Proof
Now we prove (1.1). Suppose that \(\lambda (f(z)-R(z))<\rho (f)\). Next we will prove \(\lambda (\Delta ^{n} _{c}f(z)-R(z))=\rho (f)\).
Since \(\lambda (f(z)-a)<\rho (f)\), and R is a nonconstant rational function, we have
$$\begin{aligned} \frac{f(z)-R(z)}{f(z)-a}=\alpha (z)e^{p(z)}, \end{aligned}$$
(3.1)
where \(\alpha\ (\not \equiv 0, \infty )\) is a meromorphic function such that \(\rho (\alpha )<\rho (f)\), and p is a nonconstant polynomial with \(\deg p=\rho (f)\).
Hence we have
$$\begin{aligned} T(r, \alpha )=S\bigl(r, e^{p}\bigr),\qquad T(r, f)=T\bigl(r, e^{p}\bigr)+S(r, f). \end{aligned}$$
(3.2)
By (3.1), we obtain
$$\begin{aligned} f(z)=a+\frac{a-R(z)}{\alpha (z)e^{p(z)}-1}. \end{aligned}$$
(3.3)
Thus we have
$$\begin{aligned} \Delta ^{n} _{c}f(z)&= \Delta ^{n} _{c} \bigl(f(z)-a\bigr) \\ &=\sum_{i=0}^{n}(-1)^{i}C_{n}^{i} \bigl(f\bigl(z+(n-i)c\bigr)-a\bigr) \\ &=\sum_{i=0}^{n}(-1)^{i}C_{n}^{i} \frac{a-R(z+(n-i)c)}{\alpha (z+(n-i)c)e^{p(z+(n-i)c)}-1} \end{aligned}$$
(3.4)
$$\begin{aligned} &=\frac{\sum_{i=0}^{n}(-1)^{i}C_{n}^{i}(a-R(z+(n-i)c))\Pi _{j\neq i}^{n}(\alpha (z+(n-j)c)e^{p(z+(n-j)c)}-1)}{\Pi _{i=0}^{n}(\alpha (z+(n-i)c)e^{p(z+(n-i)c)}-1)} \\ &=\frac{B_{m}(z)(e^{p(z)})^{m}+B_{m-1}(z)(e^{p(z)})^{m-1}+ \cdots +B_{1}(z)e^{p(z)}+B_{0}(z)}{A_{n+1}(z)(e^{p(z)})^{n+1}+A_{n}(z) (e^{p(z)})^{n}+\cdots +A_{1}(z)e^{p(z)}+(-1)^{n+1}}, \end{aligned}$$
(3.5)
where \(A_{1}(z), A_{2}(z), \ldots , A_{n+1}(z), B_{0}(z), B_{1}(z), \ldots , B_{m}(z)\) are small functions of \(e^{p(z)}\), and \(A_{n+1}(z)=\alpha (z)\alpha (z+c)e^{p(z+c)-p(z)}\cdots \alpha (z+nc)e^{p(z+nc)-p(z)} \not \equiv 0\), \(B_{m}(z)\not \equiv 0\), \(B_{0}(z)=(-1)^{n+1}\times \Delta _{c}^{n}R(z), m\le n\).
It follows from (3.4) that \(\Delta ^{n} _{c}f \not \equiv 0\). Suppose that \(\Delta ^{n} _{c}f \equiv 0\). Next we consider two cases.
Case 1. \(e^{p(z+nc)-p(z+(n-i)c)}-\frac{\alpha (z+(n-i)c)}{\alpha (z+nc)}\not \equiv 0, i=1, 2, \ldots, n\). Then, by (3.4) and Nevanlinna’s second fundamental theorem [1, 17, 18], we obtain
$$\begin{aligned} T\bigl(r, e^{p}\bigr)={}&T\biggl(r, \frac{\alpha (z+nc)e^{p(z+nc)}}{\alpha (z+nc)e^{p(z+nc)-p(z)}}\biggr) \\ \le{} &T\bigl(r, \alpha (z+nc)e^{p(z+nc)}\bigr)+S\bigl(r, e^{p} \bigr) \le N\bigl(r, \alpha (z+nc)e^{p(z+nc)}\bigr) \\ &{}+N\biggl(r, \frac{1}{\alpha (z+nc)e^{p(z+nc)}-1} \biggr)+N\biggl(r, \frac{1}{\alpha (z+nc)e^{p(z+nc)}}\biggr)+S \bigl(r, e^{p}\bigr) \\ \le {}&N\biggl(r, \frac{1}{\alpha (z+nc)e^{p(z+nc)}-1} \biggr)+S\bigl(r, e^{p}\bigr) \\ ={}&N\biggl(r, \frac{1}{\alpha (z+nc)e^{p(z+nc)}-1}, \Delta _{c}^{n} f=0 \biggr)+S\bigl(r, e^{p}\bigr) \\ \le {}&\sum_{i=1}^{n}N\biggl(r, \frac{1}{(\alpha (z+nc)e^{p(z+nc)}-1)-(\alpha (z+(n-i)c)e^{p(z+(n-i)c)}-1)}\biggr)+S\bigl(r, e^{p}\bigr) \\ ={}&\sum_{i=1}^{n}N\biggl(r, \frac{1}{\alpha (z+nc)e^{p(z+nc)}-\alpha (z+(n-i)c)e^{p(z+(n-i)c)}}\biggr)+S\bigl(r, e^{p}\bigr) \\ \le{} &\sum_{i=1}^{n}N\biggl(r, \frac{1}{e^{p(z+nc)-p(z+(n-i)c)}-\frac{\alpha (z+(n-i)c)}{\alpha (z+nc)}}\biggr)+S\bigl(r, e^{p}\bigr) \\ \le {}&\sum_{i=1}^{n}\biggl[T\bigl(r, e^{p(z+nc)-p(z+(n-i)c)}\bigr)+T\biggl(r, \frac{\alpha (z+(n-i)c)}{\alpha (z+nc)}\biggr)\biggr]+S\bigl(r, e^{p}\bigr) \\ ={}&S\bigl(r, e^{p}\bigr), \end{aligned}$$
(3.6)
which is a contradiction.
Case 2. \(e^{p(z+nc)-p(z+(n-i)c)}-\frac{\alpha (z+(n-i)c)}{\alpha (z+nc)} \equiv 0 \), for some \(i\in \{1, 2, \ldots, n\}\). If \(\deg p\ge 2\), then by \(\rho (\alpha )<\rho (f)=\deg p\) and Lemma 1 we get
$$ \rho \biggl(\frac{\alpha (z+(n-i)c)}{\alpha (z+nc)}\biggr)\le \rho (\alpha ) -1+ \frac{\rho (f)-\rho (\alpha )}{2}< \rho (f)-1=\rho \bigl(e^{p(z+nc)-p(z+(n-i)c)}\bigr), $$
which is a contradiction. Hence \(\deg p=1\). So, \(\rho (\alpha )<1\). Let \(p(z)=Az+B\), then we have \(\alpha (z+(n-i)c)=e^{iAc}\alpha (z+nc)\). Without loss of generality, we assume that \(i=1\), then we obtain
$$ \alpha (z)=e^{Ac}\alpha (z+c). $$
We claim that \(\alpha \neq \infty \). Suppose that there exists \(z_{0}\) such that \(\alpha (z_{0})=\infty \), without loss of generality, let \(z_{0}=0\), then we deduce that for all positive integers j, \(\alpha (jc)=\infty \). Thus for sufficiently large r, and \(2n|c|\le r<(2n+1)|c|\), we have
$$\begin{aligned} T(r, \alpha )\ge{}& N(r, \alpha )= \int _{0}^{r} \frac{n(t, \alpha )-n(0, \alpha )}{t}\,dt+n(0, \alpha ) \log r \\ \ge {}& \sum_{j=1}^{2n-1}j \int _{j \vert c \vert }^{(j+1) \vert c \vert }\frac{dt}{t}=\sum _{j=1}^{2n-1}j \log \biggl(1+\frac{1}{j}\biggr) \\ \ge {}& \sum_{j=1}^{2n-1}j\log \biggl(1+ \frac{1}{2n-1}\biggr)=n\log \biggl(1+ \frac{1}{2n-1}\biggr)^{2n-1} \\ \ge {}& n\log 2>\frac{\log 2}{4 \vert c \vert }r. \end{aligned}$$
It follows that \(\rho (\alpha )\ge 1 \), which is a contradiction. Similarly, we obtain \(\alpha \neq 0\). Thus we deduce that α is a nonzero constant C. By (3.1), we obtain
$$\begin{aligned} f(z)=a+\frac{a-R(z)}{Ce^{Az+B}-1}=a+\frac{a-R(z)}{De^{Az}-1}, \quad D=Ce^{B} \neq0. \end{aligned}$$
(3.7)
If \(R=q/p\) be a nonconstant rational function, where \(p, q\) are two polynomials with \(\deg p\ge 1\), or R is a polynomial with \(\deg R\ge n\), then, by Lemma 6, (3.5) and (3.7), we obtain \(\Delta _{c}^{n}f\not \equiv 0\). If R is a polynomial with \(1\le \deg R< n\), and \(e^{Ac}\neq 1\), then, by (3.7) and some computation, we deduce that \(\Delta _{c}^{n}f\not \equiv 0\).
In fact, by (3.4) and (3.7) we have
$$\begin{aligned} \Delta ^{n} _{c}f(z)={}&\sum_{i=0}^{n}(-1)^{i}C_{n}^{i} \frac{a-R(z+(n-i)c)}{De^{A(z+(n-i)c)}-1} \\ ={}&\frac{\sum_{i=0}^{n}(-1)^{i}C_{n}^{i}(a-R(z+(n-i)c))\Pi _{j\neq i}^{n}(De^{A(z+(n-j)c)}-1)}{\Pi _{i=0}^{n}(De^{p(z+(n-i)c)}-1)} \\ ={}&\frac{[D^{n}(e^{Ac})^{1+2+\cdots +(n-1)}\sum_{i=0}^{n}(-1)^{i}C_{n}^{i}(a-R(z+(n-i)c))(e^{Ac})^{i}](e^{Az})^{n}+\cdots }{A_{n+1}(z)(e^{Az})^{n+1}+A_{n}(z)(e^{Az})^{n} +\cdots +A_{1}(z)e^{Az}+(-1)^{n+1}} \\ ={}&\frac{[-D^{n}(e^{Ac})^{1+2+\cdots +(n-1)}a_{s}\sum_{i=0}^{n}(-1)^{i}C_{n}^{i}(e^{Ac})^{i}z^{s}+\cdots ](e^{Az})^{n}+\cdots }{A_{n+1}(z)(e^{Az})^{n+1}+A_{n}(z)(e^{Az})^{n} +\cdots +A_{1}(z)e^{Az}+(-1)^{n+1}} \\ ={}&\frac{[-D^{n}(e^{Ac})^{1+2+\cdots +(n-1)}a_{s}(e^{Ac}-1)^{n}z^{s}+\cdots ](e^{Az})^{n}+\cdots }{A_{n+1}(z)(e^{Az})^{n+1}+A_{n}(z)(e^{Az})^{n}+\cdots +A_{1}(z)e^{Az}+(-1)^{n+1}}, \end{aligned}$$
where \(R(z)=a_{s}z^{s}+a_{s-1}z^{s-1}+\cdots +a_{1}z+a_{0}\), \(1\le s=\deg R< n\). It follows from \(a_{s}\neq 0, D\neq 0, e^{Ac}-1\neq 0\) that \(\Delta _{c}^{n}f\not \equiv 0\).
Hence by the above discussion we prove that \(\Delta _{c}^{n}f\not \equiv 0\). Set
$$\begin{aligned} &B\bigl(e^{p}\bigr)=B_{m}\bigl(e^{p} \bigr)^{m}+B_{m-1}\bigl(e^{p}\bigr)^{m-1}+ \cdots +B_{1}e^{p}+B_{0}, \\ &A\bigl(e^{p}\bigr)=A_{n+1}\bigl(e^{p} \bigr)^{n+1}+A_{n}\bigl(e^{p}\bigr)^{n}+ \cdots +A_{1}e^{p}+(-1)^{n+1}. \end{aligned}$$
By the method of successive division, there exist \(D(e^{p})\), \(E(e^{p})\), \(F(e^{p})\), \(G(e^{p})\), \(H(e^{p})\) satisfying
$$\begin{aligned} &A\bigl(e^{p}\bigr)=D\bigl(e^{p}\bigr)E \bigl(e^{p}\bigr),\qquad B\bigl(e^{p}\bigr)=D\bigl(e^{p} \bigr)F\bigl(e^{p}\bigr), \end{aligned}$$
(3.8)
$$\begin{aligned} &E\bigl(e^{p}\bigr)G\bigl(e^{p}\bigr)+F \bigl(e^{p}\bigr)H\bigl(e^{p}\bigr)\equiv 1, \end{aligned}$$
(3.9)
where
$$\begin{aligned} &D\bigl(e^{p}\bigr)=D_{l}\bigl(e^{p} \bigr)^{l}+D_{l-1}\bigl(e^{p}\bigr)^{l-1}+ \cdots +D_{1}e^{p}+D_{0}, \\ &E\bigl(e^{p}\bigr)=E_{n+1-l}\bigl(e^{p} \bigr)^{n+1-l}+E_{n-l}\bigl(e^{p}\bigr)^{n-l}+ \cdots +E_{1}e^{p}+E_{0}, \\ &F\bigl(e^{p}\bigr)=F_{m-l}\bigl(e^{p} \bigr)^{m-l}+E_{m-1-l}\bigl(e^{p}\bigr)^{m-1-l}+ \cdots +F_{1}e^{p}+F_{0}, \end{aligned}$$
and \(D_{0}, D_{1}, \ldots,D_{l}\), \(E_{0}, E_{1}, \ldots,E_{n+1-l}\), \(F_{0}, F_{1}, \ldots,F_{m-l}\) are small functions of \(e^{p}\), and \(D_{l}\not \equiv 0\), \(E_{n+1-l}\not \equiv 0\), \(F_{m-l}\not \equiv 0\), \(E_{0}D_{0}=(-1)^{n+1}, 0\le l\le m\).
So by (3.5) and (3.8), we have
$$\begin{aligned} \Delta ^{n} _{c}f(z)-R(z)={}&\frac{B(e^{p})}{A(e^{p})}-R(z)= \frac{B(e^{p})-R(z)A(e^{p})}{A(e^{p})} \\ ={}&\frac{F(e^{p})D(e^{p})-R(z)E(e^{p})D(e^{p})}{E(e^{p})D(e^{p})}, \\ ={}&\frac{F(e^{p})-R(z)E(e^{p})}{E(e^{p})}. \end{aligned}$$
(3.10)
It follows that
$$ \bigl[F_{0}-R(z)E_{0}\bigr]D_{0}=B_{0}-R(z)A_{0}=(-1)^{n+1} \bigl[\Delta _{c}^{n}R(z)-R(z)\bigr]. $$
By Lemma 5, we know that \(\Delta _{c}^{n}R-R\not \equiv 0\). Thus we deduce that \(F_{0}-R(z)E_{0} \not \equiv 0\). By this and (3.9)–(3.10), we obtain
$$\begin{aligned} N \biggl(r, \frac{1}{\Delta _{c}^{n}f(z)-R(z)} \biggr)=N \biggl(r, \frac{1}{F(e^{P})-R(z)E(e^{p})} \biggr)+S \bigl(r, e^{p}\bigr). \end{aligned}$$
(3.11)
By (3.10) and Lemma 2, we obtain
$$\begin{aligned} T\bigl(r, \Delta _{c}^{n}f(z)\bigr)=(n+1-l)T\bigl(r, e^{p}\bigr)+S(r, f). \end{aligned}$$
(3.12)
Let \(F(e^{p})-R(z)E(e^{p})=R_{n+1-l}(e^{p})^{n+1-l}+\cdots +R_{1}(z)e^{p}+R_{0}(z)\), then \(R_{n+1-l}=-R(z)E_{n+1-l}(z)(\not \equiv 0)\), \(R_{n-l}(z), \ldots , R_{1}(z), R_{0}(z)=F_{0}-R(z)E_{0}\ (\not \equiv 0)\) are small functions of \(e^{p}\). Thus by Lemma 2, Lemma 4 and Nevanlinna’s second fundamental theorem, we obtain
$$\begin{aligned} &(n+1-l)T\bigl(r, e^{p}\bigr) \\ &\quad =T\bigl(r, R_{n+1-l} \bigl(e^{p}\bigr)^{n+1-l}+\cdots +R_{1}e^{p} \bigr) \\ &\quad\le\overline{N}\biggl(r, \frac{1}{R_{n+1-l}(e^{p})^{n+1-l}+\cdots +R_{1}e^{p}}\biggr)+\overline{N}\bigl(r, R_{n+1-l}\bigl(e^{p}\bigr)^{n+1-l}+ \cdots +R_{1}e^{p}\bigr) \\ &\qquad{}+\overline{N}\biggl(r, \frac{1}{R_{n+1-l}(e^{p})^{n+1-l}+\cdots +R_{1}e^{p}+R_{0}}\biggr)+S\bigl(r, e^{p} \bigr) \\ &\quad\le(n-l)T\bigl(r, e^{p}\bigr)+\overline{N}\biggl(r, \frac{1}{R_{n+1-l}(e^{p})^{n+1-l}+\cdots +R_{1}e^{p}+R_{0}}\biggr)+S\bigl(r, e^{p}\bigr). \end{aligned}$$
It follows that
$$ T\bigl(r, e^{p}\bigr)\le \overline{N}\biggl(r, \frac{1}{R_{n+1-l}(e^{p})^{n+1-l}+\cdots +R_{1}e^{p}+R_{0}} \biggr)+S\bigl(r, e^{p}\bigr). $$
From this together with (3.11) we deduce that
$$\begin{aligned} N \biggl(r, \frac{1}{\Delta _{c}^{n}f(z)-R(z)} \biggr)\ge T\bigl(r, e^{p}\bigr)+S \bigl(r, e^{p}\bigr). \end{aligned}$$
(3.13)
By (3.2), (3.12) and (3.13), we deduce that \(\lambda (\Delta ^{n} _{c}f(z)-R(z))=\rho (f)\). Thus we prove (1.1).
Now we prove (1.2). Suppose that
$$ \max \bigl\{ \lambda \bigl(f(z)-R(z)\bigr), \lambda \bigl(f(z+nc)-R(z)\bigr)\bigr\} < \rho (f). $$
Then, by Lemma 3, we have \(\lambda (f(z)-R(z-nc))<\rho (f)\). Hence there exist \(\varepsilon _{0}>0, T>0\), for \(r>T\), we have
$$\begin{aligned} &N \biggl(r, \frac{1}{f-a} \biggr)\le r^{\rho (f)-\varepsilon _{0}},\qquad N \biggl(r, \frac{1}{f-R} \biggr)\le r^{\rho (f)-\varepsilon _{0}}, \\ &N \biggl(r, \frac{1}{f-R(z-nc)} \biggr)\le r^{\rho (f)-\varepsilon _{0}}. \end{aligned}$$
By the above formulas and Lemma 4, we obtain
$$\begin{aligned} T(r, f)&\leq \overline{N} \biggl(r, \frac{1}{f-a} \biggr)+\overline{N} \biggl(r, \frac{1}{f-R} \biggr) +\overline{N} \biggl(r, \frac{1}{f-R(z-nc)} \biggr)+S(r,f) \\ &\le 3r^{\rho (f)-\varepsilon _{0}}+O(\log r). \end{aligned}$$
Thus we deduce that \(\rho (f) \le \rho (f)-\varepsilon _{0}\), a contradiction. Hence we prove (1.2).
Next we prove (1.3). Suppose that \(\lambda (f(z+nc)-R(z))<\rho (f)\). Then, by Lemma 3, we deduce that \(\lambda (f(z)-R(z-nc))<\rho (f)\). In the following, using the same methods as used in the proof of (1.1), we obtain \(\lambda (\Delta ^{n} _{c}f(z)-R(z))=\rho (f)\). Thus we prove (1.3). Therefore, Theorem 1 is proved. □