We first introduce some new notation. Let \(z_{0}\) be a zero of \(f-1\) with multiplicity p and a zero of \(g-1\) with multiplicity q. We denote by \(N_{E}^{1)}(r,\frac{1}{f-1})\) the counting function of the zeros of \(f-1\) with \(p=q=1\), by \(\overline{N}_{E}^{(2}(r,\frac{1}{f-1})\) the counting function of the zeros of \(f-1\) satisfying \(p=q\geq2\), and by \(\overline{N}_{L}(r,\frac{1}{f-1})\) the counting function of the zeros of \(f-1\) with \(p>q\geq1\), where each point in these counting functions is counted only once.
We set
$$ H(z)= \biggl(\frac{f''}{f'}-2\frac{f'}{f-1} \biggr)- \biggl( \frac{g''}{g'}-2\frac {g'}{g-1} \biggr). $$
(2.2)
Suppose that
\(H(z)\not\equiv0\). Clearly,
\(m(r,H)=S(r,f)+S(r,g)\). If
\(z_{0}\) is a common simple zero of
\(f-1\) and
\(g-1\), then a simple computation on local expansions shows that
\(H(z_{0})=0\), and then
$$ N_{E}^{1)}\biggl(r,\frac{1}{f-1}\biggr)\leq N\biggl(r, \frac{1}{H}\biggr)\leq N(r,H)+S(r,f)+S(r,g). $$
(2.3)
The poles of
\(H(z)\) only come from the zeros of
\(f'\) and
\(g'\), the multiple poles of
f and
g, and the zeros of
\(f-1\) and
\(g-1\) with different multiplicity. By analysis we can deduce that
$$\begin{aligned} N(r,H) \leq&\overline{N}_{(2}(r,f)+\overline{N}_{(2}(r,g)+ \overline {N}_{(2}\biggl(r,\frac{1}{f}\biggr)+ \overline{N}_{(2}\biggl(r,\frac{1}{g}\biggr) \\ &{}+\overline {N}_{L}\biggl(r,\frac{1}{f-1}\biggr)+\overline{N}_{L} \biggl(r,\frac{1}{g-1}\biggr) \\ &{}+\overline{N}_{0}\biggl(r,\frac{1}{f'}\biggr)+ \overline{N}_{0}\biggl(r,\frac {1}{g'}\biggr)+S(r,f)+S(r,g), \end{aligned}$$
(2.4)
where
\(N_{0}(r,\frac{1}{f'})\) denotes the counting function of the zeros of
\(f'\) but not that of
\(f(f-1)\),
\(\overline{N}_{0}(r,\frac{1}{f'})\) denotes the corresponding reduced counting function, and
\(N_{0}(r,\frac{1}{g'})\) and
\(\overline{N}_{0}(r,\frac{1}{g'})\) are defined similarly. At the same time, obviously,
$$\overline{N}\biggl(r,\frac{1}{f-1}\biggr)=N_{E}^{1)} \biggl(r,\frac{1}{f-1}\biggr)+\overline {N}_{E}^{(2} \biggl(r,\frac{1}{f-1}\biggr)+\overline{N}_{L}\biggl(r, \frac{1}{f-1}\biggr)+\overline {N}_{L}\biggl(r,\frac{1}{g-1} \biggr). $$
Combining this with (
2.3) and (
2.4) yields
$$\begin{aligned} \overline{N}\biggl(r,\frac{1}{f-1}\biggr) \leq&\overline{N}_{(2}(r,f)+ \overline {N}_{(2}(r,g)+\overline{N}_{(2}\biggl(r, \frac{1}{f}\biggr)+\overline{N}_{(2}\biggl(r,\frac {1}{g} \biggr) \\ &{}+2\overline{N}_{L}\biggl(r,\frac{1}{f-1}\biggr)+2 \overline{N}_{L}\biggl(r,\frac{1}{g-1}\biggr)+\overline{N}_{0}\biggl(r,\frac{1}{f'}\biggr) \\ &{}+ \overline{N}_{0}\biggl(r,\frac{1}{g'}\biggr)+\overline {N}_{E}^{(2}\biggl(r,\frac{1}{f-1}\biggr)+ S(r,f)+S(r,g). \end{aligned}$$
(2.5)
Since
$$\overline{N}\biggl(r,\frac{1}{g-1}\biggr)+\overline{N}_{L} \biggl(r,\frac{1}{g-1}\biggr)+\overline {N}_{E}^{(2} \biggl(r,\frac{1}{f-1}\biggr)\leq N\biggl(r,\frac{1}{g-1}\biggr)\leq T(r,g)+S(r,g), $$
combining this with (
2.5), we have
$$\begin{aligned}& \overline{N}\biggl(r,\frac{1}{f-1}\biggr)+\overline{N}\biggl(r, \frac{1}{g-1}\biggr) \\& \quad \leq \overline{N}_{(2}(r,f)+ \overline{N}_{(2}(r,g)+\overline{N}_{(2}\biggl(r, \frac {1}{f}\biggr)+\overline{N}_{(2}\biggl(r,\frac{1}{g} \biggr)+2\overline{N}_{L}\biggl(r,\frac{1}{f-1}\biggr) \\& \qquad {}+\overline{N}_{L}\biggl(r,\frac{1}{g-1}\biggr)+ \overline{N}_{0}\biggl(r,\frac {1}{f'}\biggr)+ \overline{N}_{0}\biggl(r,\frac{1}{g'}\biggr)+T(r,g)+ S(r,f)+S(r,g). \end{aligned}$$
We apply the second fundamental theorem to
f and
g and consider the above inequality. Then
$$\begin{aligned} T(r,f)+T(r,g) \leq& \overline{N}(r,f)+\overline{N}(r,g)+\overline{N}\biggl(r, \frac {1}{f}\biggr)+\overline{N}\biggl(r,\frac{1}{g}\biggr)+ \overline{N}\biggl(r,\frac {1}{f-1}\biggr) \\ &{}+\overline{N}\biggl(r, \frac{1}{g-1}\biggr) -N_{0}\biggl(r,\frac{1}{f'}\biggr)-N_{0} \biggl(r,\frac{1}{g'}\biggr)+S(r,f)+S(r,g) \\ \leq& N_{2}(r,f)+N_{2}(r,g)+N_{2}\biggl(r, \frac{1}{f}\biggr)+N_{2}\biggl(r,\frac{1}{g}\biggr)+2 \overline {N}_{L}\biggl(r,\frac{1}{f-1}\biggr) \\ &{} +T(r,g)+\overline{N}_{L}\biggl(r,\frac{1}{g-1} \biggr)+S(r,f)+S(r,g). \end{aligned}$$
Clearly, this leads to
$$\begin{aligned} T(r,f) \leq& N_{2}(r,f)+N_{2}(r,g)+N_{2}\biggl(r, \frac{1}{f}\biggr)+N_{2}\biggl(r,\frac {1}{g}\biggr)+2 \overline{N}_{L}\biggl(r,\frac{1}{f-1}\biggr) \\ &{}+\overline{N}_{L}\biggl(r,\frac{1}{g-1}\biggr)+S(r,f)+S(r,g). \end{aligned}$$
(2.6)
By Lemma
2.1 we have
$$\overline{N}_{L}\biggl(r,\frac{1}{f-1}\biggr)+ \overline{N}_{(2}\biggl(r,\frac{1}{f}\biggr)\leq \overline{N} \biggl(r,\frac{1}{f'}\biggr)\leq N_{2}\biggl(r,\frac{1}{f} \biggr)+\overline{N}(r,f)+S(r,f). $$
Then, using this inequality, we get
$$\begin{aligned} 2\overline{N}_{L}\biggl(r,\frac{1}{f-1}\biggr)+N_{2} \biggl(r,\frac{1}{f}\biggr) \leq&2N_{2}\biggl(r, \frac {1}{f}\biggr)+N_{1)}\biggl(r,\frac{1}{f}\biggr)+2 \overline{N}(r,f)+S(r,f) \\ \leq& N_{2}\biggl(r,\frac{1}{f}\biggr)+2\overline{N}\biggl(r, \frac{1}{f}\biggr)+2\overline {N}(r,f)+S(r,f), \end{aligned}$$
(2.7)
where
\(N_{1)}(r,\frac{1}{f})\) denotes the counting function of simple zeros of
f. Similarly, we obtain
$$ \overline{N}_{L}\biggl(r,\frac{1}{g-1}\biggr)+N_{2} \biggl(r,\frac{1}{g}\biggr)\leq N_{2}\biggl(r,\frac {1}{g} \biggr)+\overline{N}\biggl(r,\frac{1}{g}\biggr) +\overline{N}(r,g)+S(r,g). $$
(2.8)
Substituting (
2.7) and (
2.8) into (
2.6), this yields Case (i).
It remains to treat the case
\(H(z)\equiv 0\). Integrating twice results in
$$ \frac{1}{f-1}=A\frac{1}{g-1}+B, $$
(2.9)
where
\(A\neq0\) and
B are two constants. If now
\(B\neq0,-1\), then we rewrite (
2.9) as
$$A\frac{1}{g-1}=-\frac{B(f-\frac{1+B}{B})}{f-1}, $$
and then
$$\overline{N}\biggl(r,\frac{1}{f-\frac{1+B}{B}}\biggr)=\overline{N}(r,g). $$
By the second fundamental theorem we obtain
$$\begin{aligned} T(r,f)&\leq \overline{N}(r,f)+\overline{N}\biggl(r,\frac{1}{f}\biggr)+ \overline{N}\biggl(r,\frac {1}{f-\frac{1+B}{B}}\biggr)+S(r,f) \\ &=\overline{N}(r,f)+\overline{N}\biggl(r,\frac{1}{f}\biggr)+ \overline{N}(r,g)+S(r,f), \end{aligned}$$
which leads to Case (i). A similar reasoning results in Case (i) again, unless either
\(A=1\) and
\(B=0\) or
\(A=-1\) and
\(B=-1\). Hence, if
\(A=1\) and
\(B=0\), then
\(f\equiv g\), that is, Case (ii). If
\(A=-1\) and
\(B=-1\), then
\(f\cdot g\equiv1\), which is Case (iii). □