, which is based on theory of normal and quasinormal families.

For the conjecture, f ′ ≠ 1 and f ′ ≠ c ( c ≠ 0 ) are essentially equivalent. In fact, if f ′ ≠ c ( c ≠ 0 ) , then f ′ c ≠ 1 and f c is also transcendental meromorphic function; the zeros of f and the zeros of f c are the same and M f is unbounded if and only if M f c is unbounded.

Considering the discussion above, it’s natural to research the problem that whether the conclusion is true if f ′ ≠ 0 but not f ′ ≠ c ( c ≠ 0 ) , the problem is radically different to the conjecture and gives a important supplement. We can give a example to show that the problem is significant. Let f = e e z − 1 , it’s obvious that f ′ ≠ 0 and M f is unbounded. In details, we have

Theorem 1. Let f be meromorphic in ℂ and the order is greater than 2. If f ′ has finitely many zeros and f takes a finite non-zero value finitely many times, then M f is unbounded.

Theorem 2. Let f be entire in ℂ and the order is greater than 1. If f ′ has finitely many zeros and f takes a finite non-zero value finitely many times, then M f is unbounded.

Lemma 1. Let f be a meromorphic function. If the spherical derivative f # ( z ) of f ( z ) is bounded. Then the order of f ( z ) is at most 2.

For details of lemma 1, can see [3]

Remark: Let f n ( z ) be a sequence of meromorphic functions, f n ( z ) ⇉ l o c . g ( z ) means f n ( z ) locally uniformly convergence to meromorphic function g ( z ) ; for a meromorphic function f in ℂ , let D ( z 0 , M ) = { z : | z − z 0 | ≤ M } .

Lemma 2. Let F be a family of functions meromorphic in a domain D. Suppose that there exist K > 0 such that M g ⊂ D ( 0, K ) ¯ for all g ∈ F . For any given α satisfying − 1 < α ≤ 1 , if F is not normal, then there exist a sequence { f n } n = 1 ∞ in F , a sequence { z n } n = 1 ∞ in D, a sequence { ρ n } n = 1 ∞ of positive real numbers and a non-constant finite order function f which is meromorphic in ℂ such that z n → z 0 for some z 0 ∈ D , ρ n → 0 and

f ( z n + ρ n z ) ρ n α ⇉ l o c . f ( z ) ( z ∈ ℂ ; n → ∞ ) .

Moreover, the spherical derivative f # ( z ) of f satisfies f # ( z ) ≤ f # ( 0 ) = K + 1 for all z ∈ ℂ .

For details of lemma 2, can see [4]

In the case no hypothesis on M g is required, the case α = 0 is due to Zalcman [5], and the case − 1 < α < 1 is due to Pang [6] [7]. In the hypothesis of Lemma 2, if all g ∈ F have no zero in D, then M g = ∅ and the conclusion is still right according to the proof of Lemma 2 in which K can take 0.

Lemma 3. Let f be a meromorphic function, D be an bounded domain and c be constant in ℂ , if f ( z ) − c has l ( l ≥ 2 ) zeros in D and f ′ ( z ) has l − 1 zeros in D which are all the zeros of f ( z ) − c . Then each discriminating zero of f ( z ) − c and f ′ ( z ) in D is the same one.

Proof. Let f ( z ) − c = ∏ j = 1 l ( z − z j ) g ( z ) = R ( z ) g ( z ) with g ( z ) be a meromorphic function which have no zero in D satisfying g ( z j ) ≠ ∞   ( j = 1 , ⋯ , l ) and R ( z ) = ∏ j = 1 l ( z − z j ) in which z j ⊂ D   ( j = 1 , ⋯ , l ) .

Because f ′ ( z ) has l − 1 zeros in D which are all the zeros of f ( z ) − c , there exist l − 1 points in z j ( j = 1 , ⋯ , l ) be zeros of f ′ ( z ) and without loss of generality we may assume f ′ ( z j ) = 0   ( j = 1 , ⋯ , l − 1 ) .

As f ′ ( z ) = R ′ ( z ) g ( z ) + R ( z ) g ′ ( z ) , we can deduce that

R ′ ( z j ) = 0 ( j = 1 , ⋯ , l − 1 ) , R ′ ( z ) = l ⋅ ∏ j = 1 l ( z − z j ) and R ′ ( z ) R ( z ) = l z − z l

from the above it follows that R ( z ) = ( z − z l ) l and the proof of Lemma 3 is complete.

Lemma 4. Let f be a holomorphic function, if the spherical derivative of f is bounded. Then the order of f is at most 1.

For details of lemma 4, can see [8].

Proof. we apply Lemma 1 to obtain a sequence { ω n } n = 1 ∞ , ω n → ∞ ( n → ∞ ) such that f # ( ω n ) → ∞ , ( n → ∞ ) . ∀ n ∈ ℕ , let f n ( z ) = f ( z + ω n ) , it’s easy to apply Marty’s theorem to know { f n ( z ) } n = 1 ∞ is not normal at 0. Suppose f ( z ) − λ only have finitely many zeros ( λ ≠ 0 ), there exist a subsequence of { f n ( z ) } n = 1 ∞ we still suppose it’s { f n ( z ) } n = 1 ∞ such that f ( z ) − λ have no zero in ℂ , thus according to Lemma 2, there exist a sequence { z n } n = 1 ∞ , a sequence { ρ n } n = 1 ∞ of positive real numbers and a non-constant finite order function g ( z ) such that when n → ∞ , z n → 0, ρ n → 0 and

f n ( z n + ρ n z ) − λ ρ n = f ( ω n + z n + ρ n z ) − λ ρ n ⇉ l o c . g (z)

in ℂ and g ( z ) satisfies g # ( z ) ≤ g # ( 0 ) = 1 for all z ∈ ℂ .

∀ n ∈ ℕ , let τ n = ω n + z n , there exist entire functions F ( z ) and H ( z ) such that F ( z ) and H ( z ) have no common non-trivial divisor and f ( z ) = H ( z ) F ( z ) , then

f n ( z n + ρ n z ) − λ ρ n = f ( τ n + ρ n z ) − λ ρ n = H ( τ n + ρ n z ) F ( τ n + ρ n z ) − λ ρ n ⇉ l o c . g (z)

ρ n f ( τ n + ρ n z ) − λ ⇉ l o c . 1 g ( z ) ( z ∈ ℂ ) (1)

For f ( τ n + ρ n z ) − λ ≠ 0 and g ( z ) ≠ 0 , the derivative of (1) is

ρ n 2 f ′ ( τ n + ρ n z ) { f ( τ n + ρ n z ) − λ } 2 ⇉ l o c . g ′ ( z ) g 2 ( z ) ( z ∈ ℂ ) (2)

here we divide two cases:

Case 1: g ′ ( z ) g 2 ( z ) have no zero in ℂ .

Because ( 1 g ( z ) ) # = g # ( z ) is bounded, then we apply Lemma 4 to have that the order of 1 g ( z ) and g ′ ( z ) g 2 ( z ) are at most 1. On the other hand, g ′ ( z ) g 2 ( z ) ≠ 0 , we can deduce g ′ ( z ) g 2 ( z ) = e A z + B ( A , B ∈ ℂ , A ≠ 0 ) or constant α ( α ≠ 0 ) and 1 g ( z ) = e A z + B A + d or 1 g ( z ) = α z + β   ( d , β ∈ ℂ ) .

here we first proof that d ≠ 0 , if d = 0 , we have

f ( τ n + ρ n z ) − λ ρ n ⇉ l o c . A e A z + B ( z ∈ ℂ )

then we have that

f ( τ n + ρ n z ) = H ( τ n + ρ n z ) F ( τ n + ρ n z ) ⇉ l o c . λ ( z ∈ ℂ ) (3)

H ( τ n + ρ n z ) − λ F ( τ n + ρ n z ) F ( τ n + ρ n z ) ⇉ l o c . 0 ( z ∈ ℂ ) (4)

From (4) it can be deduced that because H ( τ n + ρ n z ) − λ F ( τ n + ρ n z ) have no zero, F ( τ n + ρ n z ) have no zero in any bounded domain when n is large enough,. Then so are H ( τ n + ρ n z ) unite (3). What’ more, f ( τ n + ρ n z ) has no zero and pole and cannot take λ in any bounded domain, which contradict with Picard’s Theorem, therefore d ≠ 0 .

From (1) we have

ρ n f ( τ n + ρ n z ) − λ − ρ n − λ ⇉ l o c . 1 g ( z ) ( z ∈ ℂ )

Because 1 g have zero in ℂ , from above it can be deduced that f ( τ n + ρ n z ) have to have zeros which convergence to the zeros of 1 g ( z ) . From (2) when f ( τ n + ρ n z ) = 0 then ρ n f ′ ( τ n + ρ n z ) λ 2 → − A d or α and f ′ ( τ n + ρ n z )   ( n ∈ ℕ ) have to be unbounded.

Case 2: g ′ ( z ) g 2 ( z ) have zero in ℂ . We will proof the case is impossible.

Take a finite zero c of g ′ ( z ) g 2 ( z ) and it’s multiple is k ( k ∈ ℕ , k ≥ 1 ) then there exist some sufficiently small neighborhood D c of c such that D c only have one zero of g ′ ( z ) g 2 ( z ) . (2) can be expressed as

ρ n 2 ⋅ F ′ ( τ n + ρ n z ) H ( τ n + ρ n z ) − H ′ ( τ n + ρ n z ) F ( τ n + ρ n z ) { H ( τ n + ρ n z ) − λ F ( τ n + ρ n z ) } 2 ⇉ l o c . g ′ ( z ) g 2 ( z ) ( z ∈ ℂ ) .

Because f ( z ) take λ finitely many times and from (2) H ( τ n + ρ n z ) − λ F ( τ n + ρ n z ) have no zero in D c then when n is large enough, F ′ ( τ n + ρ n z ) H ( τ n + ρ n z ) − H ′ ( τ n + ρ n z ) F ( τ n + ρ n z ) have k zeros in D c , which are all the zeros of F ( τ n + ρ n z ) due to that

f ′ = H ′ ( z ) F ( z ) − H ( z ) F ′ ( z ) F 2 ( z ) only has finitely many zeros; therefore, c is the zero of 1 g ( z ) with k + 1 multiple and F ( τ n + ρ n z ) have k + 1 zeros in D c .

(1) can be expressed as

ρ n F ( τ n + ρ n z ) H ( τ n + ρ n z ) 1 − λ F ( τ n + ρ n z ) H ( τ n + ρ n z ) ⇉ l o c . 1 g ( z ) ( z ∈ ℂ ) .

From (1) we have that

λ F ( τ n + ρ n z ) H ( τ n + ρ n z ) ⇉ l o c . 1 ( z ∈ D c \ { c } ) . (5)

Here we divide two cases for (5):

Subcase 2.1: If { λ F ( τ n + ρ n z ) H ( τ n + ρ n z ) } n = 1 ∞ is normal in D c .

From (5) we have

λ F ( τ n + ρ n z ) H ( τ n + ρ n z ) ⇉ l o c . 1 ( z ∈ D c ) . (6)

λ F ( τ n + ρ n z ) − H ( τ n + ρ n z ) H ( τ n + ρ n z ) ⇉ l o c . 0 ( z ∈ D c ) .

when n is large enough, notice that H ( τ n + ρ n z ) − λ F ( τ n + ρ n z ) have no zero in D c , therefore H ( τ n + ρ n z ) have no zero in D c and according to (6), F ( τ n + ρ n z ) have no zero in D c contradict with F ( τ n + ρ n z ) have k + 1 zeros in D c .

Subcase 2.2: If { λ F ( τ n + ρ n z ) H ( τ n + ρ n z ) } n = 1 ∞ is not normal in D c . Let

φ n ( z ) = λ F ( τ n + ρ n z ) − H ( τ n + ρ n z ) H ( τ n + ρ n z ) .

Then { φ n ( z ) } n = 1 ∞ is not normal and have no zero in D c , we apply Lemma 2 to obtain { ν n } n = 1 ∞ ∈ ℂ and { ρ n * } n = 1 ∞ of positive real numbers and a non-constant finite order function ψ ( ξ ) such that ν n → c , ρ n * → 0 , and

ψ n ( ξ ) = φ n ( ν n + ρ n * ξ ) ρ n * ⇉ l o c . ψ ( ξ ) ( n → ∞ , ξ ∈ ℂ ) with ψ # ( ξ ) ≤ ψ # ( 0 ) = 1.

here we will prove that ψ ( ξ ) has no simple pole if it exist; let ξ 0 be the pole of ψ ( ξ ) , for ψ ( ξ ) cannot always be ∞ , there exist closed disc D ¯ ( ξ 0 , δ ) such that 1 / ψ ( ξ ) and 1 / ψ n ( ξ ) are holomorphic in D ¯ ( ξ 0 , δ ) and 1 / ψ n ( ξ ) ⇉ 1 / ψ ( ξ ) uniformly in D ¯ ( ξ 0 , δ ) and so are 1 / ψ n ( ξ ) + ρ n * .

Notice that 1 / ψ n ( ξ ) cannot be constant, there exist { ξ n } n = 1 ∞ , ξ n → ξ 0 ( n → ∞ ) such that

1 ψ n ( ξ n ) + ρ n * = ρ n * φ n ( ν n + ρ n * ξ n ) + ρ n * = 0 ,   φ n ( ν n + ρ n * ξ n ) + 1 = 0.

We firstly show that the discriminating zeros of φ n ( z ) + 1 in D c are all the zeros of φ ′ n ( z ) in D c when n is large enough. In fact, we have that the k zeros of F ′ ( τ n + ρ n z ) H ( τ n + ρ n z ) − H ′ ( τ n + ρ n z ) F ( τ n + ρ n z ) as same as φ ′ n ( z ) , which are all belong to the k + 1 zeros of F ( τ n + ρ n z ) and φ n ( z ) + 1 in D c , then Lemma 3 can be used to prove the conclusion and we further have

( 1 ψ ( ξ ) ) ′ | ξ = ξ 0 = − ψ ′ ( ξ 0 ) { ψ ( ξ 0 ) } 2 = − lim n → ∞ ψ ′ n ( ξ n ) { ψ n ( ξ n ) } 2 = − lim n → ∞ { ρ n * } 2 ⋅ 0 = 0

which means ψ ( ξ ) has to have multiple pole if it exist.

Notice H ′ ( z ) F ( z ) − H ( z ) F ′ ( z ) F 2 ( z ) only has finitely many zeros, then H ( τ n + ρ n z ) have no multiple zero in ℂ when n is large enough.

Considering 1 / ψ n ( ξ ) ⇉ l o c . 1 / ψ ( ξ ) ( ξ ∈ ℂ ) , since λ F ( τ n + ρ n z ) − H ( τ n + ρ n z ) have no zero in D c when n is large enough, 1 / ψ n ( ξ ) are analytic in D c and according to Hurwitz’s Theorem, ψ ( ξ ) has no multiple pole. With the assert above, ψ ( ξ ) have no pole and be entire.

Notice that ψ # ( ξ ) ≤ ψ # ( 0 ) = 1 ( ξ ∈ ℂ ) and Lemma 4, the order of ψ ( ξ ) is at most 1. Since φ n ( z ) have no zero in D c when n is large enough, then φ n ( ν n + ρ n * ξ ) have no zero in any bounded domain, from the above it follows that ψ ( ξ ) ≠ 0 ( ξ ∈ ℂ ) and ψ ( ξ ) = e A ξ + B ( A , B ∈ ℂ , A ≠ 0 ) .

1 / ψ n ( ξ ) ⇉ l o c . 1 / ψ ( ξ ) ( ξ ∈ ℂ ) is

ρ n * H ( τ n + ρ n ν n + ρ n ρ n * ξ ) λ F ( τ n + ρ n ν n + ρ n ρ n * ξ ) − H ( τ n + ρ n ν n + ρ n ρ n * ξ ) ⇉ l o c . e − A ξ − B ( ξ ∈ ℂ ) . (7)

Let

η n ( z ) = F ( τ n + ρ n z ) H ′ ( τ n + ρ n z ) − F ′ ( τ n + ρ n z ) H ( τ n + ρ n z ) { λ F ( τ n + ρ n z ) − H ( τ n + ρ n z ) } 2

then the derivative of (7) is

( ρ n * ) 2 ρ n λ η n ( ν n + ρ n * ξ ) ⇉ l o c .   −   A e − A ξ − B ( n → ∞ , ξ ∈ ℂ ) . (8)

(2) can be expressed as

ρ n 2 η n ( z ) ⇉ l o c . g ′ ( z ) g 2 ( z ) ( n → ∞ , z ∈ ℂ ) .

∀ n ∈ ℕ , let h n ( z ) be the k order derivative of η n ( z ) , then the k order derivative of (2) is

ρ n 2 h n ( z ) ⇉ l o c . ( 1 g ( z ) ) ( k + 1 ) ( n → ∞ , z ∈ ℂ ) .

with ( 1 g ( z ) ) ( k + 1 ) have no zero in D c . Let ( 1 g ( z ) ) ( k + 1 ) | z = c = G c ( ≠ 0 ) then we have

ρ n 2 h n ( ν n + ρ n * ξ ) ⇉ l o c . G c ( n → ∞ , ξ ∈ ℂ ) . (9)

The k order derivative of (8) is

( ρ n * ) k + 2 ρ n λ h n ( ν n + ρ n * ξ ) ⇉ l o c . ( − A ) k + 1 e − A ξ − B ( n → ∞ , ξ ∈ ℂ ) , (10)

(9) + (10) is

{ ρ n + λ ( ρ n * ) k + 2 } ρ n h n ( ν n + ρ n * ξ ) ⇉ l o c . G c + ( − A ) k + 1 e − A ξ − B ( n → ∞ , ξ ∈ ℂ ) .

It shows that h n ( ν n + ρ n * ξ ) have to have zeros in ℂ when n is large enough, however, from (10) and Hurwitz’s theorem, it is impossible; this gives a contradiction and the proof of Theorem 1 is complete.

It follows from the proof of Theorem 1 that the hypothesis for order can be replaced by greater than 1 for entire functions. In fact, from Lemma 4, we can obtain a sequence { ω n } n = 1 ∞ , ω n → ∞ ( n → ∞ ) such that f # ( ω n ) → ∞ , ( n → ∞ ) . Then using the start point of proof of Theorem 1, ∀ n ∈ ℕ , let f n ( z ) = f ( z + ω n ) , it’s easy to apply Marty’s theorem to know { f n ( z ) } n = 1 ∞ is not normal at 0. Suppose f ( z ) − λ only has finitely many zeros ( λ ≠ 0 ), there exist a subsequence of { f n ( z ) } n = 1 ∞ . We still suppose it’s { f n ( z ) } n = 1 ∞ such that f n ( z ) − λ has no zero in ℂ . Thus according to Lemma 2, there exist a sequence { z n } n = 1 ∞ , a sequence { ρ n } n = 1 ∞ of positive real numbers and a non-constant finite order function g ( z ) such that when n → ∞ , z n → 0, ρ n → 0 and

f n ( z n + ρ n z ) − λ = f ( ω n + z n + ρ n z ) − λ ⇉ l o c . g (z)

in ℂ and g ( z ) satisfies g # ( z ) ≤ g # ( 0 ) = 1 for all z ∈ ℂ . For g ′ ≠ 0 and g ≠ 0 we apply Lemma 6 to have the order of g at most 1, and g = e A z + B ( A ≠ 0 ) and we have

f ( ω n + z n + ρ n z ) ⇉ l o c . e A z + B + λ ( n → ∞ , z ∈ ℂ ) .

and the first order derivative is

ρ n f ′ ( ω n + z n + ρ n z ) ⇉ l o c . A e A z + B ( n → ∞ , z ∈ ℂ ) .

If f ( ω n + z n + ρ n z ) = 0 , then ρ n f ′ ( ω n + z n + ρ n z ) → − A λ and f ′ ( ω n + z n + ρ n z )   ( n ∈ ℕ ) is unbounded. Then the proof of Theorem 2 is complete.

The requirement for order in Theorem 2 is sharp, let f = e z − 1 , then M f = { 1 } .

By the equivalence between the conjecture 1 and 2, we can have two corollaries from Theorem 1 and 2.

Corollary 1. Let g be meromorphic in ℂ and the order is greater than 2. If g ′ − 1 has finitely many zeros and g − z takes a finite non-zero value finitely many times, then g has a sequence { z n } n = 1 ∞ of fixed points such that g ′ ( z n ) → ∞ , ( n → ∞ ) .

Corollary 2. Let g be entire in ℂ and the order is greater than 1. If g ′ − 1 has finitely many zeros and g − z takes a finite non-zero value finitely many times, then g has a sequence { z n } n = 1 ∞ of fixed points such that g ′ ( z n ) → ∞ , ( n → ∞ ) .

I thank the Editor and the referee for their comments. Research of F. Guo is funded by the Yunnan province Science Foundation grant 2016FD015. This support is greatly appreciated.

The author declares no conflicts of interest regarding the publication of this paper.

Guo, F. (2019) Transcendental Meromorphic Functions Whose First Order Derivatives Have Finitely Many Zeros. Advances in Pure Mathematics, 9, 925-933. https://doi.org/10.4236/apm.2019.911045