On the value distribution of the Riemann zeta-function and the Euler gamma-function

TL;DR

This paper investigates the value distribution of the Riemann zeta-function and the Euler gamma-function, establishing uniqueness results through shared values and applying value distribution theory.

Contribution

It introduces new uniqueness theorems for these functions based on shared values, advancing the understanding of their value distribution properties.

Findings

Proved uniqueness results for the Riemann zeta-function.

Established new theorems for the Euler gamma-function.

Applied value distribution theory to derive these results.

Abstract

We prove some uniqueness results for the Riemann zeta-function and the Euler gamma-function by virtue of shared values using the value distribution theory.

Equations

$$\zeta (s)=\sum _{n=1}^{+\infty }\frac{1}{n^s},$$

$$Z_V:=\left\{s:\frac{1}{4}<\sigma <\frac{3}{4},t>0\right\}and{Z}_H:=\left\{s:-2<t<2,\sigma <0\right\},$$

$$F:=\frac{\zeta -a}{f-a}\cdot \frac{f-b}{\zeta -b}.$$

$$F=\frac{\zeta -a}{f-a}\cdot \frac{f-b}{\zeta -b}=R{e}^g.$$

$$T(r,\zeta )=\frac{1}{\pi }r\log r+O(r)$$

$$T(r,f)$$

$$\displaystyle=\bar{N}\Big{(}r,\frac{1}{\zeta-a}\Big{)}+\bar{N}\Big{(}r,\frac{1}{\zeta-b}\Big{)}+\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}+O(\log r)+S(r,f)$$

$$\displaystyle\leq 2\hskip 0.56905ptT(r,\zeta)+\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}+O(\log r)+S(r,f).$$

$$\left(1-\frac{1+\iota }{2}-o(1)\right)T(r,f)\le 2T(r,\zeta )+O(\log r);$$

$$\left(1-o(1)\right)T(r,f)\le O(T(r,\zeta ))+O(\log r).$$

$$1\equiv R({\varpi }_n)e^{A{\varpi }_n}=R({\beta }_n+\mathtt{i}{\gamma }_n)e^{(B+\mathtt{i}C)({\beta }_n+\mathtt{i}{\gamma }_n)},$$

$$1\equiv R({\omega }_n)e^{A{\omega }_n}=R({\mu }_n+\mathtt{i}{\nu }_n)e^{(B+\mathtt{i}C)({\mu }_n+\mathtt{i}{\nu }_n)}.$$

$$F=\frac{\zeta -a}{f-a}\cdot \frac{f-b}{\zeta -b}=R.$$

$$F_1:=\frac{\zeta -a}{f-a}.$$

$$F_1=\frac{\zeta -a}{f-a}=R{e}^g.$$

$$F_1=\frac{\zeta -a}{f-a}=R.$$

$$T(r,\Gamma )=\frac{1}{\pi }r\log r+O(r),$$

$$N(r,\Gamma )=r+O(\log r),$$

$$N\Big{(}r,\frac{1}{\Gamma-c}\Big{)}=\frac{1}{\pi}\,r\log r+O(r)\hskip 5.69054pt\mathrm{for}\hskip 5.69054ptc\neq 0,\infty.$$

$$T(r,\eta )=K_1{r}^p{\log }^{q}r+O(r^p{\log }^{q-1}r),$$

$$\bar{N}\Big{(}r,\frac{1}{\eta-c}\Big{)}\geq K_{2}\,r^{p}\log^{q}r+O(r^{p}\log^{q-1}r).$$

$$\bar{N}\Big{(}r,\frac{1}{\zeta}\Big{)}\geq\frac{1}{6\hskip 0.56905pt\pi}\,r\log r+O(r),$$

$$\bar{N}\Big{(}r,\frac{1}{\Gamma-c}\Big{)}+N\Big{(}r,\frac{1}{\Gamma^{\prime}}\Big{)}\geq N\Big{(}r,\frac{1}{\Gamma-c}\Big{)}+O(1)$$

$$\displaystyle\,N\Big{(}r,\frac{1}{\Gamma^{\prime}}\Big{)}=N\Big{(}r,\frac{1}{\Gamma^{\prime}/\Gamma}\Big{)}\leq T\Big{(}r,\frac{\Gamma^{\prime}}{\Gamma}\Big{)}+O(1)$$

$$\le$$

$$\bar{N}\Big{(}r,\frac{1}{\Gamma-c}\Big{)}=\frac{1}{\pi}\,r\log r+O(r),$$

$$G:=\frac{\eta -a}{f-a}\cdot \frac{f-b}{\eta -b}.$$

$${\Pi }_1(z):=\prod _{k=1}^{+\infty }E\left(\frac{z}{a_k},p\right)\mathrm{and}{\Pi }_2(z):=\prod _{k=1}^{+\infty }E\left(\frac{z}{b_k},p\right),$$

$$\sum _{k=1}^{+\infty }\frac{1}{{\left|a_k\right|}^{p+1}}={\int }_{|a_1|}^{+\infty }\frac{d(n(t,{\mathcal{E}}_1))}{t^{p+1}}=\underset{r\to +\infty }{\lim }{\int }_{|a_1|}^r\frac{d(n(t,{\mathcal{E}}_1))}{t^{p+1}}$$

$$=$$

$$\le$$

$$\log M(r,{\Pi }_1)\le C_p\left\{r^p{\int }_{|a_1|}^r\frac{n(t,{\mathcal{E}}_1)}{t^{p+1}}dt+r^{p+1}{\int }_r^{+\infty }\frac{n(t,{\mathcal{E}}_1)}{t^{p+2}}dt\right\}.$$

$$T(r,{\Pi }_1)$$

Full text

*†*Qi Han is the corresponding author of this research work.

2010 Mathematics Subject Classification. 30D30, 30D35, 11M06, 33B15.

Keywords. Euler gamma-function, Riemann zeta-function, value distribution, uniqueness.

1. Introduction

Recall the Riemann zeta-function $\zeta$ is originally defined through the Dirichlet series

[TABLE]

with $\Re(s)>1$, that can be analytically continued to be a meromorphic function in the complex variable $s=\sigma+\mathtt{i}\hskip 0.56905ptt\in\mathbf{C}$ having only a simple pole at $s=1$. The famous, yet unproven, Riemann hypothesis asserts that all the non-trivial zeros of $\zeta$ lie on the line $\Re(s)=\sigma=\frac{1}{2}$.

Value distribution of the Riemann zeta-function has been studied extensively. See the classic by Titchmarsh [19] and a recent monograph from Steuding [18]; results more closely related to Nevanlinna theory can be found in Liao-Yang [13] and Ye [21]. It is well-known by Nevanlinna [15] that a non-constant meromorphic function $f$ in $\mathbf{C}$ is completely determined by “$5$ IM” value sharing condition (ignoring multiplicity), whereas $f$ is completely determined by “$3$ CM” value sharing condition (counting multiplicity) when $f$ is further assumed to be of finite non-integral order. Han [7] recently proved the mixed “$1$ CM + $3$ IM” value sharing condition sufficient if $f$ has finite non-integral order, or $f$ has integral order yet is of maximal growth type; this result particularly applies to the Euler gamma-function and the Riemann zeta-function.

In this paper, we discuss some uniqueness results primarily about the Riemann zeta-function and the Euler gamma-function in light of their nice properties. Specifically, we will prove some results for the Riemann zeta-function that extends Gao-Li [4] in section 2, and will prove some results for the Euler gamma-function that extends Liao-Yang [13] in section 3; finally, in section 4, we will revisit the Riemann zeta-function and discuss some other related results.

Below, we assume the reader is familiar with the basics of Nevanlinna theory of meromorphic functions in $\mathbf{C}$ such as the first and second main theorems, and the standard notations such as the characteristic function $T(r,f)$, the proximity function $m(r,f)$, and the integrated counting functions $N(r,f)$ (counting multiplicity) and $\bar{N}(r,f)$ (ignoring multiplicity). $S(r,f)$ denotes a quantity satisfying $\displaystyle{\displaystyle{S(r,f)=o(T(r,f))}}$ as $r\to+\infty$, since $f$ is (always assumed to be) of finite order. Here, the order of $f$ is defined as $\displaystyle{\rho(f):=\limsup\limits_{r\to+\infty}\frac{\log T(r,f)}{\log r}}$. Excellent references of this theory are certainly Nevanlinna [16], Hayman [8], Yang [20], and Cherry-Ye [2] etc.

2. Results regarding the Riemann zeta-function I

The first result that we shall need is essentially due to Levinson [10], while for convenience of the reader we reproduce it as the following proposition.

Proposition 2.1**.**

Given $a\in\mathbf{C}$, $\zeta(s)-a$ has infinitely many zeros in the strips

[TABLE]

respectively, where $t\to+\infty$ in $Z_{V}$ and $\sigma\to-\infty$ in $Z_{H}$.

Now, we can formulate our main results of this section as follows.

Theorem 2.2**.**

Let $a,b,c\in\mathbf{C}$ be finite and distinct, and let $f$ be a meromorphic function in $\mathbf{C}$ such that either \displaystyle{\limsup\limits_{r\to+\infty}\frac{\bar{N}\big{(}r,\frac{1}{f-c}\big{)}}{T(r,f)}<1} or \displaystyle{\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}=O(T(r,\zeta))} with $\zeta$ being the Riemann zeta-function. When $f,\zeta$ share the finite values $a,b$ counting multiplicity, with $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ ignoring multiplicity, except possibly at finitely many points, then $f=\zeta$.

Proof.

Consider an auxiliary function

[TABLE]

Since $f,\zeta$ share $a,b$ CM except possibly at finitely many points, one knows $F$ is a meromorphic function having only finitely many zeros and poles. Thus, there exists an entire function $g$ and a rational function $R$ such that

[TABLE]

We next claim $g$ is linear. In fact, from [13, Lemma 2.7] or [21, Theorem 1], one has

[TABLE]

so that $\rho(\zeta)=1$. Using Nevanlinna’s first and second main theorems, we observe that

[TABLE]

Now, when \displaystyle{\iota:=\limsup\limits_{r\to+\infty}\frac{\bar{N}\big{(}r,\frac{1}{f-c}\big{)}}{T(r,f)}<1} holds, one has from (2.4) that

[TABLE]

while when \displaystyle{\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}=O(T(r,\zeta))} holds, one sees from (2.4) that

[TABLE]

Therefore, via (2.3), one derives that $\displaystyle{T(r,f)\leq O(r\log r)+O(r)}$ (for all $r$ outside of a possible set of finite Lebesgue measure), which implies $\rho(f)\leq 1$ (so there is no exceptional set). Hence, it is routine to note $\rho(F)\leq 1$ as well. Through the renowned Hadamard factorization theorem (see Berenstein-Gay [1, Section 4.6.15]), we know $g$ is linear as claimed earlier.

We further assert $g\equiv 0$; otherwise, suppose that $g=As$ where $A\neq 0$ is a complex number. Recall $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ except possibly at finitely many points. Via Proposition 2.1, there are infinitely many zeros of $\zeta-c$ in both the strips $Z_{V}$ and $Z_{H}$ (since $c\neq\infty$). Note these zeros of $\zeta-c$ are zeros of $R\hskip 0.56905pte^{As}-1$. Denote these zeros by $\varpi_{n}=\beta_{n}+\mathtt{i}\hskip 0.56905pt\gamma_{n}$ in the vertical strip $Z_{V}$ and $\omega_{n}=\mu_{n}+\mathtt{i}\hskip 0.56905pt\nu_{n}$ in the horizontal strip $Z_{H}$, respectively. Obviously, $\gamma_{n}\to+\infty$ and $\mu_{n}\to-\infty$ when $n\to+\infty$, and $\beta_{n}$’s and $\nu_{n}$’s are uniformly bounded. Write $A=B+\mathtt{i}\hskip 0.56905ptC$ with $B,C$ real numbers. Then, we get

[TABLE]

[TABLE]

We consider two different cases.

Case 1. $B\neq 0$. Then, we apply (2.8) to deduce \displaystyle{1\equiv\left|R(\omega_{n})\right|e^{B\mu_{n}-C\nu_{n}}\to\left\{\begin{array}[]{ll}0&\mathrm{if}\hskip 5.69054ptB>0\vskip 6.0pt plus 2.0pt minus 2.0pt\\ +\infty&\mathrm{if}\hskip 5.69054ptB<0\end{array}\right.} with $\mu_{n}\to-\infty$ and bounded $\nu_{n}$’s, since $R(s)$ is rational and $e^{As}$ is exponential.

Case 2. $B=0$. If $C\neq 0$, we then use (2.7) to show \displaystyle{1\equiv\left|R(\varpi_{n})\right|e^{-C\gamma_{n}}\to\left\{\begin{array}[]{ll}0&\mathrm{if}\hskip 5.69054ptC>0\vskip 6.0pt plus 2.0pt minus 2.0pt\\ +\infty&\mathrm{if}\hskip 5.69054ptC<0\end{array}\right.} with $\gamma_{n}\to+\infty$ and bounded $\beta_{n}$’s, since $R(s)$ is rational and $e^{As}$ is exponential.

As a result, $B=C=0$ so that the assertion $g\equiv 0$ is proved, which leads to

[TABLE]

As $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ except possibly at finitely many points, and $\zeta-c=0$ has infinitely many roots, we must have $R(s)\equiv 1$ noticing $R$ is rational. It then follows that $f=\zeta$. ∎

Theorem 2.3**.**

Let $a\neq c\in\mathbf{C}$ be finite, and let $f$ be a meromorphic function in $\mathbf{C}$ having only finitely many poles such that either \displaystyle{\limsup\limits_{r\to+\infty}\frac{\bar{N}\big{(}r,\frac{1}{f-c}\big{)}}{T(r,f)}<1} or \displaystyle{\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}=O(T(r,\zeta))} where $\zeta$ is the Riemann zeta-function. When $f,\zeta$ share $a$ counting multiplicity, with $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ ignoring multiplicity, except possibly at finitely many points, then $f=\zeta$.

Proof.

Recall $\zeta$ has its unique simple pole at $s=1$. Consider an auxiliary function

[TABLE]

In view of the assumptions that $f,\zeta$ share $a$ CM except possibly at finitely many points and $f$ has only finitely many poles, $F_{1}$ is a meromorphic function with finitely many zeros and poles. Hence, there is an entire function $g$ and a rational function $R$ such that

[TABLE]

Next, one can exploit exactly the same analyses as described in the proof of Theorem 2.2 to deduce that $g\equiv 0$. We then have

[TABLE]

As $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ except possibly at finitely many points, and $\zeta-c=0$ has infinitely many roots, we must have $R(s)\equiv 1$ noticing $R$ is rational. It then follows that $f=\zeta$. ∎

Remark 2.4**.**

Li [11] considered the uniqueness of an $L$-function in the extended Selberg class regarding a general meromorphic function in $\mathbf{C}$ having only finitely many poles, and proved the the combined “$1$ CM + $1$ IM” value sharing condition sufficient. Extensions of this nice result were given by Garunkštis-Grahl-Steuding [5] and Han [7]. On the other hand, when we focus on the Riemann zeta-function $\zeta$, results using 3 general value sharing condition were discussed by Gao-Li [4] which answered an open question of Liao-Yang [13]. In [4], the authors also provided an example to show that $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ ignoring multiplicity is not adequate in proving $f=\zeta$ besides the other hypothesis that $f,\zeta$ share $0,1$ CM, with \displaystyle{f=-\frac{\zeta}{\big{(}e^{(\zeta-c)(s-1)}-1\big{)}\zeta-e^{(\zeta-c)(s-1)}}} and $c\neq 0,1$. For this $f$, one has \displaystyle{T(r,f)=\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}+O(1)} and \displaystyle{\liminf\limits_{r\to+\infty}\frac{\bar{N}\big{(}r,\frac{1}{f-c}\big{)}}{T(r,\zeta)}=+\infty}; this observation leads to the growth constraints about $f-c$ in Theorems 2.2 and 2.3.

3. Results regarding the Euler gamma-function

Denote $\Gamma(z)$ the Euler gamma-function. Recall [13, Section 1] and [21, Theorem 2]

[TABLE]

[TABLE]

[TABLE]

This inspires us to introduce the following family of meromorphic functions in $\mathbf{C}$.

Definition 3.1**.**

Let $\mathscr{X}_{p,q}$ be the family of meromorphic functions $\eta$ in $\mathbf{C}$ such that

[TABLE]

[TABLE]

Here, $c\in\mathbf{C}$ is a finite value, $p,q\geq 1$ are integers, and $K_{1}\geq K_{2}>0$ are real numbers.

One observes $\zeta,\Gamma\in\mathscr{X}_{1,1}$. In fact, by Conrey [3, Theorem 1] and the Riemann-von Mangoldt formula [19, Theorem 9.4] (see the proof of [4, Theorem 1.2]), it follows that

[TABLE]

which together with (2.3) implies $\zeta\in\mathscr{X}_{1,1}$ for $c=0$, $K_{1}=\frac{1}{\pi}$ and $K_{2}=\frac{1}{6\pi}$; on the other hand, seeing that

[TABLE]

and recalling that [math] is the only Picard value of $\Gamma$ (yet both $0,\infty$ are Nevanlinna’s defect values of $\Gamma$), one has

[TABLE]

in light of the lemma of logarithmic derivative and (3.2) so that

[TABLE]

which along with (3.1) yields $\Gamma\in\mathscr{X}_{1,1}$ for all $c\neq 0,\infty$ and $K_{1}=K_{2}=\frac{1}{\pi}$ with equality.

Now, we can formulate our main result of this section as follows.

Theorem 3.2**.**

Assume $\eta\in\mathscr{X}_{p,q}$ is a meromorphic function associated with the numbers $c\in\mathbf{C}$, $p,q\geq 1$, and $K_{1}\geq K_{2}>0$. Let $\displaystyle{a\neq b\in\mathbf{C}\cup\left\{\infty\right\}}$ be distinct from $c$, and let $f$ be a meromorphic function in $\mathbf{C}$ such that either \displaystyle{\limsup\limits_{r\to+\infty}\frac{\bar{N}\big{(}r,\frac{1}{f-c}\big{)}}{T(r,f)}<1} or \displaystyle{\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}=O(T(r,\eta))}. When $f,\eta$ share the values $a,b$ counting multiplicity, with $\displaystyle{\eta^{-1}(c)\subseteq f^{-1}(c)}$ ignoring multiplicity, except possibly at a set $\mathcal{E}$ of points with $\displaystyle{n(r,\mathcal{E})=o(r^{p}\log^{q-1}r)}$, then $f=\eta$. Here, $n(r,\mathcal{E})$ denotes the counting function of $\mathcal{E}$, i.e., the number of points in the set $\displaystyle{\mathcal{E}\cap\left\{z\in\mathbf{C}:\left|z\right|<r\right\}}$.

Proof.

First, consider finite values $a,b$ and define an auxiliary function

[TABLE]

Suppose $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ are the zero and pole sets of $G$. Then, it follows from our assumption that $\displaystyle{n(r,\mathcal{E}_{1})=o(r^{p}\log^{q-1}r)}$ and $\displaystyle{n(r,\mathcal{E}_{2})=o(r^{p}\log^{q-1}r)}$.

Below, we follow closely the method from Li [12] (see also Han [7] and Lü [14]), and assume that $\left\{a_{k}:k\geq 1\right\}$ and $\left\{b_{k}:k\geq 1\right\}$ are the non-zero elements of $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ arranged in ascending orders, respectively, i.e., $\left|a_{k}\right|\leq\left|a_{k+1}\right|$ and $\left|b_{k}\right|\leq\left|b_{k+1}\right|$. Construct two infinite products

[TABLE]

where, for the integer $p\geq 1$, we write $\displaystyle{E(z,p):=(1-z)\hskip 0.56905pte^{z+\frac{1}{2}z^{2}+\cdots+\frac{1}{p}z^{p}}}$. Then, $\Pi_{1}$ and $\Pi_{2}$ are entire functions defined in $\mathbf{C}$ having $a_{k}$’s and $b_{k}$’s as their zeros, respectively.

Actually, using the Stieltjes integral and seeing $\displaystyle{n(r,\mathcal{E}_{1})=o(r^{p}\log^{q-1}r)}$, one has

[TABLE]

with $r_{0}\gg 1$ satisfying $\displaystyle{\log^{q-1}r\leq 2\hskip 0.56905ptr^{0.8}}$ for all $r\geq r_{0}$; this combined with Hayman [8, Theorem 1.11] and Goldberg-Ostrovskii [6, p.56, Remark 1] shows that $\Pi_{1}(z)$ is an entire function in $\mathbf{C}$ having zeros $a_{k}$’s, which in addition satisfies for $\displaystyle{C_{p}:=2\hskip 0.56905pt(p+1)(2+\log p)>0}$,

[TABLE]

Now, for each $\epsilon>0$, there exists an $r_{1}\gg 1$ such that $\displaystyle{n(r,\mathcal{E}_{1})\leq\epsilon\hskip 0.56905ptr^{p}\log^{q-1}r}$ for all $r\geq r_{1}$. Combing this and (3.9) with [6, p.37, Theorem 7.1], it follows that

[TABLE]

for all $r\geq r_{1}$, with $\epsilon>0$ taken smaller if necessary. Here, routine substitution and integration by parts were used. Similarly, we have for some $r_{2}\gg 1$ and all $r\geq r_{2}$,

[TABLE]

Next, for the possible exceptional set $\mathcal{E}$ with $\displaystyle{n(r,\mathcal{E})=o(r^{p}\log^{q-1}r)}$, one has

[TABLE]

Like (2.4), using Nevanlinna’s first and second main theorems, it yields that

[TABLE]

Now, when \displaystyle{\iota:=\limsup\limits_{r\to+\infty}\frac{\bar{N}\big{(}r,\frac{1}{f-c}\big{)}}{T(r,f)}<1} holds, one has from (3.12) and (3.13) that

[TABLE]

while when \displaystyle{\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}=O(T(r,\eta))} holds, one sees from (3.12) and (3.13) that

[TABLE]

Therefore, by (3.4), one arrives at $\displaystyle{T(r,f)\leq O(r^{p}\log^{q}r)+O(r^{p}\log^{q-1}r)}$ (for all $r$ outside of a possible set of finite Lebesgue measure), which gives $\rho(f)\leq p$ (so there is no exceptional set). Thus, $\rho(G)\leq p$ and by virtue of Hadamard factorization theorem, it leads to

[TABLE]

Here, $l$ is an integer that is the multiplicity of zero or pole of $G$ at $z=0$ and $P$ is a polynomial with $\deg P\leq\max\left\{\rho(\Pi_{1}),\rho(\Pi_{2}),\rho(G)\right\}\leq p$ in view of (3.10) and (3.11).

Observe that $G\equiv 1$. In fact, as $\displaystyle{\eta^{-1}(c)\subseteq f^{-1}(c)}$ except possibly at the set $\mathcal{E}$ of points with $\displaystyle{n(r,\mathcal{E})=o(r^{p}\log^{q-1}r)}$, for the given $\epsilon>0$, there is an $\tilde{r}\geq\max\left\{r_{1},r_{2}\right\}$ such that

[TABLE]

provided $r\geq\tilde{r}$. This contradicts our hypothesis (3.5). As a consequence, we must have $G\equiv 1$, which further implies $f=\eta$.

Finally, if one of $a,b$ is $\infty$, assume without loss of generality $a$ is finite and $b=\infty$. Consider $\displaystyle{\tilde{f}:=\frac{1}{f-d},\tilde{\eta}:=\frac{1}{\eta-d}}$ that share the finite values $\displaystyle{\tilde{a}:=\frac{1}{a-d},\tilde{b}:=0}$ counting multiplicity, plus the finite value $\displaystyle{\tilde{c}:=\frac{1}{c-d}}$ such that they satisfy all the given conditions, with $d\in\mathbf{C}$ finite and distinct from $a,c$. Exactly the same analyses as above imply $\tilde{f}=\tilde{\eta}$. So, $f=\eta$. ∎

Note [math] is the unique Picard value of $\Gamma(z)$ and $s=1$ is the unique, simple pole of $\zeta(s)$. Using these nice properties of $\Gamma(z)$ and $\zeta(s)$, we consider a sub-family of $\mathscr{X}_{p,q}$ below.

Definition 3.3**.**

Let $\displaystyle{\widetilde{\mathscr{X}}_{p,q}\subsetneq\mathscr{X}_{p,q}}$ be the sub-family of meromorphic functions $\eta$ in $\mathbf{C}$ such that $\displaystyle{n(r,\mathcal{B}_{\eta})=o(r^{p}\log^{q-1}r)}$ for the $b$-points of each $\eta$ with some value $\displaystyle{b\in\mathbf{C}\cup\left\{\infty\right\}\setminus\left\{c\right\}}$.

Theorem 3.4**.**

Suppose that $\eta\in\widetilde{\mathscr{X}}_{p,q}$ is a meromorphic function associated with the numbers $c\in\mathbf{C}$, $\displaystyle{b\in\mathbf{C}\cup\left\{\infty\right\}\setminus\left\{c\right\}}$, $p,q\geq 1$, and $K_{1}\geq K_{2}>0$. Choose $\displaystyle{a\in\mathbf{C}\cup\left\{\infty\right\}\setminus\left\{b,c\right\}}$, and let $f$ be a meromorphic function in $\mathbf{C}$ whose $b$-points satisfies $\displaystyle{n(r,\mathcal{B}_{f})=o(r^{p}\log^{q-1}r)}$ as well such that either \displaystyle{\limsup\limits_{r\to+\infty}\frac{\bar{N}\big{(}r,\frac{1}{f-c}\big{)}}{T(r,f)}<1} or \displaystyle{\bar{N}\Big{(}r,\frac{1}{f-c}\Big{)}=O(T(r,\eta))}. When $f,\eta$ share $a$ counting multiplicity, with $\displaystyle{\eta^{-1}(c)\subseteq f^{-1}(c)}$ ignoring multiplicity, except possibly at a set $\mathcal{E}$ of points with $\displaystyle{n(r,\mathcal{E})=o(r^{p}\log^{q-1}r)}$, then $f=\eta$.

Proof.

Without loss of generality, assume $b=\infty$; otherwise, set $\displaystyle{\hat{f}:=\frac{1}{f-b},\hat{\eta}:=\frac{1}{\eta-b}}$ as well as $\displaystyle{\hat{a}:=\frac{1}{a-b},\hat{b}:=\infty,\hat{c}:=\frac{1}{c-b}}$. Consider an auxiliary function

[TABLE]

Denote by $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ the zero and pole sets of $G_{1}$. Then, one sees $\displaystyle{\mathcal{E}_{1}\cup\mathcal{E}_{2}=\mathcal{E}\cup\mathcal{E}_{\eta}\cup\mathcal{E}_{f}}$ and $\displaystyle{n(r,\mathcal{E}_{1}),n(r,\mathcal{E}_{2})=o(r^{p}\log^{q-1}r)}$. Write the non-zero elements of $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ as $\left\{a_{k}:k\geq 1\right\}$ and $\left\{b_{k}:k\geq 1\right\}$ with $\left|a_{k}\right|\leq\left|a_{k+1}\right|$ and $\left|b_{k}\right|\leq\left|b_{k+1}\right|$, and construct two infinite products

[TABLE]

Then, $\Pi_{1}$ and $\Pi_{2}$ are entire functions in $\mathbf{C}$ having $a_{k}$’s and $b_{k}$’s as their zeros, respectively, and (3.10) and (3.11) follow analogously. Moreover, like (3.12), one observes

[TABLE]

All these modifications plus the proof of Theorem 3.2 lead to

[TABLE]

Here, $l$ is an integer that is the multiplicity of zero or pole of $G_{1}$ at $z=0$ and $P$ is a polynomial with $\deg P\leq\max\left\{\rho(\Pi_{1}),\rho(\Pi_{2}),\rho(G_{1})\right\}\leq p$ in view of (3.10) and (3.11). Using the hypothesis $\displaystyle{\eta^{-1}(c)\subseteq f^{-1}(c)}$ and (3.18), one similarly has $G_{1}\equiv 1$ that further implies $f=\eta$. ∎

Remark 3.5**.**

Seeing (3.2), the preceding Theorem 3.4 in particular implies when a meromorphic function $f$ in $\mathbf{C}$ having only finitely many zeros shares two values $\displaystyle{a\in\mathbf{C}\cup\left\{\infty\right\}\setminus\left\{0\right\}}$ CM and $\displaystyle{b\in\mathbf{C}\setminus\left\{a,0\right\}}$ IM with the Euler gamma-function $\Gamma$, then $f=\Gamma$. On the other hand, the preceding Theorem 2.3 in particular implies when a meromorphic function $f$ in $\mathbf{C}$ having only finitely many poles shares two values $\displaystyle{a\neq b\in\mathbf{C}}$ $a$ CM and $b$ IM with the Riemann zeta-function $\zeta$, then $f=\zeta$. These results consist with Li [11] and Han [7, Theorem 2.1].

4. Results regarding the Riemann zeta-function II

Denote $\mathscr{M}$ the space of meromorphic functions in $\mathbf{C}$, and $\mathscr{H}$ the space of entire functions in $\mathbf{C}$; denote $\mathscr{M}_{1}$ the space of meromorphic functions in $\mathbf{C}$ that have finite non-integral order or have integral order yet are of maximal growth type, and $\mathscr{H}_{1}$ the space of such entire functions in $\mathbf{C}$. Then, from the classical result of Nevanlinna [15], and Han [7], one has

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\left\{\begin{array}[]{lll}f,g\in\mathscr{M}\ {\rm are}\ {\rm identical}\ {\rm under}\ {\rm``5\ IM"}\ {\rm value}\ {\rm sharing}\ {\rm condition}.\\ \\ f\in\mathscr{M}_{1}\ {\rm and}\ g\in\mathscr{M}\ {\rm are}\ {\rm identical}\ {\rm under}\ {\rm``1\ CM+3\ IM"}\ {\rm value}\ {\rm sharing}\ {\rm condition}.\\ \\ f\in\mathscr{M}_{1}\ {\rm and}\ g\in\mathscr{M}\ {\rm are}\ {\rm identical}\ {\rm under}\ {\rm``3\ CM"}\ {\rm value}\ {\rm sharing}\ {\rm condition}.\end{array}\right.

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\left\{\begin{array}[]{lll}f,g\in\mathscr{H}\ {\rm are}\ {\rm identical}\ {\rm under}\ {\rm``4\ IM"}\ {\rm value}\ {\rm sharing}\ {\rm condition}.\\ \\ f\in\mathscr{H}_{1}\ {\rm and}\ g\in\mathscr{H}\ {\rm are}\ {\rm identical}\ {\rm under}\ {\rm``3\ IM"}\ {\rm value}\ {\rm sharing}\ {\rm condition}.\\ \\ f\in\mathscr{H}_{1}\ {\rm and}\ g\in\mathscr{H}\ {\rm are}\ {\rm identical}\ {\rm under}\ {\rm``1\ CM+1\ IM"}\ {\rm value}\ {\rm sharing}\ {\rm condition}.\end{array}\right.

Below, we follow Hu and Li [9] to discuss the remanding case regarding the number “$2$” for meromorphic functions in $\mathbf{C}$ and the number “$1$” for entire functions in $\mathbf{C}$; an earlier result on meromorphic functions $f$ in $\mathbf{C}$ with $\rho(f)<1$ was given by Rao [17, Theorem 4]. Here, we try to describe some more general results under possibly the minimum requirement.

Theorem 4.1**.**

Let $f,g$ be meromorphic functions in $\mathbf{C}$ of finite order, and let $\displaystyle{a,b,c\in\mathbf{C}\cup\left\{\infty\right\}}$ be values pairwise distinct from each other. When $f,g$ share the values $a,b$ counting multiplicity with $\displaystyle{\lim_{x\to+\infty}f(z)=\lim_{x\to+\infty}g(z)=c}$ uniformly in $y$ for $z=x+\mathtt{i}\hskip 0.56905pty\in\mathbf{C}$, then $f=g$.

Proof.

Without loss of generality, assume $b=\infty$; otherwise, set $\displaystyle{\hat{f}:=\frac{1}{f-b},\hat{g}:=\frac{1}{g-b}}$ as well as $\displaystyle{\hat{a}:=\frac{1}{a-b},\hat{b}:=\infty,\hat{c}:=\frac{1}{c-b}}$. Consider an auxiliary function

[TABLE]

By virtue of our hypotheses, one finds a polynomial $P$ such that $L=e^{P}$. Write

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with $\displaystyle{a_{m}(y),a_{m-1}(y),\ldots,a_{0}(y)}$ real polynomials in $y$. If $a_{m}(y)\not\equiv 0$ and $m\geq 1$, one may take some $y_{0}$ such that $a_{m}(y_{0})\neq 0$; then, a straightforward analysis leads to

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that is absurd. This shows $\Re\hskip 0.56905pt(P(x+\mathtt{i}\hskip 0.56905pty))=a_{0}(y)$, and hence $\left|L(z)\right|=e^{a_{0}(y)}$, independent of $x$. Letting $x\to+\infty$ again, we have $e^{a_{0}(y)}=1$ for all $y$; that is, $\left|L\right|\equiv 1$ or $L$ is simply a constant. Letting $x\to+\infty$ again, one derives $L\equiv 1$ so that $f=g$. ∎

Corollary 4.2**.**

Assume $f,g$ are finite order meromorphic functions in $\mathbf{C}$ having $b$ as a common Picard value, and take $\displaystyle{a\neq c\in\mathbf{C}\cup\left\{\infty\right\}\setminus\left\{b\right\}}$. When $f,g$ share the value $a$ counting multiplicity with $\displaystyle{\lim_{x\to+\infty}f(z)=\lim_{x\to+\infty}g(z)=c}$ uniformly in $y$ for $z=x+\mathtt{i}\hskip 0.56905pty\in\mathbf{C}$, then $f=g$.

Proof.

Without loss of generality, assume $b=\infty$ and consider the function $L(z)$ in (4.1); then, parallel discussion as that for Theorem 4.1 implies $f=g$. ∎

Remark 4.3**.**

Upon a slight modification, the preceding Corollary 4.2 in particular applies for the Euler gamma-function $\Gamma(z)$ (as [math] is the unique Picard value of $\Gamma(z)$ and $\displaystyle{\lim_{x\to+\infty}\Gamma^{-1}(z)=0}$ uniformly in $y$ for $z=x+\mathtt{i}\hskip 0.56905pty\in\mathbf{C}$) and the Riemann zeta-function $\zeta(s)$ (as $s=1$ is the unique, simple pole of $\zeta(s)$ and $\displaystyle{\lim_{\sigma\to+\infty}\zeta(s)=1}$ uniformly in $t$ for $s=\sigma+\mathtt{i}\hskip 0.56905ptt\in\mathbf{C}$).

Denote by $\mathscr{M}_{2}$ the space of meromorphic functions in $\mathbf{C}$ that are of finite order and satisfy $\displaystyle{\lim_{x\to+\infty}f(z)=c}$ with $c\in\mathbf{C}\cup\left\{\infty\right\}$ uniformly in $y$ for $z=x+\mathtt{i}\hskip 0.56905pty\in\mathbf{C}$, and $\mathscr{H}_{2}$ the space of such entire functions in $\mathbf{C}$. Then, we have just observed a general result as follows.

Proposition 4.4**.**

$f,g\in\mathscr{M}_{2}$* are identical under “$2$ CM” value sharing condition and $f,g\in\mathscr{H}_{2}$ are identical under “$1$ CM” value sharing condition.*

Finally, we shall focus on the Riemann zeta-function $\zeta(s)$ and reconsider Theorems 2.2 and 2.3; our results will cover the case $c=0$ left open in Theorem 1.2 of [14].

Theorem 4.5**.**

Let $a,b,c\in\mathbf{C}$ be finite and distinct, and let $f$ be a meromorphic function in $\mathbf{C}$ of finite order. When $f,\zeta$ share the finite values $a,b$ counting multiplicity, with $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ ignoring multiplicity, except possibly at finitely many points, then $f=\zeta$.

Proof.

Consider the auxiliary function $F(z)$ in (2.1); since $f,\zeta$ share $a,b$ CM except possibly at finitely many points and $f$ has finite order, there exists a polynomial $P$ and a rational function $R$ such that

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We claim $\deg P=0$. Otherwise, put $\displaystyle{P(s)=a_{m}s^{m}+a_{m-1}s^{m-1}+\cdots+a_{0}}$ with $m\cdot a_{m}\neq 0$, where $a_{j}=\alpha_{j}+\mathtt{i}\hskip 0.56905pt\tilde{\alpha}_{j}\in\mathbf{C}$ with real numbers $\alpha_{j},\tilde{\alpha}_{j}$ for $j=0,1,\ldots,m$. Then, one has

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By Proposition 2.1, $\zeta-c$ has infinitely many zeros $\displaystyle{\varpi_{n}=r_{n}e^{\mathtt{i}\hskip 0.56905pt\theta_{n}}}$ in the vertical strip $Z_{V}$. We assume without loss of generality $f(\varpi_{n})=c$, since $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ except possibly at finitely many points. Substitute $\varpi_{n}$ into (4.4) and take absolute value to observe

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Noticing the special form of $Z_{V}$, one may assume that $r_{n}\to+\infty$ and $\theta_{n}\to\frac{\pi}{2}$ when $n\to+\infty$. So, (4.6) further leads to

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On the other hand, $\zeta-c$ has infinitely many zeros $\displaystyle{\omega_{n}=\tilde{r}_{n}e^{\mathtt{i}\hskip 0.56905pt\tilde{\theta}_{n}}}$ in the horizontal strip $Z_{H}$. We without loss of generality assume $f(\omega_{n})=c$, so that one similarly derives

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Noticing the special form of $Z_{H}$, one may assume that $\tilde{r}_{n}\to+\infty$ and $\tilde{\theta}_{n}\to\pi$ when $n\to+\infty$. So, (4.8) further leads to

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(4.7) and (4.9) combined yields that $\alpha_{m}=0$ and $m\geq 2$ is an even integer.

The same argument as conducted above implies that for $\varpi_{n}\in Z_{V}$, one has

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while for $\omega_{n}\in Z_{H}$, one has

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(4.10) and (4.11) combined shows $\alpha_{m-1}=\tilde{\alpha}_{m-1}=0$ for the odd integer $m-1$.

Using induction for the analyses as described in (4.6)-(4.11) leads to

[TABLE]

for $j=1,2,\ldots,m$. As a result, it follows that $\alpha_{j}=0$ for $j=1,2,\ldots,m$ and $\tilde{\alpha}_{1}=\tilde{\alpha}_{3}=\cdots=\tilde{\alpha}_{m-3}=\tilde{\alpha}_{m-1}=0$. Hence, we can rewrite $P(s)$ as

[TABLE]

Write $\varpi_{n}=\beta_{n}+\mathtt{i}\hskip 0.56905pt\gamma_{n}\in Z_{V}$ for real $\beta_{n},\gamma_{n}$. Then, one may assume that $\beta_{n}\to\beta_{0}\in\left[\frac{1}{4},\frac{3}{4}\right]$ and $\gamma_{n}\to+\infty$ as $n\to+\infty$. Thus, recalling $m\geq 2$ is an even integer, one has

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which further implies $m(-1)^{\frac{m}{2}}\tilde{\alpha}_{m}\beta_{0}=0$ that is impossible because $m\cdot a_{m}\neq 0$ yet $\alpha_{m}=0$ (so $\tilde{\alpha}_{m}\neq 0$) and $\beta_{0}\in\left[\frac{1}{4},\frac{3}{4}\right]$. This contradiction can be interpreted as saying that $a_{m}=a_{m-1}=\cdots=a_{1}=0$; that is, $P=a_{0}$ or $\deg P=0$. As a consequence, we have

[TABLE]

As $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ except possibly at finitely many points, and $\zeta-c=0$ has infinitely many roots, we must have $R(s)\equiv 1$ noticing $R$ is rational. It then follows that $f=\zeta$. ∎

Corollary 4.6**.**

Let $a\neq c\in\mathbf{C}$ be finite, and let $f$ be a finite order meromorphic function in $\mathbf{C}$ having only finitely many poles. When $f,\zeta$ share $a$ counting multiplicity, with $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ ignoring multiplicity, except possibly at finitely many points, then $f=\zeta$.

Proof.

Consider the auxiliary function $F_{1}(z)$ in (2.10). Then, one has

[TABLE]

Here, $P$ is a polynomial and $R$ is a rational function.

Next, one can exploit exactly the same analyses as described in the proof of Theorem 4.5 to deduce that $P$ is simply a constant. We then have

[TABLE]

As $\displaystyle{\zeta^{-1}(c)\subseteq f^{-1}(c)}$ except possibly at finitely many points, and $\zeta-c=0$ has infinitely many roots, we must have $R(s)\equiv 1$ noticing $R$ is rational. It then follows that $f=\zeta$. ∎