Normal functions concerning derivatives and shared sets

Abstract

Let f(z) be meromorphic in ${\varDelta }$, $E_1=\{a_1, a_2, a_3\}$ and $E_2=\{b_1, b_2, b_3\}$ be two sets in ${\mathbb {C}}$, $k\in Z^+$. Suppose that $f (z)\in E_1 \Leftrightarrow f^{(k)}(z) \in E_2$ and $\max \limits _{0\le i\le k-1}|f^{(i)}(z)|=0$ whenever $f (z)\in E_1$, then f(z) is a normal function.

1 Introduction and main results

By the Bloch principle, the criteria of normal functions are studied, which are consistent with the known criteria of normal families. For example, corresponding to the well-known Montel’s theorem of normal families, Lehto and Virtanen [4] showed that if f(z) is meromorphic in ${\varDelta }$ and $f(z)\ne a, b, c$, then f(z) is a normal function, where a, b, c are three distinct points in $\overline{{\mathbb {C}}}$. But, they are not always right when the criteria of normal functions are related to derivatives. For instance, corresponding to the well-known Miranda criterion for the family of holomorphic functions, Hayman and Storvick [1] proved that there exists a non-normal function f(z) satisfying $f(z)\ne 0$ and $f'(z)\ne 1$ in ${\varDelta }$. In the following, we focus on the criteria of normal functions concerning derivatives.

The normality of families of meromorphic functions concerning shared values was proved by Schwick [8] in 1992, which is listed below.

Theorem A

[8, Theorem 2] Let ${\mathcal {F}}$ be a family of meromorphic in D, and let $a_1, a_2, a_3$ be three distinct complex numbers in ${\mathbb {C}}$. Suppose that, f(z) and $f'(z)$ share $a_1, a_2, a_3$, for every function $f(z) \in {\mathcal {F}}$, then ${\mathcal {F}}$ is normal in D.

In 2007, Liu-Pang [5] improved Schwick’s result, by means of substituting sharing the set $\{a_1, a_2, a_3\}$ for sharing three values $a_1, a_2, a_3$ in Theorem A, as follows:

Theorem B

[5, Theorem 1] Let ${\mathcal {F}}$ be a family of meromorphic in D, and let $E=\{a_1, a_2, a_3\}$ be a set in ${\mathbb {C}}$. Suppose that f(z) and $f'(z)$ share E, for every function $f \in {\mathcal {F}}$; then ${\mathcal {F}}$ is normal in D.

Recently, we consider the question about normal functions concerning derivatives and shared sets. And, we get the following results:

Theorem 1

Let f(z) be meromorphic in ${\varDelta }$, $E_1=\{a_1, a_2, a_3\}$ and $E_2=\{b_1, b_2, b_3\}$ be two sets in ${\mathbb {C}}$, $k\in Z^+$. Suppose that $f (z)\in E_1 \Leftrightarrow f^{(k)}(z) \in E_2$ and $\max \limits _{0\le i\le k-1}|f^{(i)}(z)|=0$ whenever $f (z)\in E_1$; then f(z) is a normal function.

And for the case of holomorphic functions, the following result can be obtained.

Theorem 2

Let f(z) be holomorphic in ${\varDelta }$, $a_1, a_2, a_3$ be three distinct complex numbers in ${\mathbb {C}}$, and $A>0$. Suppose that $|f'(z)|\le A$ whenever $f(z)=a_i (i=1,2,3)$; then f(z) is a normal function.

Remark 1

The condition “$\max \limits _{0\le i\le k-1}|f^{(i)}(z)|=0$ whenever $f (z)\in E_1$” in Theorem 1 holds naturally for $k=1$. Or more accurately, if $k=1$, then $i=0$ and the condition “$\max \limits _{0\le i\le k-1}|f^{(i)}(z)|=0$ whenever $f (z)\in E_1$” is removed. And if $k\ge 2$, then $1\le i \le k-1$ and the condition is needed. Furthermore, in the latter case, the multiplicities of zeros of $f(z)-a$ are at least k, where $a\in E_1$.

The following example shows that the condition “$\max \limits _{0\le i\le k-1}|f^{(i)}(z)|=0$ whenever $f (z)\in E_1$” in Theorem 1 is necessary.

Example 1

Let ${\mathcal {F}}=\{f_n(z)|f_n(z)= n^2(e^{a_1 z}-e^{a_2 z}), z\in {\varDelta }, n=1, 2, \ldots \}$, $E_1=E_2=\{-1, 0, 1\}$, where $a_1 \ne a_2$ satisfying $a_1^k=a_2^k=1$, $k\in Z^+$.

By calculating, it yields that $f_n(z)=f_n^{(k)}(z)$, and $f_n(z) \in E_1 \Leftrightarrow f_n^{(k)}\in E_2.$$\max \limits _{0\le i\le k-1}|f_n^{(i)}(z)|\ne 0$ whenever $f_n (z)\in E_1$.

However, $(1-|z|^2)f_n^{\sharp }(z)|_{z=0}=(1-|0|^2)\frac{|f_n^{\prime }(0)|}{1+|f_n(0)|^2}= n^2 |a_1-a_2|\rightarrow \infty$, as $n \rightarrow \infty$. Thus, as $n \rightarrow \infty$, $f_n(z)$ is not a normal function.

Here and in the sequel, ${\mathbb {C}}$ is the complex plane and $\overline{{\mathbb {C}}}$ is the extended complex plane. D is a domain in ${\mathbb {C}}$. ${\varDelta }(z_0, r) = \{z: |z - z_0| <r\}$, ${\varDelta }'(z_0, r) = \{z: 0< |z - z_0| < r\}$, where $z_0\in {\mathbb {C}}, r>0$. The unit disc is marked as ${\varDelta }={\varDelta }(0, 1)$. $f_n(z)\overset{\chi }{\Rightarrow }f(z)$ in D shows that the sequence $\{f_n(z)\}$ converges to f(z) in the spherical metric uniformly in compact subsets of D and $f_n(z)\Rightarrow f(z)$ in D if the convergence is in the Euclidean metric. $f^\sharp (z)=|f^\prime (z)|/(1+|f(z)|^2)$ is the spherical derivative of f(z).

Let f(z), L(z) be meromorphic in D, a, b be two complex numbers in ${\mathbb {C}}$, and $E_1$, $E_2$ be two sets in ${\mathbb {C}}$. $f(z)=a\Rightarrow L(z)=b$ if $L(z) = b$ whenever $f(z)=a$, and $f(z)=a\Leftrightarrow L(z)=b$ if $f(z)=a\Rightarrow L(z)=b$ and $L(z)=b \Rightarrow f(z)=a$. When $f(z)=a\Leftrightarrow L(z)=a$ in D, we say that f(z) and L(z) share the value a. $D(f, E_1):=\cup _{a\in E_1}\{z\in D: f(z)=a\}$, $D(L, E_2):=\cup _{a\in E_2}\{z\in D: L(z)=a\}$. If $D(f, E_1)\subset D(L, E_2)$, we write $f(z)\in E_1\Rightarrow L(z)\in E_2$ in D. Furthermore, if $D(f, E_1)\subset D(L,E_2)$ and $D(L,E_2)\subset D(f,E_1)$, that is, $D(f,E_1)= D(L,E_2)$, we write $f (z)\in E_1 \Leftrightarrow L(z) \in E_2$ in D. If $f (z)\in E \Leftrightarrow L(z) \in E$ in D, we say that f(z) and L(z) share the set E in D.

Recall that a family ${\mathcal {F}}$ of meromorphic in D is said to be a normal family in D, if each sequence $\{f_{n}(z)\}\subset {\mathcal {F}}$ contains a subsequence which converges spherically locally in D. The subtracted set may depend on the subsequence. See [2, 3]. A function f(z) meromorphic in ${\varDelta }$, it is said to be a normal function if and only if the family $\{f(L(z))\}$ is normal (see [4]), where L(z) shows an arbitrary one-one conformal mapping of ${\varDelta }$ onto itself.

2 Preliminary results

First, we introduce some lemmas which will be used in the proofs of main results.

Lemma 1

[7, Lemma 2] Let ${\mathcal {F}}$ be a family of meromorphic in D, all of whose zeros have multiplicities at least $k (\in Z^+)$, and suppose that there exists $M \ge 1$ such that $|f^{(k)}(z)|\le M$ whenever $f(z)=0$ and $f(z)\in {\mathcal {F}}$. If ${\mathcal {F}}$ is not normal at $z_0\in D$, then for each $\alpha , 0\le \alpha \le k$, there exist a sequence of complex numbers $z_n\in D, z_n\rightarrow z_0$, positive numbers $\rho _n\rightarrow 0$, and a sequence of functions $f_n\in {\mathcal {F}}$ such that

$$\begin{aligned} \begin{aligned} L_n(\xi )= \frac{f_n(z_n + \rho _n \xi )}{\rho _n^\alpha } \overset{\chi }{\Rightarrow }L(\xi ), \end{aligned} \end{aligned}$$

where $L(\xi )$ is nonconstant and meromorphic in ${\mathbb {C}}$, all of whose zeros have multiplicities at least k, such that $L^\sharp (\xi )\le L^\sharp (0)=kM+1$. Moreover, $L(\xi )$ has order at most 2.

Lemma 2

[6] Let f(z) be meromorphic in ${\varDelta }$. Suppose that f(z) is not a normal function, then there exists a sequence of points $z_n\in {\varDelta }$ and positive numbers $\rho _n \rightarrow 0$ such that

$$\begin{aligned} L_n(z)=f(z_n + \rho _n z)\overset{\chi }{\Rightarrow }L(z) \end{aligned}$$

in ${\mathbb {C}}$, where L(z) is nonconstant and meromorphic in ${\mathbb {C}}$.

Lemma 3

[3, Theorem 1.5] Let f(z) be nonconstant and meromorphic, $a_1, a_2,\cdots , a_q$ be $q(>2)$ distinct complex numbers in ${\mathbb {C}}$. Then

$$\begin{aligned} m(r,f)+\sum _{i=1}^{q} m\left( r,\frac{1}{f-a_i}\right) \le 2T(r,f)-N_1(r)+S(r,f), \end{aligned}$$

where $N_1(r)=2N(r,f)-N(r,f')+N(r,\frac{1}{f'})$ and $S(r,f)=o(T(r,f),$ as $r\rightarrow \infty$ possibly outside a set of finite measure.

Lemma 4

[2, Corollary to Theorem 3.5] Let f(z) be a transcendental meromorphic function in ${\mathbb {C}}$ and $k \in Z^+$. Then f(z) or $f^{(k)}(z) - 1$ has infinitely many zeros.

3 Proof of theorems

Proof of Theorem 1

Suppose, to the contrary, that f(z) is not a normal function in ${\varDelta }$. Then, based on Lemma 2, there exist points $z_n \in {\varDelta }$, $\rho _n \rightarrow 0^+$ such that

$$\begin{aligned} \begin{aligned} L_n(\eta )= f(z_n + \rho _n \eta )\overset{\chi }{\Rightarrow }L(\eta ), \end{aligned} \end{aligned}$$

(1)

where $L(\eta )$ is nonconstant and meromorphic in ${\mathbb {C}}$.

And we claim that the multiplicities of all zeros of $L(\eta )-a_i$ (for each $a_i\in E_1$ ) are no less than $k+1$. In fact, assume that $L(\eta _0)-a_1=0$, it gets that there exist points $\eta _n \rightarrow \eta _0$ such that $f(z_n+\rho _n \eta _n)=a_1$ by Hurwitz’s theorem and $L(\eta )$ is nonconstant. According to $f (z)\in E_1 \Leftrightarrow f^{(k)}(z) \in E_2$ and $\max \limits _{0\le i\le k-1}|f^{(i)}(z)|=0$ whenever $f (z)\in E_1$, it obtains $f^{(k)}(z_n+\rho _n \eta _n)\in E_2$ and $f^{(i)}(z_n+\rho _n \eta _n)=0 (0\le i\le k-1)$. Then $L_n^{(i)}(\eta _n)=\rho _n^i f^{(i)}(z_n+\rho _n \eta _n)= \rho _n^i \cdot 0=0, (0\le i\le k-1)$, and $L_n^{(k)}(\eta _n)=\rho _n^k f^{(k)}(z_n+\rho _n \eta _n)=\rho _n^k b_i$. Clearly, $L^{(i)}(\eta _0)=\lim \limits _{n\rightarrow \infty } L_n^{(i)}(\eta _n)=0(0\le i\le k)$. Thus, the multiplicities of all zeros of $L(\eta )-a_1$ are no less than $k+1$. Similar results that the multiplicities of all zeros of $L(\eta )-a_i (i=2,3)$ are no less than $k+1$ can be obtained. So, the claim is proved.

Among $L(\eta )-a_i (i=1,2,3)$, there exists at least one term which has zeros, according to Lemma 3 and $L(\eta )$ is nonconstant. Suppose that $\eta _0$ is a zero of $L(\eta )-a_1$ with multiplicities l. It can find $\delta >0$, for large enough n, such that $L_n(\eta )$ is holomorphic in ${\varDelta }(\eta _0, \delta )$.

Let

$$\begin{aligned} \begin{aligned} \varphi _n(\eta )= \frac{L_n(\eta )-a_1}{\rho _n^k}. \end{aligned} \end{aligned}$$

(2)

Then $\{\varphi _n(\eta )\}$ is holomorphic in ${\varDelta }(\eta _0, \delta )$.

It is asserted that (i)$\{\varphi _n(\eta )\}$ is not normal at $\eta _0$, and (ii)$|\varphi _n^{(k)}(\eta )|\le |b_1|+|b_2|+|b_3|$ whenever $\varphi _n(\eta )=0$.

First of all , we prove the claim (i). Suppose, to the contrary, that $\{\varphi _n(\eta )\}$ is normal at $\eta _0$. According to the definition of normal family, there exist $0< \delta _1 <\delta$ and a subsequence of $\{\varphi _n(\eta )\}$ (still denoted by $\{\varphi _n(\eta )\}$), such that

$$\begin{aligned} \varphi _n(\eta )\Rightarrow \varphi (\eta ) \end{aligned}$$

in ${\varDelta }(\eta _0, \delta _1)$, where $\varphi (\eta )$ is holomorphic or identical to infinity in ${\varDelta }(\eta _0, \delta _1)$. Since $L(\eta _0)=a_1$ and $L(\eta )$ is nonconstant, there exist points $\eta _n \rightarrow \eta _0$, for large enough n, such that $L_n(\eta _n)-a_1=0$ by Hurwitz’s theorem. And hence

$$\begin{aligned} \begin{aligned} \varphi (\eta _0)= \lim _{n \rightarrow \infty }\varphi _n(\eta _n) =\lim _{n \rightarrow \infty } \frac{L_n(\eta _n)-a_1}{\rho _n^k}=0. \end{aligned} \end{aligned}$$

(3)

On the other side, it can find that $\eta _0^\prime \in {\varDelta }(\eta _0, \delta _1)$ such that $\eta _0^\prime \ne \eta _0$ and $L(\eta _0^\prime )\ne a_1$ according to the isolated of zeros. Thus, it gets, for large enough n, $|L_n(\eta _0^\prime )-a_1|>|L(\eta _0^\prime )-a_1|/2>0$. Hence

$$\begin{aligned} \begin{aligned} |\varphi _n(\eta _0^\prime )|= \frac{|L_n(\eta _0^\prime )-a_1|}{\rho _n^k}> \frac{|L(\eta _0^\prime )-a_1|}{2 \rho _n^k} \rightarrow \infty . \end{aligned} \end{aligned}$$

Then, $\varphi _n(\eta ) \Rightarrow \infty$ in ${\varDelta }(\eta _0, \delta _1)$, which contradicts Eq. (3). So the claim (i) is proved.

Next, we prove the claim (ii). Indeed, assume that $\varphi _n(\eta )=0$, it gets that $f(z_n+\rho _n \eta )= a_1 \in E_1$ by Eqs. (1) and (2). By the condition $f(z) \in E_1 \Rightarrow f^{(k)} (z)\in E_2$, it gets $f^{(k)}(z_n + \rho _n \eta ) \in E_2$. Hence, $|\varphi _n^{(k)}(\eta )|=|f^{(k)}(z_n+\rho _n \eta )| \le |b_1|+|b_2|+|b_3|$. Thus the claim (ii) holds immediately.

Then, according to the claim (i) and Lemma 1, there exist a subsequence of $\{\varphi _n(\eta )\}$ (still marked as $\{\varphi _n(\eta )\}$), points $\eta _n \rightarrow \eta _0$, and $\zeta _n \rightarrow 0^+$ such that

$$\begin{aligned} \begin{aligned} {\varPhi }_n(\xi )=\frac{\varphi _n(\eta _n+ \zeta _n \xi )}{\zeta _n^k} =\frac{L_n(\eta _n+ \zeta _n \xi )-a_1}{\zeta _n^k \rho _n^k} \overset{\chi }{\Rightarrow }{\varPhi }(\xi ), \end{aligned} \end{aligned}$$

(4)

where ${\varPhi }(\xi )$ is meromorphic and nonconstant in ${\mathbb {C}}$. What is more, based on the claim (ii), it has ${\varPhi }^\sharp (\xi ) \le {\varPhi }^\sharp (0)=k(|b_1|+|b_2|+|b_3|+1)+1$.

It can be asserted that (iii) ${\varPhi }(\xi )$ is an entire function in ${\mathbb {C}}$, (iv) ${\varPhi }(\xi )$ has no more than l distinct zeros, and (v) ${\varPhi }(\xi )=0$ if and only if ${\varPhi }^{(k)}(\xi )\in E_2$.

We first prove the claim (iii). Indeed, based on the fact that $\{{\varPhi }_n(\xi )\}$ is holomorphic in ${\varDelta }(\eta _0, \delta )$, and $\eta _n+\zeta _n \xi \rightarrow \eta _0$ for each $\xi \in {\mathbb {C}}$, it obtains that ${\varPhi }(\xi )$ is an entire function in ${\mathbb {C}}$ by Eq. (4). Thus, the claim (iii) is proved.

In the following, we prove the claim (iv). Suppose, to the contrary, that ${\varPhi }(\xi )$ has (no less than) $l + 1$ distinct zeros: $\xi _1, \xi _2, \ldots , \xi _{l+1}$. Then there exist $l + 1$ distinct sequences $\{\xi _{nj}\}$ such that $\xi _{nj}\rightarrow \xi _j$ and ${\varPhi }_n(\xi _{nj})=0 (j=1, 2, \ldots , l+1)$ by Hurwitz’s theorem and Eq. (4). Hence $L_n(\eta _n+\zeta _n \xi _{nj})-a_1 = 0$. Obviously, $\eta _n+\zeta _n \xi _{nj}\rightarrow \eta _0$ and $\eta _n+\zeta _n \xi _{ni}\ne \eta _n+\zeta _n \xi _{nj}$ for $1\le i < j \le l+1$. It gets that $\eta _0$ is a zero of $L(\eta )-a_1$ with multiplicities at least $l+1$ by Eq. (1). But, this contradicts the fact that $\eta _0$ is a zero of $L(\eta )-a_1$ with multiplicities l. So, the claim (iv) is true.

Last, we prove the claim (v). On the one hand, set ${\varPhi }(\xi _0)=0$. Combining ${\varPhi }(\xi )\not \equiv 0$ and Eq. (4) with Hurwitz’s theorem, it follows that there exist points $\xi _n \rightarrow \xi _0$ such that ${\varPhi }_n(\xi _n)=0$. Then $L_n(\eta _n + \zeta _n \xi _n)-a_1 = 0$. It gets that

$$\begin{aligned} f(z_n+\rho _n \eta _n + \rho _n \zeta _n \xi _n )=a_1 \in E_1. \end{aligned}$$

According to the condition that $f(z)\in E_1 \Rightarrow f^{(k)}(z) \in E_2$, and $\max \limits _{0\le i\le k-1}|f^{(i)}(z)|=0$ whenever $f (z)\in E_1$, it yields

$$\begin{aligned} {\varPhi }_n^{(i)} (\xi _n) = \frac{f^{(i)} (z_n + \rho _n \eta _n +\rho _n \zeta _n \xi _n)}{\zeta _n^{k-i} \rho _n^{k-i}} =0~(0\le i\le k-1) \end{aligned}$$

and

$$\begin{aligned} {\varPhi }_n^{(k)} (\xi _n) = f^{(k)} (z_n + \rho _n \eta _n +\rho _n \zeta _n \xi _n) \in E_2. \end{aligned}$$

It gets that ${\varPhi }^{(i)}(\xi _0)=0 ~ (0\le i \le k-1)$, ${\varPhi }^{(k)}(\xi _0)\in E_2$ according to ${\varPhi }_n^{(i)}(\xi _n) \rightarrow {\varPhi }^{(i)}(\xi _0)$. Further, we obtain that all zeros of ${\varPhi }(\xi )$ have multiplicities at least k.

On the other hand, assume that ${\varPhi }^{(k)}(\xi _0) \in E_2$, and ${\varPhi }^{(k)}(\xi _0)=b_1$. It can be concluded that ${\varPhi }^{(k)}(\xi ) \not \equiv b_1$. Otherwise, ${\varPhi }(\xi )=b_1\frac{(\xi -\xi _1)^k}{k!}$ since all zeros of ${\varPhi }(\xi )$ have multiplicities at least k. By simple calculation,

$$\begin{aligned} {\varPhi }^\sharp (0)\le \left\{ \begin{array}{rl} \frac{k}{2},&{} \quad \text {if} \, |\xi _1|\ge 1,\\ |b_1|, &{} \quad \text{ if } \, |\xi _1|< 1. \end{array} \right. \end{aligned}$$

That is, ${\varPhi }^\sharp (0)<k(|b_1|+|b_2|+|b_3|+1)+1$ which contradicts the fact that ${\varPhi }^\sharp (0)=k(|b_1|+|b_2|+|b_3|+1)+1$. Thus, ${\varPhi }^{(k)}(\xi ) \not \equiv b_1$. Then, there exist $\xi _n \rightarrow \xi _0$ such that ${\varPhi }_n^{(k)}(\xi _n)=b_1$ by Hurwitz’s theorem. Clearly,

$$\begin{aligned} f^{(k)}(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n) =b_1 \in E_2. \end{aligned}$$

It follows from $f(z) \in E_1 \Leftarrow f^{(k)}(z) \in E_2$ that $f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)\in E_1$. And we claim that there exist a subsequence of $\{f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)\}$ (still marked as $\{f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)\}$) such that

$$\begin{aligned} f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)=a_1. \end{aligned}$$

Otherwise, there exists $f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)=a_2$ for large enough n. Then, Eq. (4) deduces

$$\begin{aligned} \begin{aligned} {\varPhi }(\xi _0)= \lim _{n \rightarrow \infty }{\varPhi }_n(\xi _n) =\lim _{n \rightarrow \infty } \frac{ f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)-a_1 }{\zeta _n^k \rho _n^k} =\lim _{n \rightarrow \infty } \frac{a_2-a_1 }{\zeta _n^k \rho _n^k}=\infty , \end{aligned} \end{aligned}$$

which contradicts the fact that ${\varPhi }^{(k)}(\xi _0)=b_1$. Thus, there exist a subsequence of $\{f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)\}$ (still marked as $\{f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)\}$) such that $f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)=a_1.$ Obviously,

$$\begin{aligned} \begin{aligned} {\varPhi }(\xi _0)= \lim _{n \rightarrow \infty }{\varPhi }_n(\xi _n) =\lim _{n \rightarrow \infty } \frac{ f(z_n +\rho _n \eta _n + \rho _n \zeta _n \xi _n)-a_1 }{\zeta _n^k \rho _n^k} =0. \end{aligned} \end{aligned}$$

Hence, the claim (v) is proved.

The claims (iv) and (v) imply that ${\varPhi }^{(k)}(\xi )-b_1$, ${\varPhi }^{(k)}(\xi )-b_2$ and ${\varPhi }^{(k)}(\xi )-b_3$ have only finitely many zeros. Thus, ${\varPhi }(\xi )$ is a polynomial according to the claim (iii) and Lemma 4.

Set

$$\begin{aligned} {\varPhi }(\xi )=c_p \xi ^p +c_{p-1}\xi ^{p-1}+\cdots + c_0, \end{aligned}$$

where $p\in Z^+$, $c_0, c_1, \ldots , c_p (\ne 0)$ are complex constants. And Lemma 3 yields

$$\begin{aligned} \begin{aligned} 2T(r, {\varPhi }^{(k)}) \le \sum _{i=1}^3 N\left( r, \frac{1}{{\varPhi }^{(k)}- b_i}\right) +S(r, {\varPhi }^{(k)}). \end{aligned} \end{aligned}$$

(5)

Furthermore, the claim (v) deduces

$$\begin{aligned} \begin{aligned} \sum _{i=1}^3 N\left( r, \frac{1}{{\varPhi }^{(k)}- b_i}\right) \le N\left( r, \frac{1}{{\varPhi }}\right) =p \log r. \end{aligned} \end{aligned}$$

(6)

Obviously, $T(r, {\varPhi }^{(k)})=(p-k)\log r$ and $S(r, {\varPhi }^{(k)})=O(1)$. Combining this fact and Eq. (6) with Eq. (5), it yields

$$\begin{aligned} (p-2k)\log r \le O(1), \text {~as~} r \rightarrow \infty . \end{aligned}$$

So, $p \le 2k$. Moreover, we know that all zeros of ${\varPhi }(\xi )$ have multiplicity at least k. Thus, four cases are divided as follows:

1. When $1\le p\le k-1$. It obtains that ${\varPhi }(\xi )$ is a polynomial with degree at most $k-1$. This contradicts the fact that all zeros of ${\varPhi }(\xi )$ have multiplicities at least k.

2. When $p=k$. It gets that ${\varPhi }(\xi )=c_k\frac{(\xi -\xi _1)^k}{k!}$ and ${\varPhi }^{(k)}(\xi )\equiv c_k$, where $c_k\ne 0$. It follows from the claim v that $c_k\in E_2$. Thus, ${\varPhi }^{(k)}(\xi )\in E_2$ for each $\xi \in {\mathbb {C}}$. However, ${\varPhi }(\xi )$ has only one distinct zero. This contradicts the claim v.

3. When $k+1\le p \le 2k-1$. It yields that ${\varPhi }(\xi )=c_p\frac{(\xi -\xi _1)^p}{p!}$. And ${\varPhi }^{(k)}(\xi )$ is a polynomial with degree $p-k~(\ge 1)$. Thus, ${\varPhi }^{(k)}(\xi )=a_i (i=1,2,3)$ has at least three distinct zeros. But, ${\varPhi }(\xi )$ has only one distinct zero. This contradicts the claim v.

4. When $p=2k$, it obtains that ${\varPhi }(\xi )=c_{2k}\frac{(\xi -\xi _1)^k (\xi -\xi _2)^k}{(2k)!}$ or ${\varPhi }(\xi )=c_{2k}\frac{(\xi -\xi _1)^{2k} }{(2k)!}$, and ${\varPhi }^{(k)}(\xi )$ is a polynomial with degree k. Thus, ${\varPhi }^{(k)}(\xi )=a_i (i=1,2,3)$ has at least three distinct zeros. However, ${\varPhi }(\xi )$ has at most two distinct zeros. This contradicts the claim v. Therefore, Theorem 1 is proved.

Proof of Theorem 2

Suppose, to the contrary, f(z) is not a normal function in ${\varDelta }$. Then, based on Lemma 2, there exist points $z_n \in {\varDelta }$, positive numbers $\rho _n \rightarrow 0$ such that

$$\begin{aligned} \begin{aligned} L_n(\eta )= f(z_n + \rho _n \eta )\overset{\chi }{\Rightarrow }L(\eta ), \end{aligned} \end{aligned}$$

where $L(\eta )$ is a nonconstant holomorphic function in ${\mathbb {C}}$.

It is asserted that $L'(\eta )=0$ whenever $L(\eta )=a_1, a_2, a_3$. In fact, assume that $L(\eta _0)=a_1$; then there exist points $\eta _n \rightarrow \eta _0$, for large enough n, such that $a_1=L_n(\eta _n)=f(z_n + \rho _n \eta _n)$ by Hurwitz$'$s theorem and $L(\eta )$ is nonconstant. Thus $|f'(z_n + \rho _n \eta _n)|\le A$ according to the condition that $|f'(z)|\le A$ whenever $f(z)=a_i (i=1,2,3)$. Then $|L'_n(\eta _n)|=|\rho _n f'(z_n + \rho _n \eta _n|\le \rho _n A$. Clearly, $L'(\eta _0)=\lim \limits _{n\rightarrow \infty } L'_n(\eta _n)=0$. Then $L'(\eta )=0$ whenever $L(\eta )=a_1$. Similarly, it gets that $L'(\eta )=0$ whenever $L(\eta )=a_2$ or $a_3$. Then, the claim is proved.

Referring to the fact that $L(\eta )$ is nonconstant, one may easily get $L'(\eta )\not \equiv 0$. And based on the fact $L(\eta )$ is nonconstant and Lemma 3, it follows that

$$\begin{aligned} \begin{aligned} 2T(r, L)&\le \sum _{i=1}^3 {\overline{N}}(r, \frac{1}{L- a_i})+S(r, L)\\&\le N(r,\frac{1}{L'})+S(r, L)\\&\le T(r,\frac{1}{L'})+S(r, L)\\&\le T(r,L')+S(r, L)\\&\le T(r,L)+S(r, L).\\ \end{aligned} \end{aligned}$$

One can then obtain $T(r, L)=S(r, L)$, which is a contradiction. Thus, f(z) is a normal function in ${\varDelta }$. This completes the proof of Theorem 2. $\square$