1 Introduction and main results
In this paper, we focus on the zero distributions of differential polynomials in a meromorphic function f with small meromorphic coefficients. We assume that the reader is familiar with the standard notations and some basic results in Nevanlinna theory (see [Reference Hayman13, Reference Laine16]).
In 1959, Hayman [Reference Hayman12, Reference Hayman and Lingham14] conjectured that if f is a transcendental meromorphic function and
$n\geq 2$
is an integer, then
$(f^n)'$
assumes every nonzero complex number infinitely often. He proved this conjecture for
$n\geq 4$
. The case for
$n=3$
was settled by Mues [Reference Mues18] in 1979, and the remaining for
$n=2$
was obtained by Bergweiler and Eremenko [Reference Bergweiler and Eremenko5], Chen and Fang [Reference Chen and Fang7] and Pang and Zalcman [Reference Pang and Zalcman20]. One principal extension was studied in [Reference Bergweiler and Eremenko5], the authors showed the Hayman conjecture is valid if
$(f^n)'$
is replaced by
$(f^n)^{(k)}$
for
$n> k\geq 1$
.
Two related questions arising in connection with Hayman’s problem are as follows.
The first one is to consider the zero distribution of
$(f^n)'-p$
where p is a small function of f. In fact, this extension originates from the study of the zero distribution of
$ff'-p$
, proposed by Chiang in 1994 (see [Reference Bergweiler4]). Bergweiler [Reference Bergweiler4] gave a positive answer when p is a nonzero polynomial and f is of finite order. This result (and a more general form
$f'f^n-R$
, where R is a rational function) was completely solved by Bergweiler and Pang (see [Reference Bergweiler and Pang6, Theorem 1.1]) from the perspectives of normal families and dynamic arguments [Reference Bergweiler and Eremenko5].
The second question given by Eremenko and Langley [Reference Eremenko and Langley8] is whether one can consider a more general differential polynomial of f such as a linear differential polynomial
$$ \begin{align*}F:=f^{(k)}+a_{k-1}f^{(k-1)}+\ldots+a_0f\end{align*} $$
with suitable small function coefficients
$a_j$
instead of
$f^{(k)}$
only.
All known proofs of generalizations of Hayman’s conjecture rely on the method of either the dynamic argument in [Reference Bergweiler and Eremenko5], or a much later result of Yamanoi [Reference Yamanoi21]. Indeed, the deep and highly technical result of Yamanoi is crucial to the resolution of some difficult conjectures in value distribution theory, for example, the Gol’dberg conjecture [Reference Gol’dberg and Ostrovskii11, p. 456 (B.4)] and Mues’ conjecture [Reference Mues19]. Very recently, applying Yamanoi’s result [Reference Yamanoi21, Theorem 1.2], An and Phuong [Reference An and Phuong2] investigated this question for a differential polynomial
$Q(f)$
with some restrictive conditions.
Theorem 1 ([Reference An and Phuong2], Theorem 1)
Let f be a transcendental meromorphic function, and
$Q(z)=b(z-a_1)^{m_1}(z-a_2)^{m_2}\ldots (z-a_l)^{m_l}$
be a polynomial of degree q, where
$b\in \mathbb {C}^*$
and
$a_1, \dots , a_l\in \mathbb {C}$
. If
$q\geq l+1$
, then
$(Q(f))^{(k)}$
takes every finite nonzero value infinitely often, for any positive integer k.
In 2013, Fang and Wang [Reference Fang and Wang10] relied on the Yamanoi result to derive some new consequences, one thing they obtained is that the pole order of a transcendental meromorphic function f can be controlled by the zero order of
$f^{(k)}-p$
(see Lemma 2 below), where p is a polynomial.
In this paper, by Fang and Wang’s result (an outgrowth of generalizations of Yamanoi’s result), we generalize An and Phuong’s result. In fact, we consider more general situations in which the coefficients of the polynomial Q are allowed to be functions meromorphic in
$\mathbb {C}$
, and the nonzero value is replaced by a polynomial. Our main result is the following.
Theorem 2 Let f be a transcendental meromorphic function, p be a nonzero polynomial and
$Q(z)=b(z-a_1)^{m_1}(z-a_2)^{m_2}\ldots (z-a_l)^{m_l}$
be a polynomial of degree q, where
$b\not \equiv 0, a_1, \dots , a_l$
are small functions of f. If
$q\geq l+1$
, then
$(Q(f))^{(k)}-p$
has infinitely many zeros, and particularly
$(Q(f))^{(k)}$
has infinitely many fixed points, for any positive integer k.
Remark 1 Recently, based on Yamanoi’s result [Reference Yamanoi21, Theorem 1.2], Fang et al. [Reference Fang, He and Wang9] proved that Theorem 1 is still true if the condition
$q\geq l+1$
is replaced by
$q\geq 2$
. Hence, it is natural to ask if the above theorem also works under such a condition
$q\geq 2$
. In general, this question is not true, the condition on
$q\geq l+1$
is necessary. For example, let
$f=\exp (z)$
,
$Q(w)=(w-z)(w+z)$
and
$p=-2z$
, in this case, we have
$$ \begin{align*}Q(f)'-p=(f^2-z^2)'+2z=2ff'=2\exp(2z)\end{align*} $$
which has no zeros.
As a consequence of Theorem 2, we also give an affirmative answer to a question of Chiang when p is a polynomial, without any growth restriction on f.
Corollary 1 (Hayman’s problem for polynomials)
If f is a transcendental meromorphic function and p is a nonzero polynomial, then
$f^nf'-p$
has infinitely many zeros, for any positive integer n.
The corollary follows immediately if one takes
$k=1$
, and
$Q(z)=z^{n+1}$
. Moreover, when
$Q(z)=z^n$
,
$n\geq 2$
, one can extend a result of Hayman in [Reference Hayman12, Theorem 2]. Note that Hayman considered the value distribution of
$(f^n)^{(k)}-c$
for any nonzero complex number c, while we can take c to be a polynomial.
Corollary 2 Let f be a transcendental meromorphic function and p be a nonzero polynomial. Then for
$n\geq 2, k\in \mathbb {N}$
,
$(f^n)^{(k)}-p$
has infinitely many zeros.
Remark 2 Corollaries 1 and 2 are contained in the results of the paper by Bergweiler and Pang (see [Reference Bergweiler and Pang6, Theorem 1.1]) who used the method of normal families and dynamic arguments, while we obtain these results from the other viewpoint mentioned before, which is an implication of Yamanoi’s result.
As a consequence, we also obtain the following corollary, which could be regarded as a precursor of Theorem 3.
Corollary 3 Let f be a transcendental meromorphic function and P be a nonzero polynomial, then
$f'-Pf^n$
has infinitely many zeros for any
$n\geq 3$
.
Proof Let
$f=1/g$
, then
$$ \begin{align*}Pf^n-f'=\frac{P}{g^n}+\frac{g'}{g^2}=\frac{P+g'g^{n-2}}{g^n}.\end{align*} $$
Hence, the result follows from Corollary 1.
Remark 3 Indeed, if P is a nonzero constant, Corollary 3 was proved by Pang for
$n\geq 4$
and by Chen and Fang [Reference Chen and Fang7, Theorem 3] for
$n=3$
from the point of view of the normal family.
In view of the above results, we only consider the polynomial case, it is natural to ask what happens if one replaces the polynomial p in the above results with some small functions. This question is in general not easy to answer. However, using some ideas from Liao and Ye [Reference Liao and Ye17], the classical Logarithmic Derivative Lemma, and the Clunie Lemma, we give some partial results as follows.
To describe our result, we need to introduce some classes of meromorphic functions.
Let
$T(r, f)$
be the Nevanlinna characteristic function of f. We denote by
$S(r, f)$
any quantity which is of growth
$o(T(r, f))$
as
$r\rightarrow \infty $
outside a set
$E\subset (0, \infty )$
of finite measure. A meromorphic function y is called a small function of f if it satisfies that
$T(r, y)=S(r, f)$
. The family of small functions of f is defined by
$\mathcal {S}_f$
. By
$\mathcal {N}_0$
and
$\mathcal {S}_0$
, we mean that the family of meromorphic functions y with finitely many poles and
$\mathcal {S}_0=\mathcal {S}_f\cap \mathcal {N}_0$
, respectively. Clearly, the field
$\mathbb {C}(z)$
of rational functions and the ring of entire functions are contained in
$\mathcal {N}_0$
. We say a differential polynomial in w is non-degenerate if it is not a polynomial in w.
Theorem 3 Let
$$ \begin{align*}P(z, w)=\sum_{j=m}^{n} b_j (z)w^j, \quad b_m\neq0\end{align*} $$
be a polynomial in w with coefficients
$b_j(z)$
in the family
$\mathcal {S}_0$
, and
$$ \begin{align*}L(z, w)=\sum_{|I|=0}^{k} a_{I} (z)w^{i_0}w^{{\prime} i_1}\cdots (w^{(q)})^{i_q}\end{align*} $$
be a non-degenerate differential polynomial in w over
$\mathcal {S}_0$
, where
$I=(i_0, \dots , i_q)$
is a multi-index with length
$|I|=i_0+\ldots +i_q$
. If f is a transcendental meromorphic function in
$\mathcal {N}_0$
such that
$L(z, f)\not \equiv 0$
, then
$P(z, f)+L(z, f)$
has infinitely many zeros for any
$m\geq k+2$
.
Remark 4 In general, a meromorphic function that does not satisfy any nontrivial algebraic differential equation
$L(z, w)=0$
is said to be hypertranscendental. Well-known examples of such meromorphic functions are the Euler gamma function and the Riemann zeta function. The study of hypertranscendental functions can be found in [Reference Adamczewski, Dreyfus and Hardouin1, Reference Bank and Kaufman3, Reference Huang and Ng15].
In the special case that
$P(z, w)=Q_1(z)w^n$
and
$L(z, w)=Q_2(z)w^{(q)}-R(z)$
, where
$n, q\in \mathbb {N}$
,
$Q_1(z)$
and
$Q_2(z)$
are nonzero rational functions, and
$R(z)$
is a rational function, we obtain the following which is a generalization of [Reference Hayman12, Theorem 8] and also partially answer a question proposed in [Reference Chen and Fang7, Proposition 1].
Corollary 4 Let f be a transcendental entire function and let R be a rational function. If
$n\geq 3$
, then
$Q_1f^n+Q_2f^{(q)}-R$
has infinitely many zeros, and hence
$Q_1f^n-Q_2f^{(q)}$
has infinitely many fixed points for all
$q\in \mathbb {N}$
.
2 Some Lemmas
We first recall some useful lemmas.
Lemma 1 ([Reference Laine16], Theorem 2.2.5)
Let f be a meromorphic function in
$\mathbb {C}$
and
$a\in \mathcal {S}_f$
, then
$$ \begin{align*}T(r, a, f)=T(r, f)+S(r, f)\end{align*} $$
and
$$ \begin{align*}T(r, af)=T(r, f)+S(r, f), \quad a\neq 0.\end{align*} $$
Lemma 2 ([Reference Fang and Wang10], Proposition 3)
Let g be a transcendental meromorphic function in
$\mathbb {C}$
, k be a positive integer, and
$p(\not \equiv 0)$
be a polynomial. Then for any
$\epsilon>0$
,
$$ \begin{align} (1-\epsilon)T(r, g)+(k-1)\overline{N}(r, g)\leq\overline{N}\left(r, \frac{1}{g}\right)+N \left(r, \frac{1}{g^{(k)}-p}\right)+S(r, f). \end{align} $$
Lemma 3 ([Reference Laine16], Theorem 2.3.3)
Let f be a transcendental meromorphic function and
$k\geq 1$
be an integer. Then
$$ \begin{align*}m\left(r, \frac{f^{(k)}}{f}\right)=S(r, f).\end{align*} $$
We also need the following Clunie lemma which plays an essential role in the proof of main results.
Lemma 4 ([Reference Hayman13], Lemma 3.3)
Let f be a transcendental meromorphic function in the complex plane such that
$$ \begin{align*}f^nP(z, f)=Q(z,f),\end{align*} $$
where
$P(z,f)$
and
$Q(z,f)$
are polynomials in f and its derivatives with meromorphic coefficients, say
$\{a_{\lambda }|\lambda \in \Lambda \}$
, such that
$T(r, a_{\lambda })=S(r, f)$
for all
$\lambda \in \Lambda $
. If the total degree of
$Q(z, f)$
as a polynomial in f and its derivative is at most n, then
$$ \begin{align*}m(r, P(z,f))=S(r, f)\quad \mathrm{as}\quad r\rightarrow\infty.\end{align*} $$
3 Proof of Theorems 2 and 3
Proof of Theorem 2
Applying Lemma 2 to
$g=Q(f)$
, we have
$$ \begin{align*} &\ (1-\epsilon)T(r, Q(f))+(k-1)\overline{N}(r, Q(f))\\ &\quad \leq \overline{N}\left(r, \frac{1}{Q(f)}\right)+N\left(r, \frac{1}{Q(f)^{(k)}-p}\right)+S(r, f)\\ &\quad \leq \sum_{i=1}^l\overline{N}\left(r, \frac{1}{f-a_i}\right)+\overline{N}\left(r, \frac{1}{b}\right)+N\left(r, \frac{1}{Q(f)^{(k)}-p}\right)+S(r, f)\\ &\quad \leq lT(r, f)+N\left(r, \frac{1}{Q(f)^{(k)}-p}\right)+S(r, f), \end{align*} $$
and then from the Mohon’ko result [Reference Laine16, Theorem 2.2.5], it follows that
$$ \begin{align*}(q-l-\epsilon)T(r, f)+(k-1)\overline{N}(r, f)\leq N\left(r, \frac{1}{Q(f)^{(k)}-p}\right)+S(r, f)\end{align*} $$
for any
$\epsilon>0$
. Therefore,
$Q(f)^{(k)}=p$
has infinitely many solutions when
$q\geq l+1$
, which means that
$Q(f)^{(k)}-p$
has infinitely many zeros.
Proof of Theorem 3
Let
$P(f)=P(z, f)$
and
$L(f)=L(z, f)$
. Suppose that
$P(f)+L(f)$
takes zero finitely many. As f has finitely many poles and the coefficients of P and L are in
$\mathcal {S}_0$
, it follows that
$$ \begin{align} \sum_{j=m}^nb_jf^j+L(f)=Ae^h \end{align} $$
where A is a nonzero rational function and h is an entire function with
$T(r, h)=S(r, f)$
. By differentiating both sides of (3.1), we obtain that
$$ \begin{align} \sum_{j=m}^nB_jf^{j-1}+L(f)'=A^*e^h \end{align} $$
where
$B_j=b_j'f+jb_jf'$
is a linear differential polynomial of f over
$\mathcal {S}_0$
and
$A^*=A'+Ah'$
with
$T(r, A^*)=S(r, f)$
. It follows from (3.1) and (3.2) that
$$ \begin{align} f^{m-1}H(z, f)=Q(z, f) \end{align} $$
where
$$ \begin{align*}H(z, f)=\sum_{j=m}^n(AB_jf^{j-m}-A^*b_jf^{j-m+1})\end{align*} $$
is a differential polynomial in f with coefficients in
$\mathcal {S}_f$
and
$$ \begin{align} Q(z, f)&=A^2\left(\frac{-L(f)}{A}\right)'+Ah'L(f) \end{align} $$
is also an
$\mathcal {S}_f$
–differential polynomial with total degree at most
$k(<m-1)$
. We claim that
$H(z, f)\not \equiv 0$
. Otherwise, in view of (3.3), (3.4) and
$L(f)\not \equiv 0$
, one has
$Q(z, f)\equiv 0$
, and then
$KL(f)=Ae^h$
with some constant K. Since f is a transcendental meromorphic function and
$b_j\in \mathcal {S}_0$
, (3.1) gives that
$K\neq 1$
and
$$ \begin{align*}f^m\left(\sum_{j=m}^nb_jf^{j-m}\right)=(K-1)L(f)\end{align*} $$
with
$\deg L=k< m$
. By Clunie’s lemma (Lemma 4), we have
$$ \begin{align*}m\left(r, \sum_{j=m}^nb_jf^{j-m}\right)=S(r, f)\end{align*} $$
Thus, by
$N(r, f)=O(\log r)$
and
$b_j\in \mathcal {S}_0$
, we have
$$ \begin{align*}T\left(r, \sum_{j=m}^nb_jf^{j-m}\right)=S(r, f)\end{align*} $$
yielding a contradiction. Hence
$H(z, f)\not \equiv 0$
.
Applying Clunie’s lemma (Lemma 4) to (3.3), we have
$$ \begin{align*}m(r, H(z, f))=S(r, f).\end{align*} $$
As
$A^*, A\in \mathcal {S}_f$
,
$B_j$
are differential polynomials of f with coefficients in
$\mathcal {S}_0$
and
$b_j\in \mathcal {S}_0$
, for
$j=m, \dots , n$
, it is not hard to see that
$$ \begin{align*}N(r, H(z, f))=S(r, f)\quad\mathrm{and\ hence}\quad T(r, H(z, f))=S(r, f).\end{align*} $$
Therefore, by the Mohon’ko result [Reference Laine16, Theorem 2.2.5] and
$T(r, H(z, f))=S(r, f)$
, we have
$$ \begin{align*}T(r, Q(z, f))=T(r, H(z, f)f^{m-1})=(m-1)T(r, f)+S(r, f).\end{align*} $$
From the form of
$Q(z, f)$
, one can rewrite it as follows
$$ \begin{align*}Q(z, f)=\sum_{l=0}^kC_l(z)f(z)^l\end{align*} $$
where
$C_l(z)$
’s are differential polynomials in
$f'/f$
and its derivatives with meromorphic coefficients in
$\mathcal {S}_f$
, and hence by the LDL,
$m(r, C_l)=S(r, f)$
for all l. Since
$$ \begin{align*} m\left(r, \sum_{l=0}^kC_lf^l\right)&\leq m\left(r, f\sum_{l=1}^kC_lf^{l-1}\right)+m(r, C_0)+O(1)\\ &\leq m(r, f)+m\left(r, \sum_{l=1}^kC_lf^{l-1}\right)+m(r, C_0)+O(1), \end{align*} $$
an immediate inductive argument implies that
$$ \begin{align*}m\left(r, \sum_{l=0}^kC_lf^l\right)\leq km(r, f)+\sum_{l=0}^km(r, C_l)+S(r, f)=km(r, f)+S(r, f).\end{align*} $$
From the form of
$Q(z, f)$
in (3.4) and
$N(r, f)=O(\log r)$
, it follows that
$$ \begin{align*}N(r, Q(z, f))=S(r, f),\end{align*} $$
thus one can obtain that
$$ \begin{align*}T(r, Q(z, f))=m(r, Q(z, f))+N(r, Q(z, f))\leq km(r, f)+S(r, f)\end{align*} $$
and hence
$$ \begin{align*}(m-1)T(r, f)+S(r, f)\leq km(r, f)+S(r, f)\leq kT(r, f)+S(r, f),\end{align*} $$
which is impossible, as
$m\geq k+2$
.
Acknowledgments
The authors would like to thank Fang Mingliang for introducing his recent related work and pointing out the counterexample in Remark 2. The authors also wish to express gratitude to the anonymous referees for a number of useful remarks that have been made for the improvement of this paper.
