Value distribution for difference operator of meromorphic functions with maximal deficiency sum

Abstract

The main purpose of this paper is to investigate the relationship between the characteristic function of a meromorphic function $f(z)$ with maximal deficiency sum and that of the exact difference ${\mathrm{\Delta }}_{c}f=f(z+c)-f(z)$. As an application, the author also establishes an inequality on the zeros and poles for ${\mathrm{\Delta }}_{c}f$ and gives an example to show that the upper bound of the inequality is accurate.

MSC:30D30, 39A05.

1 Introduction

If $f(z)$ is a meromorphic function in the complex plane ℂ and $a\in \mathbb{C}$, we use the following notations frequently used in Nevanlinna theory (see [1–3]): $m(r,f)$, $N(r,f)$, $m(r,a)=m(r,\frac{1}{f-a})$, $N(r,a)=N(r,\frac{1}{f-a})$, … . Denote by $S(r,f)$ any quantity such that $S(r,f)=o(T(r,f))$, $r\to \mathrm{\infty }$ without restriction if $f(z)$ is of finite order and otherwise except possibly for a set of values of r of finite linear measure. The Nevanlinna deficiency of f with respect to a finite complex number a is defined by

$$\delta (a,f)=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{m(r,a)}{T(r,f)}=1-\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,a)}{T(r,f)}.$$

If $a=\mathrm{\infty }$, then one should replace $m(r,a)(N(r,a))$ in the above formula by $m(r,f)(N(r,f))$. The classical second fundamental theorem of Nevanlinna theory asserts that the total deficiency of any meromorphic function $f(z)$ satisfies the inequality

$$\sum _{a\in \mathbb{C}}\delta (a,f)+\delta (\mathrm{\infty },f)\le 2.$$

If the above equality holds, then we say that f has maximal deficiency sum. The Valiron-Mo’honko identity states that if the function $R(z,f)$ is rational in f and has small meromorphic coefficients, then

$$T(r,R(z,f))={deg}_f(R)T(r,f)+S(r,f).$$

(1.1)

Certain relationship between the characteristic function of a meromorphic function $f(z)$ with maximal deficiency sum and that of derivative $f^{\prime }(z)$ plays a key role in the study of a conjecture of Nevanlinna (see [4]). The main contribution of this paper is to study the relationship between the characteristic function of a meromorphic function $f(z)$ with maximal deficiency sum and that of the exact difference ${\mathrm{\Delta }}_{c}f=f(z+c)-f(z)$, where $c\ne 0$ (see [5]).

In 1956, Shan and Singh [6] proved the following theorem.

Theorem A [6]

Suppose that $f(z)$ is a transcendental meromorphic function of finite order and ${\sum }_{a\in \mathbb{C}}\delta (a,f)=2$. Then

$$T(r,f^{\prime })\sim 2T(r,f),r\to +\mathrm{\infty }.$$

After that, Edrei [7] and Weitsman [4] proved the following theorem, respectively.

Theorem B [4, 7]

Suppose that $f(z)$ is a transcendental meromorphic function of finite order with maximal deficiency sum. Then

$$\underset{r\to +\mathrm{\infty }}{lim}\frac{T(r,f^{\prime })}{T(r,f)}=2-\delta (\mathrm{\infty },f),$$

and

$$\underset{r\to +\mathrm{\infty }}{lim}\frac{N(r,\frac{1}{f^{\prime }})}{T(r,f^{\prime })}=0.$$

Under the condition of Theorem B, Singh and Gopalakrishna [8] proved that

$$\underset{r\to +\mathrm{\infty }}{lim}\frac{N(r,a)}{T(r,f)}=1-\delta (a,f)$$

(1.2)

holds for every $a\in \mathbb{C}$.

Let $f(z)$ be a transcendental meromorphic function of order less than one. Bergweiler and Langley [9] proved that ${\mathrm{\Delta }}_{c}f(z)\sim f^{\prime }(z)$ outside some exceptional set. Motivated by this result, we extend Theorem B to the exact difference ${\mathrm{\Delta }}_{c}f$ and prove the following theorem.

Theorem 1.1 (main)

Suppose that $f(z)$ is a transcendental meromorphic function of order less than one with maximal deficiency sum. Then we have

• (1) ${lim}_{r\to +\mathrm{\infty }}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}=2-\delta (\mathrm{\infty },f)$.

• (2) ${lim}_{r\to +\mathrm{\infty }}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}=0$.

Consequently, we have that the deficiency of ${\mathrm{\Delta }}_{c}f$ with respect to 0 is 1, i.e.,

$$\delta (0,{\mathrm{\Delta }}_{c}f)=1-\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}=1.$$

For the zeros and poles involving the derivative of a transcendental meromorphic function of finite order with maximal deficiency sum, Singh and Kulkarni [10] proved the following theorem.

Theorem C [10]

Suppose that $f(z)$ is a transcendental meromorphic function of finite order with maximal deficiency sum. Then

$$\frac{1-\delta (\mathrm{\infty },f)}{2-\delta (\mathrm{\infty },f)}\le K\left(f^{\prime }\right)\le \frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)},$$

where

$$K\left(f^{\prime }\right)=\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,f^{\prime })+N(r,\frac{1}{f^{\prime }})}{T(r,f^{\prime })}.$$

In 2000, Fang [11] proved the following theorem.

Theorem D [11]

Suppose that $f(z)$ is a transcendental meromorphic function of finite order with maximal deficiency sum. Then

$$K\left(f^{\prime }\right)=\frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)}.$$

In fact, Fang [11] proved that Theorem D is valid for higher order derivatives of $f(z)$. In this paper, we shall extend Theorem D to the exact difference ${\mathrm{\Delta }}_{c}f$ and prove the following theorem.

Theorem 1.2 (main)

Suppose that $f(z)$ is a transcendental meromorphic function of order less than one with maximal deficiency sum. Then

$$K({\mathrm{\Delta }}_{c}f)\le \frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)},$$

where

$$K({\mathrm{\Delta }}_{c}f)=\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,{\mathrm{\Delta }}_{c}f)+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}.$$

The following example shows that the upper bound of the inequality in Theorem 1.2 is accurate.

Example 1.3 Let $f(z)=\frac{1}{e^z-1}$, then ${\mathrm{\Delta }}_{c}f=\frac{(1-e^c)e^z}{(e^c{e}^z-1)(e^z-1)}$. Then ${\mathrm{\Delta }}_{c}f\ne 0$, $\delta (0,f)=1$, $\delta (-1,f)=1$, $\delta (\mathrm{\infty },f)=0$. Thus $f(z)$ is a meromorphic function with maximal deficiency sum. It is obvious that $\delta (0,e^z)=1$, $\delta (\mathrm{\infty },e^z)=1$ and $N(r,{\mathrm{\Delta }}_{c}f)=N(r,e^z=1)+N(r,e^z=e^{-c})$. It follows from (1.2) that

$$N(r,e^z=1)=N(r,e^z=e^{-c})\sim T(r,e^z),r\to +\mathrm{\infty }$$

and from Valiron-Mo’honko identity (1.1) that

$$T(r,{\mathrm{\Delta }}_{c}f)\sim 2T(r,e^z),r\to +\mathrm{\infty }.$$

Therefore, $K({\mathrm{\Delta }}_{c}f)=\frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)}=1$.

By Theorem D and Example 1.3, we pose the following question.

Question 1.4 Under the condition of Theorem 1.2, can we replace $K({\mathrm{\Delta }}_{c}f)\le \frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)}$ by $K({\mathrm{\Delta }}_{c}f)=\frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)}$?

Corollary 1.5 Let $f(z)$ be a transcendental meromorphic function of order less than one with maximal deficiency sum, and assume $\delta (\mathrm{\infty },f)=1$. Then

$$\underset{r\to +\mathrm{\infty }}{lim}\frac{N(r,{\mathrm{\Delta }}_{c}f)}{T(r,{\mathrm{\Delta }}_{c}f)}=0.$$

Consequently, we have that the deficiency of ${\mathrm{\Delta }}_{c}f$ with respect to ∞ is 1, i.e.,

$$\delta (\mathrm{\infty },{\mathrm{\Delta }}_{c}f)=1-\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,{\mathrm{\Delta }}_{c}f)}{T(r,{\mathrm{\Delta }}_{c}f)}=1.$$

As the end of this paper, we shall prove the following theorem.

Theorem 1.6 (main)

Let $f(z)$ be a transcendental meromorphic function of order less than one, and assume $\delta (\mathrm{\infty },f)=1$. Then

$$\sum _{a\in \mathbb{C}}\delta (a,f)\le \delta (0,{\mathrm{\Delta }}_{c}f).$$

2 Some lemmas

Lemma 2.1 [12]

Let $f(z)$ be a meromorphic function of finite order σ, and let c be a non-zero complex number. Then, for each $\epsilon >0$, we have

$$m(r,\frac{f(z+c)}{f(z)})=O\left(r^{\sigma -1+\epsilon }\right).$$

Lemma 2.2 Let $f(z)$ be a transcendental meromorphic function of order σ (<1), and let c be a non-zero complex number. Then

$$m(r,\frac{f(z+c)}{f(z)})=o(T(r,f))=S(r,f).$$

Proof Since the order of $f(z)$ is less than one, then, for any $0<\epsilon <1-\sigma$, it follows from Lemma 2.1 that

$$m(r,\frac{f(z+c)}{f(z)})=O\left(r^{\sigma -1+\epsilon }\right)=O(1).$$

Therefore, we have

$$m(r,\frac{f(z+c)}{f(z)})=O(1)=o(T(r,f))=S(r,f).$$

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Lemma 2.3 [12]

Let $f(z)$ be a meromorphic function with the exponent of convergence of poles $\lambda (\frac{1}{f})=\lambda <+\mathrm{\infty }$, and let c be a non-zero complex number. Then, for each $\epsilon >0$, we have

$$N(r,f(z+c))=N(r,f)+O\left(r^{\lambda -1+\epsilon }\right)+O(logr).$$

From Lemma 2.3, using a similar method as that in the proof of Lemma 2.2, we can prove the following lemma.

Lemma 2.4 Let $f(z)$ be a transcendental meromorphic function of order less than one, and let c be a non-zero complex number. Then

$$N(r,f(z+c))=N(r,f)+S(r,f).$$

3 Proof of Theorem 1.1

Proof By combining the first main theorem of Nevanlinna theory and Lemmas 2.2, 2.4, we have

$$\begin{array}{rl}T(r,{\mathrm{\Delta }}_{c}f)& =m(r,{\mathrm{\Delta }}_{c}f)+N(r,{\mathrm{\Delta }}_{c}f)\\ =m(r,\frac{f{\mathrm{\Delta }}_{c}f}{f})+N(r,{\mathrm{\Delta }}_{c}f)\\ \le m(r,\frac{{\mathrm{\Delta }}_{c}f}{f})+m(r,f)+N(r,f)+N(r,f(z+c))+O(1)\\ =T(r,f)+N(r,f)+S(r,f).\end{array}$$

Hence,

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}\le 1+\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,f)}{T(r,f)}=2-\delta (\mathrm{\infty },f).$$

(3.1)

Let $\{a_i\}$ be a sequence of distinct complex numbers in ℂ containing all the finite deficient values of $f(z)$. For any positive q, define

$$F(z)=\sum _{i=1}^q\frac{1}{f-a_i}.$$

Since $T(r,f(z)-a_i)=T(r,f(z))+O(1)$ and ${\mathrm{\Delta }}_c(f(z)-a_i)={\mathrm{\Delta }}_{c}f(z)$, we deduce from Lemma 2.2 that

$$m(r,F(z){\mathrm{\Delta }}_{c}f(z))\le \sum _{i=1}^{q}m(r,\frac{{\mathrm{\Delta }}_c(f(z)-a_i)}{f(z)-a_i})+logq=S(r,f).$$

This relation yields

$$m(r,F(z))=m(r,F(z){\mathrm{\Delta }}_{c}f(z)\frac{1}{{\mathrm{\Delta }}_{c}f(z)})\le m(r,\frac{1}{{\mathrm{\Delta }}_{c}f})+S(r,f).$$

(3.2)

By combining the first main theorem of Nevanlinna theory, (3.2) and Valiron-Mo’honko identity (1.1), we have

$$\begin{array}{r}qT(r,f)+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})\\ =T(r,F(z))+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})+O(1)\\ =m(r,F(z))+N(r,F(z))+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})+O(1)\\ \le m(r,F(z))+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})+\sum _{i=1}^{q}N(r,a_i)+O(1)\\ \le m(r,\frac{1}{{\mathrm{\Delta }}_{c}f})+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})+\sum _{i=1}^{q}N(r,a_i)+S(r,f)\\ =T(r,{\mathrm{\Delta }}_{c}f)+\sum _{i=1}^{q}N(r,a_i)+S(r,f).\end{array}$$

Hence,

$$\begin{array}{rl}q& \le \underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}+\sum _{i=1}^q\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,a_i)}{T(r,f)}\\ =\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}+\sum _{i=1}^q\{1-\delta (a_i,f)\}.\end{array}$$

Thus

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}\ge \sum _{i=1}^q\delta (a_i,f).$$

Since q is arbitrary, we have

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}\ge \sum _{a\in \mathbb{C}}\delta (a,f)=2-\delta (\mathrm{\infty },f).$$

Then

$$\underset{r\to +\mathrm{\infty }}{lim}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}=2-\delta (\mathrm{\infty },f).$$

(3.3)

On the other hand, by combining the first main theorem of Nevanlinna theory and (3.2), we have

$$\begin{array}{r}\sum _{i=1}^{q}m(r,a_i)+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})\\ \le m(r,F(z))+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})+O(1)\\ \le T(r,{\mathrm{\Delta }}_{c}f)+S(r,f).\end{array}$$

Thus

$$\sum _{i=1}^q\frac{m(r,a_i)}{T(r,{\mathrm{\Delta }}_{c}f)}+\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}\le 1+\frac{S(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}.$$

We derive from (3.3) that

$$\begin{array}{r}\sum _{i=1}^q\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{m(r,a_i)}{T(r,{\mathrm{\Delta }}_{c}f)}+\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}\\ \le 1+\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{S(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}\\ \le 1+\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{S(r,f)}{T(r,f)}\frac{T(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}\\ =1.\end{array}$$

Thus

$$\begin{array}{rl}1& \ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\sum _{i=1}^q\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{m(r,a_i)}{T(r,{\mathrm{\Delta }}_{c}f)}\\ \ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\sum _{i=1}^q\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{m(r,a_i)}{T(r,f)}\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{T(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}.\end{array}$$

It follows from (3.3) that

$$1\ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\frac{{\sum }_{i=1}^q\delta (a_i,f)}{2-\delta (\mathrm{\infty },f)}.$$

Since q is arbitrary, we have

$$1\ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+1.$$

Then

$$\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}=0.$$

Therefore,

$$\underset{r\to +\mathrm{\infty }}{lim}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}=0.$$

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4 Proof of Theorem 1.2

Proof It follows from Lemma 2.4 that

$$N(r,{\mathrm{\Delta }}_{c}f)\le 2N(r,f)+S(r,f).$$

The above inequality implies that

$$\frac{N(r,{\mathrm{\Delta }}_{c}f)}{T(r,{\mathrm{\Delta }}_{c}f)}\frac{T(r,{\mathrm{\Delta }}_{c}f)}{T(r,f)}\le 2\frac{N(r,f)}{T(r,f)}+\frac{S(r,f)}{T(r,f)}.$$

By Theorem 1.1(1), we have

$$(2-\delta (\mathrm{\infty },f))\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,{\mathrm{\Delta }}_{c}f)}{T(r,{\mathrm{\Delta }}_{c}f)}\le 2(1-\delta (\mathrm{\infty },f)).$$

Therefore,

$$\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,{\mathrm{\Delta }}_{c}f)}{T(r,{\mathrm{\Delta }}_{c}f)}\le \frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)}.$$

This relation and Theorem 1.1(2) together yield

$$K({\mathrm{\Delta }}_{c}f)\le \frac{2(1-\delta (\mathrm{\infty },f))}{2-\delta (\mathrm{\infty },f)}.$$

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5 Proof of Theorem 1.6

Proof If ${\sum }_{a\in \mathbb{C}}\delta (a,f)=0$, Theorem 1.6 is valid in this case. In the following, we assume that ${\sum }_{a\in \mathbb{C}}\delta (a,f)>0$. Let $\{a_{\mu }\}$ be a sequence of distinct complex numbers in ℂ containing all the finite deficient values of $f(z)$. For any positive integer q, as we did in the proof of Theorem 1.1(2), we can get that

$$\sum _{\mu =1}^{q}m(r,a_{\mu })+N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})\le T(r,{\mathrm{\Delta }}_{c}f)+S(r,f)$$

holds for any q finite complex numbers in $\{a_{\mu }\}$. Therefore, we have

$$\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\frac{T(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}(\frac{{\sum }_{\mu =1}^{q}m(r,a_{\mu })}{T(r,f)}-o(1))\le 1,r\to +\mathrm{\infty }.$$

Hence, from (3.1) we can get

$$\begin{array}{rl}1& \ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}[\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\frac{T(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}(\frac{{\sum }_{\mu =1}^{q}m(r,a_{\mu })}{T(r,f)}-o(1))]\\ \ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{T(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}(\frac{{\sum }_{\mu =1}^{q}m(r,a_{\mu })}{T(r,f)}-o(1))\\ \ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{T(r,f)}{T(r,{\mathrm{\Delta }}_{c}f)}\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{{\sum }_{\mu =1}^{q}m(r,a_{\mu })}{T(r,f)}\\ \ge \underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{N(r,\frac{1}{{\mathrm{\Delta }}_{c}f})}{T(r,{\mathrm{\Delta }}_{c}f)}+\frac{{\sum }_{j=1}^q\delta (a_{\mu },f)}{2-\delta (\mathrm{\infty },f)}.\end{array}$$

Since q is arbitrary and $\delta (\mathrm{\infty },f)=1$, we have

$$\sum _{a\in \mathbb{C}}\delta (a)\le \delta (0,{\mathrm{\Delta }}_{c}f).$$

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