On difference equations concerning Schwarzian equation

Abstract

Consider the difference equation

$$\biggl[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2} \biggl(\frac {\Delta^{2}f(z)}{\Delta f(z)} \biggr)^{2} \biggr]^{k} =\frac{P(z,f(z))}{Q(z,f(z))}, $$

where $P(z,f)$ and $Q(z,f)$ are prime polynomials in $f(z)$ with $\deg_{f}P=p, \deg_{f}Q=q$, and $d=\max\{p,q\}>0$. We give the supremum of d, an estimation of the sum of Nevanlinna exceptional values of meromorphic solution $f(z)$ of the equation, and study the value distributions of their difference $\Delta f(z)$ and divided difference $\frac{\Delta f(z)}{f(z)}$.

1 Introduction and main results

In this paper, we use the basic notions of Nevanlinna theory, such as $T(r,f)$, $m(r,f)$, $N(r,f)$, and so on; see [1–3]. Let $S(r,f)$ denote any quantity satisfying $S(r,f) = o (T(r,f) )$ for all r outside a set of finite logarithmic measure. We call f an admissible solution of a difference (or differential) equation if all coefficients α of the equation satisfy $T(r,\alpha)=S(r,f)$. In addition, we denote by $\sigma(f)$ the order of growth of a meromorphic function $f(z)$ and by $\lambda(f)$ and $\lambda(\frac{1}{f} )$, respectively, the exponents of convergence of zeros and poles of $f(z)$, which are defined by

$$\sigma(f)=\mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r},\qquad\lambda(f)= \mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log N (r,\frac {1}{f} )}{\log r}, \qquad\lambda\biggl( \frac{1}{f} \biggr)=\mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log N(r,f)}{\log r}. $$

If $\lambda(f-a)<\sigma(f)$, then a is called a Borel exceptional value of f.

For $a\in \mathbb {C}\cup\{\infty\}$, we denote by $\delta(a, f)$ the deficiency of a to $f(z)$, which is defined by

$$\delta(a,f)=\mathop{\underline{\lim}}_{r\rightarrow\infty}\frac{m (r,\frac {1}{f-a} )}{T(r,f)}\quad(\mbox{if } a \in \mathbb {C}),\qquad\delta(\infty, f)=\mathop{\underline{\lim}}_{r\rightarrow\infty} \frac{m(r,f)}{T(r,f)}. $$

Obviously, $\delta(a,f)\geq0$. If $\delta(a,f)>0$, then a is called a Nevanlinna exceptional value of f. The forward differences $\Delta ^{n}f(z), n\in \mathbb {N}^{+}$, are defined in the standard way [4] by

$$ \Delta f(z)=f(z+1)-f(z), \qquad\Delta^{n+1} f(z)=\Delta^{n}f(z+1)- \Delta^{n}f(z). $$

Ishizaki [5] studied Schwarzian differential equations and obtained the following:

Theorem A

Let $\alpha_{1}, \alpha_{2},\ldots,\alpha_{s}$ be distinct constants. If

$$\biggl[ \biggl(\frac{f''}{f'} \biggr)'-\frac{1}{2} \biggl(\frac{f''}{f'} \biggr)^{2}=\frac{f'''}{f'}- \frac{3}{2} \biggl(\frac{f''}{f'} \biggr)^{2} \biggr]^{k}=\frac{P(z,f)}{Q(z,f)} $$

possesses an admissible solution, then we have

$$\sum_{j=1}^{s}\delta(\alpha_{j}, f)\leq2-\frac{d}{2k}, $$

where $d=\max\{\deg_{f}P(z,f), \deg_{f}Q(z,f)\}$.

Chen and Li [6] investigated difference equations and obtained the following theorem, which can be regarded as a difference analogue of Theorem A.

Theorem B

Let $f(z)$ be an admissible solution of the difference equation

$$ \biggl[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2} \biggl( \frac{\Delta ^{2}f(z)}{\Delta f(z)} \biggr)^{2} \biggr]^{k} =R\bigl(z,f(z) \bigr)=\frac{P(z,f(z))}{Q(z,f(z))} $$

(1.1)

such that $\sigma_{2}(f)<1$, where k (≥1) is an integer, $P(z,f)$ and $Q(z,f)$ are polynomials with $\deg_{f}P(z,f)=p, \deg_{f}Q(z,f)=q$, and $d=\max\{p,q\}$. Let $\alpha_{1}, \ldots, \alpha_{s}$ be s (≥2) distinct complex constants. Then

$$ \sum_{j=1}^{s}\delta( \alpha_{j}, f)\leq4-\frac{q}{2k}. $$

(1.2)

In particular, if $N(r,f)=S(r,f)$, then

$$ \sum_{j=1}^{s}\delta( \alpha_{j}, f)\leq2-\frac{d}{2k}. $$

(1.3)

Lan and Chen [7] studied the value distribution of transcendental meromorphic solutions and their differences of some difference equation concerning a Schwarzian equation. From Theorem B we do a rough calculation: if $f(z)$ has two finite Borel exceptional values, (1.2) shows that $q\leq4k$; if $f(z)$ has only one finite Borel exceptional value, (1.2) shows $q\leq6k$ ((1.3) shows that $d\leq2k$); if $f(z)$ has no finite Nevanlinna exceptional values, (1.2) shows that $q\leq8k$ ((1.3) shows that $d\leq4k$). The upper bound of d seems to bear on k. We ask the following interesting questions: what is the supremum of d? whether inequalities (1.2) and (1.3) can be improved? Once we fix d and k, can we get the number of Nevanlinna exceptional values of meromorphic solutions of (1.1)? In this paper, we answer these questions and obtain the following theorem.

Theorem 1.1

Suppose that $f(z)$ is an admissible solution of difference equation (1.1) of finite order, where k (≥1) is an integer, $\deg _{f}P=p, \deg_{f}Q=q$, and $d=\max\{p,q\}>0$. Let $\alpha_{j}$ ($j=1,\ldots, s$) be distinct constants. Then

• (i) $d\leq7k$;

• (ii) $\sum_{j=1}^{s}\delta(\alpha_{j},f)\leq\frac {7}{2}-\frac{d}{2k}$;

• (iii) if $N(r,f)=S(r,f)$, then $q=d\leq2k$, and $\sum _{j=1}^{s}\delta(\alpha_{j},f)\leq1-\frac{d}{2k}$.

Example 1.1

The function $f(z)=z\tan\frac{\pi z}{4}$ satisfies the difference equation

$$\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2} \biggl(\frac{\Delta ^{2}f(z)}{\Delta f(z)} \biggr)^{2} =\frac{P(z,f)}{Q(z,f)}, $$

where

$$\begin{aligned} P(z,f)={}&16f^{7}+2(9z+44)f^{6}+\bigl(61z^{2}+90z+82 \bigr)f^{5}+z\bigl(99z^{2}+116z+56\bigr)f^{4} \\ &{}+2z^{2}\bigl(43z^{2}+84z+26\bigr)f^{3}+4z^{3} \bigl(21z^{2}+16z+6\bigr)f^{2}+z^{4} \bigl(53z^{2}-6z-32\bigr)f \\ &{}+3z^{5}(3z+2)^{2}, \\ Q(z,f)={}&{-}8f^{2}(f+z) \bigl(f^{2}+f+z^{2}+z \bigr)^{2}. \end{aligned}$$

We see that $k=1$ and $p=q=7$. So, $d=\max\{p,q\}=7=7k$. Obviously, $f(z)=z\tan\frac{\pi z}{4}$ has no Nevanlinna exceptional value, that is, for any distinct constants $\alpha_{j}$ ($j=1,\ldots, s$), we have $\sum_{j=1}^{s}\delta(\alpha_{j},f)=0=\frac{7}{2}-\frac{d}{2k}$.

Example 1.2

The function $f(z)=e^{z}+z^{2}$ satisfies the difference equation

$$\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2} \biggl(\frac{\Delta ^{2}f(z)}{\Delta f(z)} \biggr)^{2} =\frac{P(z,f)}{Q(z,f)}, $$

where

$$\begin{aligned} P(z,f)={}&{-}(e-1)^{4}f^{2}+2(e-1)^{2} \bigl((e-1)^{2}z^{2}+2(e-1)z+e-7\bigr)f \\ &{}-(e-1)^{4}z^{2}-(e-1)^{4}z-4(e-1)^{3}z^{3}-2(e-7) (e-1)^{2}z^{2}-12, \\ Q(z,f)={}&2 \bigl((e-1)f+\bigl((1+e)z+1\bigr) \bigl((1-e)z+1\bigr) \bigr)^{2}. \end{aligned}$$

We see that $k=1$ and $p=q=2$. So, $d=\max\{p, q\}=2=2k$. Obviously, $f(z)$ has no finite Nevanlinna exceptional value, that is, for any distinct constants $\alpha_{j}$ $(j=1,\ldots, s)$, we have $\sum _{j=1}^{s}\delta(\alpha_{j},f)=0=1-\frac{d}{2k}$.

Remark 1.1

• (i) Comparing Theorem 1.1 and Theorem B, we see that the result of Theorem 1.1 is better than that of Theorem B.

• (ii) Under conditions of Theorem 1.1(iii), the result shows that $\sum_{j=1}^{s}\delta(\alpha_{j},f)\leq1-\frac {d}{2k}<1$, that is, $f(z)$ has no finite Borel exceptional value.

• (iii) Examples 1.1 and 1.2 show the equalities in Theorem 1.1 can be reached. So, the result of Theorem 1.1 is precise.

In general, $\lambda(\Delta f)\leq\sigma(f)$, where $f(z)$ is a meromorphic function. For example, if $f_{1}(z)=e^{z}+1$ and $f_{2}(z)=e^{z}+z$, by calculation we see that $\Delta f_{1}(z)=(e-1)e^{z}$ has no zeros, whereas $\Delta f_{2}(z)=(e-1)e^{z}+1$ satisfies $\lambda(\Delta f_{2})=\sigma(f_{2})$. Even for a meromorphic function of small lower order, $\Delta f(z)$ may have finitely many zeros. To show this, Bergweiler and Langley [8] constructed a meromorphic function like this and obtained the following:

Theorem C

Let $\phi(r)$ be a positive nondecreasing function on $[1,\infty)$ that satisfies $\lim_{r\rightarrow\infty}\phi (r)=\infty$. Then there exists a transcendental and meromorphic function f in the plane with

$$\varlimsup_{r\rightarrow\infty}\frac{T(r,f)}{r}< \infty, \qquad \varliminf _{r\rightarrow\infty}\frac{T(r,f)}{\phi(r)\log r}< \infty $$

such that $g(z)=\Delta f(z)$ has only one zero. Moreover, the function g satisfies

$$\varlimsup_{r\rightarrow\infty}\frac{T(r,g)}{\phi(r)\log r}< \infty. $$

Many authors tried hard to investigate zeros of difference and divided difference of meromorphic solutions of difference equation and obtained profound results (see [7, 9, 10], etc.). Fixing the connection of d and k, we get the number of Nevanlinna exceptional values of a meromorphic solution f of (1.1) and discover some very good properties of f. These results are stated as follows.

Theorem 1.2

Suppose that $f(z)$ is an admissible meromorphic solution of difference equation (1.1) of finite order, where $k\in \mathbb {N}^{+}$ and $d=\max\{ \deg_{f}P,\deg_{f}Q\}=7k$. Then

• (i) $f(z)$ has no Nevanlinna exceptional value;

• (ii) $\lambda(\Delta f)=\lambda(\frac{1}{\Delta f} )=\sigma(f)$, $\lambda(\frac{\Delta f}{f} )=\lambda(\frac{1}{\frac {\Delta f}{f}} )=\sigma(f)$;

• (iii) $T(r,\Delta f)=2T(r,f)+S(r,f)$.

Set $d=2k$ in Theorem 1.1(iii). By the proof of Theorem 1.1(iii) we easily obtain the following corollary.

Corollary 1.1

Suppose that $f(z)$ is an admissible entire solution of difference equation (1.1) of finite order, where $k\in \mathbb {N}^{+}$ and $d=\max\{ \deg_{f}P,\deg_{f}Q\}=2k$. Then

• (i) $f(z)$ has no finite Nevanlinna exceptional value;

• (ii) $T(r,\Delta f)=T(r,f)+S(r,f)$.

The following Example 1.3 satisfies the conditions and results of Theorem 1.2.

Example 1.3

In Example 1.1, we see that $f(z)=z\tan\frac{\pi z}{4}$ is also a transcendental meromorphic solution of finite order of the equation

$$f(z+1)f(z-1)=1-z^{2}. $$

By the following Lemma 2.6, $f(z)$ has no Nevanlinna exceptional value, and

$$\lambda(\Delta f)=\lambda\biggl(\frac{1}{\Delta f} \biggr)=\sigma (f), \lambda \biggl(\frac{\Delta f}{f} \biggr)=\lambda\biggl(\frac{1}{\frac {\Delta f}{f}} \biggr)= \sigma(f). $$

By calculation we have

$$\begin{aligned} \Delta f(z)=\frac{z\tan^{2}\frac{\pi z}{4}+\tan\frac{\pi z}{4}+z+1}{1-\tan\frac{\pi z}{4}}. \end{aligned}$$

Thus,

$$T(r,\Delta f)=2T \biggl(r,\tan\frac{\pi z}{4} \biggr)+S \biggl(r,\tan \frac {\pi z}{4} \biggr)=2T(r,f)+S(r,f). $$

2 Lemmas for the proofs of theorems

Lemma 2.1

[11, 12]

Let $f(z)$ be a meromorphic function of finite order σ, and let η be a nonzero complex constant. Then, for each ε ($0<\varepsilon<1$), we have

$$ m \biggl(r, \frac{f(z+\eta)}{f(z)} \biggr)+m \biggl(r, \frac {f(z)}{f(z+\eta )} \biggr)=O \bigl(r^{\sigma-1+\varepsilon}\bigr). $$

Lemma 2.2

[11]

Let $f(z)$ be a meromorphic function with exponent of convergence of poles $\lambda(\frac{1}{f} )=\lambda<\infty$, and let $\eta \neq0$ be fixed. Then, for each ε $(0<\varepsilon<1)$,

$$N\bigl(r, f(z+\eta)\bigr)=N\bigl(r, f(z)\bigr)+O\bigl(r^{\lambda -1+\varepsilon}\bigr)+O( \log r). $$

Lemma 2.3

[11]

Let $f(z)$ be a meromorphic function of order $\sigma=\sigma(f), \sigma <\infty$, and let η be a fixed nonzero complex number. Then, for each $\varepsilon>0$,

$$ T\bigl(r, f(z+\eta)\bigr)=T\bigl(r, f(z)\bigr)+O \bigl(r^{\sigma -1+\varepsilon} \bigr)+O(\log r). $$

Lemma 2.4

[2], Lemma 1.3

Suppose that $f(z)$ is a nonconstant meromorphic function in $|z|< R$. Let $a_{j}$ $(j=1, 2,\ldots, q)$ be distinct finite complex numbers. Then, for $0< r< R$, we have

$$m \Biggl(r,\sum_{j=1}^{q} \frac{1}{f-a_{j}} \Biggr)=\sum_{j=1}^{q}m \biggl(r,\frac{1}{f-a_{j}} \biggr)+O(1). $$

Lemma 2.5

[13] (Valiron-Mohon’ko)

Let $f(z)$ be a meromorphic function. Then, for all irreducible rational functions in f,

$$R\bigl(z, f(z)\bigr)=\frac{a_{n}(z)f(z)^{n}+\cdots +a_{0}(z)}{b_{m}(z)f(z)^{m}+\cdots+b_{0}(z)} $$

with meromorphic coefficients $a_{i}(z)$ and $b_{j}(z)$ being small with respect to f, the characteristic function of $R(z, f(z))$ satisfies

$$T \bigl(r, R\bigl(z, f(z)\bigr) \bigr)=\max\{m, n\} T(r, f)+S(r, f). $$

Lemma 2.6

[10]

Suppose that $h(z)$ is a nonconstant rational function. If $w(z)$ is a transcendental meromorphic solution of finite order of the equation

$$w(z+1)w(z-1)=h(z). $$

Then

• (i) $w(z)$ has no Nevanlinna exceptional value;

• (ii) $\lambda(\Delta w)=\lambda(\frac{1}{\Delta w} )=\sigma(w)$, $\lambda(\frac{\Delta w}{w} )=\lambda(\frac{1}{\frac {\Delta w}{w}} )=\sigma(w)$.

Lemma 2.7

[14]

Let $g:(0,+\infty)\rightarrow R, h:(0,+\infty)\rightarrow R$ be nondecreasing functions. Suppose that

• (i) $g(r)\leq h(r)$ outside of an exceptional set of finite linear measure, or

• (ii) $g(r)\leq h(r), r\notin H\cup(0,1]$, where $H\subset (1,\infty)$ is a set of finite logarithmic measure. Then, for any $\alpha>1$, there exists $r_{0}>0$ such that $g(r)\leq h(\alpha r)$ for all $r>r_{0}$.

From Lemma 2.7 we easily obtain the following:

Lemma 2.8

Let $g(r)$ be a nondecreasing function, and E be a set of finite logarithmic measure or finite linear measure. Then

$$\mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\in(0,\infty)}\frac{\log g(r)}{\log r}=\mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E} \frac {\log g(r)}{\log r} $$

and

$$\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\in(0,\infty)}\frac{\log g(r)}{\log r}=\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E} \frac {\log g(r)}{\log r}. $$

3 Proofs of theorems

Proof of Theorem 1.1

(i) We first prove that $d\leq 7k$. Lemma 2.1 shows that

$$ m(r,R)=km \biggl(r,\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2} \biggl(\frac{\Delta^{2}f(z)}{\Delta f(z)} \biggr)^{2} \biggr)=S(r,f). $$

Combining this with Lemma 2.5, we have

$$ T(r,R)=m(r,R)+N(r,R)=N(r,R)+S(r,f)=dT(r,f)+S(r,f). $$

(3.1)

Rewrite (1.1) in the form

$$\begin{aligned} R(z,f)&= \biggl[\frac{\Delta^{3}f(z)}{\Delta f(z)}-\frac{3}{2} \biggl( \frac {\Delta^{2}f(z)}{\Delta f(z)} \biggr)^{2} \biggr]^{k} \\ &= \biggl[\frac{\Delta f(z+2)-2\Delta f(z+1)+\Delta f(z)}{\Delta f(z)}-\frac{3}{2} \biggl(\frac{\Delta f(z+1)-\Delta f(z)}{\Delta f(z)} \biggr)^{2} \biggr]^{k} \\ &= \biggl[\frac{\Delta f(z+2)}{\Delta f(z)}+\frac{\Delta f(z+1)}{\Delta f(z)}-\frac{3}{2} \biggl( \frac{\Delta f(z+1)}{\Delta f(z)} \biggr)^{2}-\frac{1}{2} \biggr]^{k} \\ &= \biggl[\frac{f(z+3)-f(z+2)}{f(z+1)-f(z)}+\frac {f(z+2)-f(z+1)}{f(z+1)-f(z)}-\frac{3}{2} \biggl( \frac {f(z+2)-f(z+1)}{f(z+1)- f(z)} \biggr)^{2}-\frac{1}{2} \biggr]^{k} \\ &= \biggl[\frac{f(z+3)-f(z+1)}{f(z+1)-f(z)}-\frac{3}{2} \biggl(\frac {f(z+2)-f(z+1)}{f(z+1)- f(z)} \biggr)^{2}-\frac{1}{2} \biggr]^{k}. \end{aligned}$$

(3.2)

If $z_{0}$ is a pole of $f(z+1)$ and not a pole of $f(z+3), f(z+2)$, or $f(z)$, then $z_{0}$ cannot be a pole of $\frac {f(z+3)-f(z+2)}{f(z+1)-f(z)}$ or a pole of $\frac {f(z+2)-f(z+1)}{f(z+1)-f(z)}$. By (3.2) we see that $z_{0}$ is not a pole of $R(z, f(z))$. Similarly, if $z_{1}$ is a pole of $f(z)$and not a pole of $f(z+3), f(z+2)$, or $f(z+1)$, then $z_{1}$ cannot be a pole of $R(z,f(z))$. By this and (3.2), the poles of $R(z,f(z))$ come from poles of $f(z+3)$ and $f(z+2)$ and zeros of $\Delta f(z)$. By (3.1), (3.2), and Lemmas 2.2 and 2.3 we obtain

$$\begin{aligned} dT(r,f)&=N(r,R)+S(r,f) \\ &\leq kN\bigl(r,f(z+3)\bigr)+2kN\bigl(r,f(z+2)\bigr)+2kN \biggl(r, \frac{1}{\Delta f(z)} \biggr)+S(r,f) \\ &\leq3kN\bigl(r,f(z)\bigr)+2kN \biggl(r,\frac{1}{\Delta f(z)} \biggr)+S(r,f) \end{aligned}$$

(3.3)

$$\begin{aligned} &\leq3kT\bigl(r,f(z)\bigr)+2kN \biggl(r,\frac {1}{\Delta f(z)} \biggr)+S(r,f) \end{aligned}$$

(3.4)

$$\begin{aligned} &\leq3kT\bigl(r,f(z)\bigr)+2kT\bigl(r,\Delta f(z)\bigr)+S(r,f) \\ &\leq7kT(r,f)+S(r,f). \end{aligned}$$

(3.5)

From this it follows that $d\leq7k$.

(ii) By (3.4) we have

$$N \biggl(r,\frac{1}{\Delta f(z)} \biggr)\geq\frac{d-3k}{2k}T(r,f)+S(r,f). $$

Combining this with Lemma 2.3, we obtain

$$\begin{aligned} m \biggl(r,\frac{1}{\Delta f(z)} \biggr)&=T\bigl(r,\Delta f(z) \bigr)-N \biggl(r,\frac {1}{\Delta f(z)} \biggr) \\ &\leq2T\bigl(r,f(z)\bigr)-\frac{d-3k}{2k}T(r,f)+S(r,f) \\ &= \biggl(\frac{7}{2}-\frac{d}{2k} \biggr)T(r,f)+S(r,f). \end{aligned}$$

(3.6)

From (3.6) and Lemmas 2.1 and 2.4 we deduce that

$$\begin{aligned} \sum_{j=1}^{s}m \biggl(r, \frac{1}{f-\alpha_{j}} \biggr) &=m \Biggl(r,\sum_{j=1}^{s} \frac{1}{f-\alpha_{j}} \Biggr)+O(1) \\ &\leq m \Biggl(r,\sum_{j=1}^{s} \frac{\Delta f}{f-\alpha_{j}} \Biggr)+m \biggl(r,\frac{1}{\Delta f} \biggr)+O(1) \\ &=m \biggl(r,\frac{1}{\Delta f} \biggr)+S(r,f) \\ &\leq\biggl(\frac{7}{2}-\frac{d}{2k} \biggr)T(r,f)+S(r,f). \end{aligned}$$

(3.7)

By the definition of $S(r,f)$, (3.7) shows that

$$\sum_{j=1}^{s}m \biggl(r,\frac{1}{f-\alpha_{j}} \biggr) \leq\biggl(\frac{7}{2}-\frac{d}{2k}+o(1) \biggr)T(r,f), \quad r\notin E, $$

where $E\subset(1, \infty)$ is a set of finite logarithmic measure. Thus,

$$\begin{aligned} \sum_{j=1}^{s}\delta(\alpha_{j},f)&= \sum_{j=1}^{s}\mathop{\underline{\lim}}_{r\rightarrow\infty} \frac{m (r,\frac{1}{f-\alpha_{j}} )}{T(r,f)} \leq \mathop{\underline{\lim}}_{r\rightarrow\infty}\frac{\sum_{j=1}^{s}m (r,\frac{1}{f-\alpha_{j}} )}{T(r,f)} \\ &\leq \mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E}\frac{\sum _{j=1}^{s}m (r,\frac{1}{f-\alpha_{j}} )}{T(r,f)} \\ &\leq \mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E}\frac{ (\frac {7}{2}-\frac{d}{2k}+o(1) )T(r,f)}{T(r,f)} \\ &=\frac{7}{2}-\frac{d}{2k}, \end{aligned}$$

that is,

$$\sum_{j=1}^{s}\delta(\alpha_{j},f) \leq\frac{7}{2}-\frac{d}{2k}. $$

(iii) We still have (3.1)-(3.3). By $N(r,f)=S(r,f)$ and Lemmas 2.1 and 2.2 we have

$$\begin{aligned} N \biggl(r,\frac{1}{\Delta f} \biggr)&\leq T(r,\Delta f)=m(r, \Delta f)+N(r,\Delta f) \\ &\leq m(r,f)+m \biggl(r,\frac{\Delta f}{f} \biggr)+N\bigl(r,f(z+1)\bigr)+N \bigl(r,f(z)\bigr) \\ &= m(r,f)+2N\bigl(r,f(z)\bigr)+S(r,f) \\ &= T(r,f)+S(r,f). \end{aligned}$$

(3.8)

By $N(r,f)=S(r,f)$, (3.3), (3.8), and Lemmas 2.1 and 2.5 we obtain

$$\begin{aligned} dT(r,f)&=N(r,R)+S(r,f) \\ &\leq2kN \biggl(r,\frac{1}{\Delta f} \biggr)+3kN\bigl(r,f(z)\bigr)+S(r,f) \\ &= 2kN \biggl(r,\frac{1}{\Delta f} \biggr)+S(r,f) \\ &\leq2kT(r,f)+S(r,f), \end{aligned}$$

that is,

$$\begin{aligned} dT(r,f)=N(r,R)+S(r,f) \leq2kN \biggl(r,\frac{1}{\Delta f} \biggr)+S(r,f) \leq2kT(r,f)+S(r,f). \end{aligned}$$

(3.9)

It follows that $d\leq2k$.

By (3.1) and Lemma 2.5 we have

$$\begin{aligned} dT(r,f)&=N(r,R)+S(r,f) \\ &=N \biggl(r,\frac{1}{Q} \biggr)+S(r,f) \\ &\leq T(r,Q)+S(r,f) \\ &=qT(r,f)+S(r,f), \end{aligned}$$

which yields $q\geq d$. So, $q=d\leq2k$.

By (3.9) we have

$$N \biggl(r,\frac{1}{\Delta f} \biggr)\geq\frac{d}{2k}T(r,f)+S(r,f). $$

Combining this with Lemma 2.1, we obtain

$$\begin{aligned} m \biggl(r,\frac{1}{\Delta f} \biggr)&=T(r,\Delta f)-N \biggl(r, \frac {1}{\Delta f} \biggr) \\ &=m(r,\Delta f)-N \biggl(r,\frac{1}{\Delta f} \biggr) \\ &\leq m(r,f)+m \biggl(r,\frac{\Delta f}{f} \biggr)-\frac {d}{2k}T(r,f)+S(r,f) \\ &= \biggl(1-\frac{d}{2k} \biggr)T(r,f)+S(r,f). \end{aligned}$$

(3.10)

By (3.10) and Lemmas 2.1 and 2.4 we have

$$\begin{aligned} \sum_{j=1}^{s}m \biggl(r,\frac{1}{f-\alpha_{j}} \biggr) &=m \Biggl(r,\sum_{j=1}^{s} \frac{1}{f-\alpha_{j}} \Biggr)+O(1) \\ &\leq m \Biggl(r,\sum_{j=1}^{s} \frac{\Delta f}{f-\alpha_{j}} \Biggr)+m \biggl(r,\frac{1}{\Delta f} \biggr)+O(1) \\ &=m \biggl(r,\frac{1}{\Delta f} \biggr)+S(r,f) \\ &\leq\biggl(1-\frac{d}{2k} \biggr)T(r,f)+S(r,f), \end{aligned}$$

that is,

$$\sum_{j=1}^{s}m \biggl(r,\frac{1}{f-\alpha_{j}} \biggr)\leq\biggl(1-\frac{d}{2k}+o(1) \biggr)T(r,f), \quad r\notin E_{1}, $$

where $E_{1}\subset(1,\infty)$ is a set of finite logarithmic measure. Thus,

$$\begin{aligned} \sum_{j=1}^{s}\delta(\alpha_{j},f)&= \sum_{j=1}^{s}\mathop{\underline{\lim}}_{r\rightarrow\infty} \frac{m (r,\frac{1}{f-\alpha_{j}} )}{T(r,f)} \leq \mathop{\underline{\lim}}_{r\rightarrow\infty}\frac{\sum_{j=1}^{s}m (r,\frac{1}{f-\alpha_{j}} )}{T(r,f)} \\ &\leq \mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E_{1}}\frac{\sum _{j=1}^{s}m (r,\frac{1}{f-\alpha_{j}} )}{T(r,f)} \leq \mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E_{1}}\frac{ (1-\frac{d}{2k}+o(1) )T(r,f)}{T(r,f)} \\ &=1-\frac{d}{2k}. \end{aligned}$$

So,

$$\sum_{j=1}^{s}\delta(\alpha_{j},f) \leq1-\frac{d}{2k}. $$

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Proof of Theorem 1.2

(i) Using the same method as in Theorem 1.1(i), we still have (3.1)-(3.5). By the fact $d=7k$ and (3.5) we have

$$\begin{aligned} 3kN(r,f)+2kN \biggl(r,\frac{1}{\Delta f(z)} \biggr) &=3kT(r,f)+2kT\bigl(r,\Delta f(z)\bigr)+S(r,f) \\ &=7k T(r,f)+S(r,f). \end{aligned}$$

From this it follows that

$$ N(r,f)=T(r,f)+S(r,f) $$

(3.11)

and

$$ N \biggl(r,\frac{1}{\Delta f(z)} \biggr)=T\bigl(r,\Delta f(z) \bigr)+S(r,f)=2T(r,f)+S(r,f). $$

(3.12)

By (3.11) we see that

$$ m(r,f)=T(r,f)-N(r,f)=S(r,f), $$

(3.13)

Hence,

$$m(r,f)=o\bigl(T(r,f)\bigr), \quad r\notin E, $$

where $E\subset(1,\infty)$ is a set of finite logarithmic measure. Therefore,

$$\delta(\infty, f)=\mathop{\underline{\lim}}_{r\rightarrow\infty}\frac {m(r,f)}{T(r,f)} \leq \mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E}\frac{m(r,f)}{T(r,f)} =\mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E} \frac{o(T(r,f))}{T(r,f)} =0. $$

So, $\delta(\infty, f)=0$.

We deduce from (3.12) that

$$ m \biggl(r,\frac{1}{\Delta f(z)} \biggr)=T\bigl(r,\Delta f(z)\bigr)-N \biggl(r,\frac {1}{\Delta f(z)} \biggr)=S(r,f). $$

(3.14)

For any given $a\in \mathbb {C}$, by (3.14) and Lemma 2.1 we have

$$ m \biggl(r,\frac{1}{f(z)-a} \biggr)\leq m \biggl(r, \frac{\Delta f(z)}{f(z)-a} \biggr)+m \biggl(r,\frac{1}{\Delta f(z)} \biggr)=S(r,f), $$

that is,

$$m \biggl(r,\frac{1}{f(z)-a} \biggr)=o\bigl(T(r,f)\bigr), \quad r\notin E_{1}, $$

where $E_{1}\subset(1,\infty)$ is a set of finite logarithmic measure. Hence,

$$\delta(a, f)=\mathop{\underline{\lim}}_{r\rightarrow\infty}\frac{ m (r,\frac {1}{f(z)-a} )}{T(r,f)} \leq \mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E_{1}}\frac{ m (r,\frac{1}{f(z)-a} )}{T(r,f)} =\mathop{\underline{\lim}}_{r\rightarrow\infty\atop r\notin E_{1}} \frac{o(T(r,f))}{T(r,f)} =0. $$

So, $\delta(a,f)=0$.

Thus, for any $a\in \mathbb {C}\cup\{\infty\}$, we have $\delta(a,f)=0$. So, $f(z)$ has no Nevanlinna exceptional value.

(ii) By (3.12) we easily obtain that $\lambda(\Delta f)=\sigma (\Delta f)=\sigma(f)$. By (3.13) and Lemma 2.1 we see

$$m(r,\Delta f)\leq m \biggl(r,\frac{\Delta f}{f} \biggr)+m(r,f)=S(r,f). $$

Together with (3.12), we have

$$N(r,\Delta f)=T(r, \Delta f)-m(r,\Delta f)=T(r,\Delta f)+S(r,f)=2T(r,f)+S(r,f) $$

or

$$N(r,\Delta f)=\bigl(2+o(1)\bigr)T(r,f), \quad r\notin E_{2}, $$

where $E_{2}\subset(1,\infty)$ is a set of finite logarithmic measure.

By Lemmas 2.7 and 2.8 we see

$$\begin{aligned} \lambda\biggl(\frac{1}{\Delta f} \biggr)&=\mathop{\overline{\lim}}_{r\rightarrow \infty} \frac{\log N(r,\Delta f)}{\log r} =\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{2}}\frac{\log N(r,\Delta f)}{\log r} \\ &=\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{2}}\frac{\log (2+o(1))T(r,f)}{\log r} =\mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r}=\sigma(f). \end{aligned}$$

So, $\lambda(\frac{1}{\Delta f} )=\sigma(f)$.

Finally, we prove that $\lambda(\frac{\Delta f}{f} )=\lambda (\frac{1}{\frac{\Delta f}{f}} )=\sigma(f)$. Set $g(z)=\frac{\Delta f(z)}{f(z)}$. Then

$$ \left \{ \textstyle\begin{array}{@{}l} f(z+1)=(g(z)+1)f(z),\\ f(z+2)=(g(z+1)+1)f(z+1)=(g(z+1)+1)(g(z)+1)f(z),\\ f(z+3)=(g(z+2)+1)f(z+2)=(g(z+2)+1)(g(z+1)+1)(g(z)+1)f(z). \end{array}\displaystyle \right . $$

Substituting this into (3.2), we have

$$\begin{aligned} R(z,f)= \biggl[\frac{(g(z)+1) ((g(z+2)+1)(g(z+1)+1)-1 )}{g(z)}-\frac{3}{2} \biggl(\frac{g(z+1)(g(z)+1)}{g(z)} \biggr)^{2} \biggr]^{k}. \end{aligned}$$

Combining this with (3.1) and Lemma 2.3, we obtain

$$\begin{aligned} dT(r,f)&=T\bigl(r,R(z,f)\bigr)+S(r,f) \\ &\leq M\bigl[T\bigl(r,g(z)\bigr)+T\bigl(r,g(z+1)\bigr)+T\bigl(r,g(z+2)\bigr) \bigr]+S(r,f) \\ &= 3MT\bigl(r,g(z)\bigr)+S(r,g)+S(r,f), \end{aligned}$$

(3.15)

where M is some nonzero constant.

By the definition of $S(r,f)$ and $S(r,g)$, (3.15) can be rewritten as

$$\bigl(d+o(1)\bigr)T(r,f)\leq\bigl(3M+o(1)\bigr)T(r, g), \quad r\notin E_{3}, $$

where $E_{3}\subset(1,\infty)$ is a set of finite logarithmic measure.

Applying Lemmas 2.7 and 2.8 to the last inequality, we have

$$\begin{aligned} \sigma(f)&=\mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r} =\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{3}}\frac{\log T(r,f)}{\log r} \\ &\leq \mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{3}}\frac{\log (\frac{3M+o(1)}{d+o(1)} )+\log T(r,g)}{\log r} \\ &=\mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log T(r,g)}{\log r} =\sigma(g), \end{aligned}$$

that is, $\sigma(f)\leq\sigma(g)$. So, $\sigma(\frac{\Delta f}{f} )=\sigma(g)=\sigma(f)$.

If the quantity $\phi(r)$ satisfies $\phi(r)=S(r,f)$, then we have $\phi(r)=S(r,g)$ by (3.15). Hence,

$$ S(r,f)=o\bigl(T(r,f)\bigr)\leq o\bigl(T(r,g)\bigr)=S(r,g)=S \biggl(r, \frac{\Delta f}{f} \biggr). $$

(3.16)

Lemma 2.1 and (3.16) show that

$$m \biggl(r,\frac{\Delta f}{f} \biggr)=S(r,f)\leq S \biggl(r, \frac {\Delta f}{f} \biggr), $$

from which it follows that

$$\begin{aligned} N \biggl(r,\frac{\Delta f}{f} \biggr) =T \biggl(r,\frac{\Delta f}{f} \biggr)-m \biggl(r,\frac{\Delta f}{f} \biggr) =T \biggl(r,\frac{\Delta f}{f} \biggr)+S \biggl(r, \frac{\Delta f}{f} \biggr). \end{aligned}$$

By the definition of $S (r, \frac{\Delta f}{f} )$, the last equality shows that

$$ N \biggl(r,\frac{\Delta f}{f} \biggr)=\bigl(1+o(1)\bigr)T \biggl(r,\frac{\Delta f}{f} \biggr), \quad r\notin E_{4}, $$

(3.17)

where $E_{4}\subset(1,\infty)$ is a set of finite logarithmic measure.

Applying Lemmas 2.7 and 2.8 to (3.17), we see that

$$\begin{aligned} \lambda\biggl(\frac{1}{\frac{\Delta f}{f}} \biggr)&=\mathop{\overline{\lim}}_{r\rightarrow\infty} \frac{\log N (r,\frac{\Delta f}{f} )}{\log r} =\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{4}}\frac {\log N (r,\frac{\Delta f}{f} )}{\log r} \\ &=\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{4}}\frac{\log (1+o(1))T (r,\frac{\Delta f}{f} )}{\log r} \\ &=\mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log T (r,\frac{\Delta f}{f} )}{\log r}=\sigma\biggl( \frac{\Delta f}{f} \biggr). \end{aligned}$$

So, $\lambda(\frac{1}{\frac{\Delta f}{f}} )=\sigma(\frac {\Delta f}{f} )=\sigma(f)$.

By (3.13), (3.14), and (3.16) we see that

$$m \biggl(r,\frac{1}{\frac{\Delta f}{f}} \biggr)=m \biggl(r,\frac {f}{\Delta f} \biggr)\leq m(r,f)+m \biggl(r,\frac{1}{\Delta f} \biggr)=S(r,f)\leq S \biggl(r, \frac{\Delta f}{f} \biggr). $$

Thus,

$$N \biggl(r,\frac{1}{\frac{\Delta f}{f}} \biggr)=T \biggl(r,\frac {\Delta f}{f} \biggr)-m \biggl(r,\frac{1}{\frac{\Delta f}{f}} \biggr)=T \biggl(r,\frac {\Delta f}{f} \biggr)+S \biggl(r,\frac{\Delta f}{f} \biggr) $$

or

$$ N \biggl(r,\frac{1}{\frac{\Delta f}{f}} \biggr)=\bigl(1+o(1)\bigr)T \biggl(r,\frac {\Delta f}{f} \biggr), \quad r\notin E_{5}, $$

(3.18)

where $E_{5}\subset(1,\infty)$ is a set of finite logarithmic measure.

Applying Lemmas 2.7 and 2.8 to (3.18), we obtain

$$\begin{aligned} \lambda\biggl(\frac{\Delta f}{f} \biggr)&=\mathop{\overline{\lim}}_{r\rightarrow \infty} \frac{\log N (r,\frac{1}{\frac{\Delta f}{f}} )}{\log r} =\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{4}}\frac{\log N (r,\frac{1}{\frac{\Delta f}{f}} )}{\log r} \\ &=\mathop{\overline{\lim}}_{r\rightarrow\infty\atop r\notin E_{4}}\frac{\log (1+o(1))T (r,\frac{\Delta f}{f} )}{\log r} \\ &=\mathop{\overline{\lim}}_{r\rightarrow\infty}\frac{\log T (r,\frac{\Delta f}{f} )}{\log r}=\sigma\biggl( \frac{\Delta f}{f} \biggr). \end{aligned}$$

So, $\lambda(\frac{\Delta f}{f} )=\sigma(\frac{\Delta f}{f} )=\sigma(f)$.

(iii) From (3.12) we see that

$$T(r,\Delta f)=2T(r,f)+S(r,f). $$

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