A Unified Proof of the Extended, Generalized, and Grand Riemann Hypothesis

1. Introduction

The Riemann Hypothesis (RH) is proved in Ref.[1] based on a new expression of the completed zeta function $\xi \left(s\right)$, which was obtained through pairing the conjugate zeros ${\rho }_i$ and ${\overline{\rho }}_i$ in the Hadamard product expression, with consideration of the multiplicities of zeros, i.e.

$$\xi \left(s\right)=\xi \left(0\right)\prod _{\rho }(1-{\displaystyle \frac{s}{\rho }})=\xi \left(0\right){\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})=\xi \left(0\right){\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}$$

where $\xi \left(0\right)={\displaystyle \frac{1}{2}}$, ${\rho }_i={\alpha }_i+j{\beta }_i$, ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$, with $0<{\alpha }_i<1$, ${\beta }_i\ne 0$, $0<|{\beta }_1|\le |{\beta }_2|\le \dots$, and $m_i\ge 1$ is the multiplicity of ${\rho }_i$.

It should be noted that in this article and Ref.[1], j is used to denote the imaginary unit ($j^2=-1$), while i serves as a natural number index.

Lemma 8 is the key lemma to the proof of the RH in Ref.[1]. The key points include the divisibility contained in the variant of functional equation $\xi \left(s\right)=\xi (1-s)$ and the uniqueness of the multiplicity $m_i$ of zero ${\rho }_i$ (although it is unknown), we actually obtain

$$\xi \left(s\right)=\xi (1-s)\iff (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i}=(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i},i=1,2,3,\dots$$

which is further equivalent to

$${\alpha }_i={\displaystyle \frac{1}{2}},0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots ,i=1,2,3,\dots$$

Here we give the details of Lemma 8 as the base of subsequent discussions.

Lemma 8^[1]: Given two entire functions represented as absolutely convergent (on the whole complex plane) infinite products of polynomial factors

$$f\left(s\right)={\displaystyle \prod _{i=1}^{\infty }}(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i}$$

(1)

and

$$f(1-s)={\displaystyle \prod _{i=1}^{\infty }}(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i}$$

(2)

where s is the complex variable, ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of the completed zeta function $\xi \left(s\right)$, $0<{\alpha }_i<1$ and ${\beta }_i\ne 0$ are real numbers, $0<|{\beta }_1|\le |{\beta }_2|\le |{\beta }_3|\le \dots$, $m_i\ge 1$ is the multiplicity of quadruplets of zeros $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$.

Then we have

$$f\left(s\right)=f(1-s)\iff \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{1}{2}}\hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots \hfill \\ & i=1,2,3,\dots \hfill \end{array}\right.$$

(3)

The other related Lemmas in Ref.[1] are also provided in the following.

Lemma 1^[1]: Non-trivial zeroes of $\zeta \left(s\right)$, noted as $\rho =\alpha +j\beta$, have the following properties

1)

The number of non-trivial zeroes is infinity;

2)

$\beta \ne 0$;

3)

$0<\alpha <1$;

4)

$\rho ,\overline{\rho },1-\overline{\rho },1-\rho$ are all non-trivial zeroes.

Lemma 2^[1]: The zeros of $\xi \left(s\right)$ coincide with the non-trivial zeros of $\zeta \left(s\right)$.

Lemma 3^[1]: Let $m\left(x\right),g_1\left(x\right),...,g_n\left(x\right)\in \mathbb{R}\left[x\right],n\ge 2$. If $m\left(x\right)$ is irreducible (prime) and divides the product $g_1\left(x\right)\dots g_n\left(x\right)$, then $m\left(x\right)$ divides one of the polynomials $g_1\left(x\right),...,g_n\left(x\right)$.

Lemma 4^[1]: Let $f\left(x\right),m\left(x\right)\in \mathbb{R}\left[x\right]$. If $m\left(x\right)$ is irreducible and $f\left(x\right)$ is any polynomial, then either $m\left(x\right)$ divides $f\left(x\right)$ or $gcd\left(m\right(x),f(x\left)\right)=1$.

Lemma 5^[1]: Let $m\left(x\right),g_1\left(x\right),g_2\left(x\right),\dots \in \mathbb{R}\left[x\right]$. If $m\left(x\right)$ is irreducible and divides the infinite product ${\displaystyle {\prod }_{i=1}^{\infty }}{g}_i\left(x\right)$, then $m\left(x\right)$ divides one of the polynomials $g_1\left(x\right),g_2\left(x\right),\dots$.

Lemma 6^[1]: Let $f\left(s\right)$ be a non-zero entire function, and let $s_0$ be a zero of $f\left(s\right)$. Then the multiplicity of $s_0$ is a finite positive integer.

Lemma 7^[1]: Let $f\left(s\right)$ be a non-zero entire function, and let $s_0$ be a zero of $f\left(s\right)$. Then the multiplicity of $s_0$ is unique.

Lemma 9^[1]: The infinite product ${\displaystyle {\prod }_{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}$ converges to a non-zero constant, given the conditions: $0<{\alpha }_i<1,{\beta }_i\ne 0,{\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{{\displaystyle {\beta }_i^2}}}<\infty$, and $m_i\ge 1$ is the multiplicity of zero ${\alpha }_i+j{\beta }_i$.

Remark: Lemmas 1-4 are the well-known results summarized from related journal papers, or textbooks/monographs. Lemmas 5-9 are proved in Ref.[1].

2. Generalization of Lemma 8

As pointed in Ref.[2] (on page 57), we can enumerate the nontrivial zeros of the zeta function in order of the increasing absolute value of their imaginary parts, where zeros whose imaginary parts have the same absolute value are arranged arbitrarily. Thus we remove the ordering of $|{\beta }_i|$, $|{\beta }_1|\le |{\beta }_2|\le \dots$, as condition hereafter for simplicity.

Generalization 1 of Lemma 8: The following Theorem 1 extends the critical strip from $0<\Re \left(s\right)<1$ in Lemma 8 to $0<\Re \left(s\right)<k$, $k>0$ is a constant real number.

Theorem 1: Given two entire functions represented as absolutely convergent (on the whole complex plane) infinite products of polynomial factors

$$F\left(s\right)={\displaystyle \prod _{i=1}^{\infty }}(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i}$$

(4)

and

$$F(k-s)={\displaystyle \prod _{i=1}^{\infty }}(1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i}$$

(5)

where s is a complex variable, ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of $F\left(s\right)$, $0<{\alpha }_i<k$, $k>0$, and ${\beta }_i\ne 0$ are real numbers, $m_i\ge 1$ is the multiplicity of quadruplets of zeros $({\rho }_i,{\overline{\rho }}_i,k-{\rho }_i,k-{\overline{\rho }}_i)$, ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$.

Then we have

$$F\left(s\right)=F(k-s)\iff \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{k}{2}},i=1,2,3,\dots \hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots \hfill \end{array}\right.$$

(6)

Proof. 

We will prove both directions of Eq.(6).

Sufficiency (⇐):

Assume that for all i, ${\alpha }_i={\displaystyle \frac{k}{2}}$ and $0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots$.

Then we have

$$(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i}=(1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}{)}^{m_i},i=1,2,3,\dots$$

(7)

Taking infinite products on both sides of Eq.(7), the sufficiency of Eq.(6) is proven.

Necessity (⇒):

According to the definition of divisibility of entire functions [3,4] (or more specifically the definition that a polynomial divides an entire function expressed as infinite product of polynomial factors ^[1]), the functional equation $F\left(s\right)=F(k-s)$ implies that each polynomial factor on either side must divide the infinite product on the opposite side, i.e.,

$$F\left(s\right)=F(k-s)\Rightarrow \left\{\begin{array}{cc}& (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}})\mid F(k-s)\hfill \\ \hfill & (1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}})\mid F\left(s\right)\hfill \\ \hfill & i=1,2,3,\dots \hfill \end{array}\right.$$

(8)

where "∣" is the divisible sign.

Since $1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}$ (with discriminant $\Delta =-4·{\displaystyle \frac{1}{{\displaystyle {\beta }_i^2}}}<0$) and $1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}$ (with discriminant $\Delta =-4·{\displaystyle \frac{1}{{\displaystyle {\beta }_i^2}}}<0$) are irreducible over the field $\mathbb{R}$, then by Lemma 5, Eq.(8) yields:

$$\left\{\begin{array}{cc}& (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}})\mid (1+{\displaystyle \frac{{(k-s-{\alpha }_l)}^2}{{\displaystyle {\beta }_l^2}}})\hfill \\ \hfill & (1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}})\mid (1+{\displaystyle \frac{{(s-{\alpha }_l)}^2}{{\displaystyle {\beta }_l^2}}})\hfill \\ \hfill & i=1,2,3,\dots ;l=1,2,3,\dots \hfill \end{array}\right.$$

(9)

For the special kind of polynomials in Eq.(9), "divisible" means "equal", which can be verified by comparing the like terms in equation $(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}})=k^{\prime }(1+{\displaystyle \frac{{(k-s-{\alpha }_l)}^2}{{\displaystyle {\beta }_l^2}}}),k^{\prime }\ne 0\in \mathbb{R}$ to get $k^{\prime }=1$. Further, due to the uniqueness of the multiplicity $m_i$, the only solution to Eq.(9) is $i=l$, otherwise, new duplicated zeros with ${\alpha }_l=k-{\alpha }_i,k-{\alpha }_l={\alpha }_i,{\displaystyle {\beta }_i^2}={\displaystyle {\beta }_l^2},l\ne i$ would be generated to change $m_i$. Therefore we have from Eq.(9):

$$(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}})=(1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\beta }_i^2}}}),i=1,2,3,\dots$$

(10)

By comparing the like terms in the above polynomial equation, we obtain ${\alpha }_i={\displaystyle \frac{k}{2}}$. Further, to ensure the uniqueness of $m_i$ while ${\alpha }_i={\displaystyle \frac{k}{2}}$, we need limit the ${\beta }_i$ values to be distinct, i.e., $0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots$.

Thus the necessity of Eq.(6) is proven.

That completes the proof of Theorem 1. □

Remark: Lemma 8 is a special case of Theorem 1 (with $k=1$). Consequently, the proof of Theorem 1 also provides a simpler alternative proof for Lemma 8.

Generalization 2 of Lemma 8: The following Theorem 2 will extend the polynomial factor expression in Lemma 8 to several other variants.

Theorem 2: Given two entire functions represented as absolutely convergent (on the whole complex plane) infinite products of polynomial factors

$$\begin{array}{cc}\hfill G\left(s\right)& ={\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}\hfill \\ \hfill & ={\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}}{)}^{m_i}(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}}{)}^{m_i}\hfill \\ \hfill & ={\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})\hfill \end{array}$$

(11)

and

$$\begin{array}{cc}\hfill G(k-s)& ={\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}\hfill \\ \hfill & ={\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{k-s}{{\rho }_i}}{)}^{m_i}(1-{\displaystyle \frac{k-s}{{\overline{\rho }}_i}}{)}^{m_i}\hfill \\ \hfill & ={\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{k-s}{{\rho }_i}})\hfill \end{array}$$

(12)

where s is a complex variable, ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of $G\left(s\right)$, $0<{\alpha }_i<k$, $k>0$, and ${\beta }_i\ne 0$ are real numbers, $m_i\ge 1$ is the multiplicity of quadruplets of zeros $({\rho }_i,{\overline{\rho }}_i,k-{\rho }_i,k-{\overline{\rho }}_i)$, ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$.

Then we have

$$G\left(s\right)=G(k-s)\iff \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{k}{2}},i=1,2,3,\dots \hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots \hfill \end{array}\right.$$

(13)

Proof. 

According to Theorem 1 and Lemma 9, we have

$$\begin{array}{cc}\hfill & F\left(s\right)=F(k-s)\hfill \\ \hfill & \iff \hfill \\ \hfill & {\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}F\left(s\right)={\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}F(k-s)\hfill \\ \hfill & \iff \hfill \\ \hfill & {\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}={\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\displaystyle {\beta }_i^2}}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}{)}^{m_i}\hfill \\ \hfill & \iff \hfill \\ \hfill & G\left(s\right)=G(k-s)\hfill \end{array}$$

(14)

Then we know that

$$G\left(s\right)=G(k-s)\iff \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{k}{2}},i=1,2,3,\dots \hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots \hfill \end{array}\right.$$

(15)

That completes the proof of Theorem 2. □

Generalization 3 of Lemma 8: In this case, we actually intend to lay a foundation for the study of zero distribution of completed L-functions.

Theorem 3: Given two entire functions represented by their Hadamard products:

$$\Lambda (\lambda ,s)=e^{A\left(\lambda \right)+B\left(\lambda \right)s}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})e^{{\displaystyle \frac{s}{{\rho }_i}}}$$

(16)

and

$$\Lambda (\overline{\lambda },k-s)=e^{A\left(\overline{\lambda }\right)+B\left(\overline{\lambda }\right)(k-s)}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{k-s}{{\rho }_i}})e^{{\displaystyle \frac{k-s}{{\rho }_i}}}$$

(17)

where s is a complex variable, $\lambda$ denotes a mathematical object (e.g., Dirichlet character, modular form, automorphic representation), $\overline{\lambda }$ is the dual of $\lambda$, ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of $\Lambda (\lambda ,s)$, $0<{\alpha }_i<k$, $k>0$, and ${\beta }_i\ne 0$ are real numbers, ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$.

Then we have

$$\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)\Rightarrow \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{k}{2}},i=1,2,3,\dots \hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots \hfill \end{array}\right.$$

(18)

where $\epsilon \left(\lambda \right)$ is a complex number of absolute 1, called the "root number" of L-function $L(\lambda ,s)$ (see Ref.[5] on page 94 for more details).

Proof. 

First, we have

$$\begin{array}{cc}\hfill \Lambda (\lambda ,s)& =e^{A\left(\lambda \right)+B\left(\lambda \right)s}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})e^{{\displaystyle \frac{s}{{\rho }_i}}}\hfill \\ \hfill & =e^{A\left(\lambda \right)+B\left(\lambda \right)s}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})e^{{\displaystyle \frac{2{\alpha }_{i}s}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}}\hfill \\ \hfill & =e^{A\left(\lambda \right)+B\left(\lambda \right)s}{e}^{s·{\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{2{\alpha }_i}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})\hfill \end{array}$$

(19)

Noticing that ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{2{\alpha }_i}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}<2k·{\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$, then we have ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{2{\alpha }_i}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}=c,c\in \mathbb{R},c\ne 0$. With further consideration of the multiplicity $m_i\ge 1$ of each zero, we obtain

$$\Lambda (\lambda ,s)=e^{A\left(\lambda \right)+\left[B\right(\lambda )+c]s}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}}{)}^{m_i}(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}}{)}^{m_i}$$

(20)

Accordingly

$$\Lambda (\overline{\lambda },k-s)=e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c](k-s)}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{k-s}{{\rho }_i}}{)}^{m_i}(1-{\displaystyle \frac{k-s}{{\overline{\rho }}_i}}{)}^{m_i}$$

Then we conclude that Theorem 3 is true according to Theorem 2, considering $e^{A\left(\lambda \right)+\left[B\right(\lambda )+c]s}$ and $\epsilon \left(\lambda \right)e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c](k-s)}$ have no zeros, thus both of them have no effect on the complex zeros related divisibility in the functional equation $\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$.

That completes the proof of Theorem 3. □

Generalization 4 of Lemma 8: In this case, we make further efforts to lay a foundation for the study of completed L-functions that possess both real and complex non-trivial zeros, denoted by $\rho \in {\mathcal{Z}}_{\mathrm{real}}$ ($\Im \left(\rho \right)=0$) and $\rho \in {\mathcal{Z}}_{\mathrm{complex}}$ ($\Im \left(\rho \right)\ne 0$), respectively. When these two zero sets have no common elements, we express their disjointness by: ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀$.

The reason we need to consider this case is that, so far, we cannot rule out the existence of exceptional zeros (or Landau-Siegel zeros), although their numbers are very limited even if they do exist.

To be specific, denote the set of real zeros in the critical strip as ${\mathcal{Z}}_{\mathrm{real}}=\left\{a_n\in \mathbb{R}\mid 0<a_n<k,n=1,2,\dots ,N\right\},$ where N is a finite natural number. This finiteness follows from the Identity Theorem, which implies that any non-zero entire function cannot have infinitely many zeros in a bounded region, as zeros of such functions are isolated and cannot accumulate.

Theorem 4: Given two entire functions represented by their Hadamard products:

$$\begin{array}{cc}\hfill \Lambda (\lambda ,s)& =e^{A\left(\lambda \right)+B\left(\lambda \right)s}\prod _{\rho }(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}\hfill \\ \hfill & =e^{A\left(\lambda \right)+B\left(\lambda \right)s}\prod _{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}\prod _{\rho \in {\mathcal{Z}}_{\mathrm{complex}}}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}\hfill \\ \hfill & =e^{A\left(\lambda \right)+\left[B\right(\lambda )+c]s}{\displaystyle \prod _{n=1}^N}(1-{\displaystyle \frac{s}{a_n}})e^{{\displaystyle \frac{s}{a_n}}}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})\hfill \end{array}$$

(21)

and

$$\begin{array}{cc}\hfill \Lambda (\overline{\lambda },k-s)& =e^{A\left(\overline{\lambda }\right)+B\left(\overline{\lambda }\right)(k-s)}\prod _{\rho }(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}\hfill \\ \hfill & =e^{A\left(\overline{\lambda }\right)+B\left(\overline{\lambda }\right)(k-s)}\prod _{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}\prod _{\rho \in {\mathcal{Z}}_{\mathrm{complex}}}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}\hfill \\ \hfill & =e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c](k-s)}{\displaystyle \prod _{n=1}^N}(1-{\displaystyle \frac{k-s}{a_n}})e^{{\displaystyle \frac{k-s}{a_n}}}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{k-s}{{\rho }_i}})(1-{\displaystyle \frac{k-s}{{\overline{\rho }}_i}})\hfill \end{array}$$

(22)

where s is a complex variable, $c={\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{2{\alpha }_i}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}$, $\lambda$ denotes a mathematical object, $\overline{\lambda }$ is the dual of $\lambda$, ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ are the complex conjugate zeros of $\Lambda (\lambda ,s)$, $0<{\alpha }_i<k$, $0<a_n<k$, $k>0$, and ${\beta }_i\ne 0$ are real numbers, ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$, ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀.$

Then we have

$$\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)\Rightarrow \left\{\begin{array}{cc}& {\alpha }_i={\displaystyle \frac{k}{2}},i=1,2,3,\dots \hfill \\ & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots \hfill \\ \hfill & a_n={\displaystyle \frac{k}{2}},n=1\hfill \end{array}\right.$$

(23)

i.e., all the zeros (both real and complex) of $\Lambda (\lambda ,s)$ lie on the critical line.

Proof. 

By Theorem 3, to determine the distribution of the complex zeros of $\Lambda (\lambda ,s)$, we only need to show that the newly added parts ${\prod }_{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}$ and ${\prod }_{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}$ do not affect the complex zeros related divisibility in the functional equation $\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$, which is an obvious fact since the given condition ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀$ means that ${\prod }_{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}$ and ${\prod }_{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}$ are co-prime according to Ref.[3] (on pages 174, 208) and Ref.[4] (see its THEOREM 4).

Thus, we conclude by Theorem 3 that

$$\left\{\begin{array}{cc}\hfill & {\alpha }_i={\displaystyle \frac{k}{2}},i=1,2,3,\dots \hfill \\ \hfill & 0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots \hfill \end{array}\right.$$

(24)

Next, we consider the real zeros of $\Lambda (\lambda ,s)$.

By canceling the complex non-trivial zeros related polynomial factors on both sides of $\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$, we have

$$e^{A\left(\lambda \right)+\left[B\right(\lambda )+c]s}{\displaystyle \prod _{n=1}^N}(1-{\displaystyle \frac{s}{a_n}})e^{{\displaystyle \frac{s}{a_n}}}=\epsilon \left(\lambda \right)e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c](k-s)}{\displaystyle \prod _{n=1}^N}(1-{\displaystyle \frac{k-s}{a_n}})e^{{\displaystyle \frac{k-s}{a_n}}}$$

(25)

Further Eq.(25) is equivalent to

$$e^{A\left(\lambda \right)+[B\left(\lambda \right)+c^{\prime }]s}{\displaystyle \prod _{n=1}^N}(s-a_n)=\epsilon \left(\lambda \right)e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c^{\prime }](k-s)}{\displaystyle \prod _{n=1}^N}(s-(k-a_n))$$

(26)

where $c^{\prime }=c+{\displaystyle {\sum }_{n=1}^N}{\displaystyle \frac{1}{a_n}}$.

Suppose the multiplicity of zero $s=a_n$ is $m_n$ ($m_n\ge 1$) that is finite and unique although unknown. Then Eq.(26) becomes

$$e^{A\left(\lambda \right)+[B\left(\lambda \right)+c^{\prime }]s}{\displaystyle \prod _{t=1}^T}(s-a_t{)}^{m_t}=\epsilon \left(\lambda \right)e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c^{\prime }](k-s)}{\displaystyle \prod _{t=1}^T}(s-(k-a_t){)}^{m_t}$$

(27)

where ${\displaystyle {\sum }_{t=1}^T}{m}_t=N$.

Considering $(s-a_t)$ and $(s-(k-a_t))$ are irreducible over $\mathbb{R}$, then by Lemma 3, Eq.(27) means

$$\left\{\begin{array}{cc}& (s-a_t)\mid (s-(k-a_m)\hfill \\ \hfill & (s-(k-a_t))\mid (s-a_m)\hfill \\ \hfill & t=1,2,3,\dots ,T;m=1,2,3,\dots ,T\hfill \end{array}\right.$$

(28)

The only solution to Eq.(28) is $t=m,a_t={\displaystyle \frac{k}{2}},t=1,\dots ,T$, otherwise the uniqueness of $m_t$ would be violated with $a_t+a_m=k,t\ne m$. To avoid changing the multiplicity of $m_t$ while $a_t={\displaystyle \frac{k}{2}}$, we need to limit $T=1$. Thus we get

$$a_t={\displaystyle \frac{k}{2}},t=1$$

(29)

where zero $s=a_1={\displaystyle \frac{k}{2}}$ with multiplicity $m_1=N$ .

Putting Eq.(24) and Eq.(29) together, we proved Eq.(23).

That completes the proof of Theorem 4. □

Remark: As pointed out in Ref.[5] (on page 102), if ${\rho }_i={\alpha }_i+j{\beta }_i$ is a zero of $\Lambda (\lambda ,s)$, then ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ is a zero of $\Lambda (\overline{\lambda },s)$. Therefore, to use Theorem 4 while $\lambda \ne \overline{\lambda }$, we need to construct a new symmetric functional equation $\Lambda (\overline{\lambda },s)\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\epsilon \left(\overline{\lambda }\right)\Lambda (\lambda ,k-s)\Lambda (\overline{\lambda },k-s)$ to ensure that the conjugate zeros appear together. For more details, see the proofs of Theorem 5 and Theorem 7.

Actually, Theorem 4 provides a unified proof framework for the Extended Riemann Hypothesis, the Generalized Riemann Hypothesis, and the Grand Riemann Hypothesis.

3. The Applications of Theorem 4

We will make use of Theorem 4 to prove the Extended Riemann Hypothesis (for Dedekind zeta function), the Generalized Riemann Hypothesis (for Dirichlet L-function), and the Grand Riemann Hypothesis (for modular form L-Function, automorphic L-function, and etc.).

To facilitate the subsequent discussion, we give the details of the Extended Riemann Hypothesis, the Generalized Riemann Hypothesis, and the Grand Riemann Hypothesis. In the following contents, the critical line means: $\mathfrak{R}\left(s\right)={\displaystyle \frac{1}{2}}$, or more generally, $\mathfrak{R}\left(s\right)={\displaystyle \frac{k}{2}}$, $k>0$ is a constant real number.

The Generalized Riemann Hypothesis: The nontrivial zeros of Dirichlet L-functions lie on the critical line.

The Extended Riemann Hypothesis: The nontrivial zeros of Dedekind zeta function lie on the critical line.

The Grand Riemann Hypothesis: The nontrivial zeros of all L-functions lie on the critical line.

To begin with, we provide a general property of L-functions, which was labeled Lemma 5.5 in Ref.[5] on page 101.

Lemma 5.5^[5]: Let $L(f,s)$ be an L-function. All zeros $\rho$ of $\Lambda (f,s)$ are in the critical strip $0\le \sigma \le 1$. For any $ϵ>0$, we have

$$\sum _{\rho \ne 0,1}{\left|\rho \right|}^{-1-ϵ}<+\infty .$$

where, $\Lambda (f,s)$ is the completed L-function corresponding to $L(f,s)$, $\sigma$ is the real part of $\rho$, and f is identical to $\lambda$ in this paper as a symbol representing a mathematical object (e.g., Dirichlet character, modular form, automorphic representation).

Another general property of L-functions is as follows.

The zeros of $\Lambda (f,s)$ are precisely the non-trivial zeros of $L(f,s)$, as the trivial zeros of $L(f,s)$ are canceled by the poles of the Gamma factors in the completion process (see Ref.[5] on page 96 for more details).

Thus, we can discuss the non trivial zeros of L-functions based on the zeros of the corresponding completed L-functions.

3.1. Dirichlet L-Function

Definition: The Dirichlet L-function associated with a Dirichlet character $\chi$ modulo q is defined for $\Re \left(s\right)>1$ by the series:

$$L(\chi ,s)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\chi \left(n\right)}{n^s}}$$

(30)

For the principal character ${\chi }_0$ (where ${\chi }_0\left(n\right)=1$ if $gcd(n,q)=1$ and ${\chi }_0\left(n\right)=0$ otherwise), the L-function is related to the Riemann zeta function by:

$$L({\chi }_0,s)=\zeta \left(s\right)\prod _{p|q}\left(1-{\displaystyle \frac{1}{p^s}}\right)$$

(31)

Completed L-function: The completed Dirichlet L-function is defined as:

$$\Lambda (\chi ,s)={\left({\displaystyle \frac{q}{\pi }}\right)}^{{\displaystyle \frac{s+a}{2}}}\Gamma \left({\displaystyle \frac{s+a}{2}}\right)L(\chi ,s)$$

(32)

where $a=0$ if $\chi (-1)=1$ (even character) and $a=1$ if $\chi (-1)=-1$ (odd character).

Functional Equation: The completed Dirichlet L-function satisfies the functional equation:

$$\Lambda (\chi ,s)=W\left(\chi \right)\Lambda (\overline{\chi },1-s)$$

(33)

where $W\left(\chi \right)$ is the Gauss sum:

$$W\left(\chi \right)={\displaystyle \frac{\tau \left(\chi \right)}{i^a\sqrt{q}}}$$

(34)

and $\tau \left(\chi \right)={\displaystyle {\sum }_{n=1}^q}\chi \left(n\right)e^{2\pi in/q}$ is the Gauss sum associated with $\chi$.

Hadamard Product: For non-principal characters $\chi$, the completed L-function $\Lambda (\chi ,s)$ is an entire function and has the Hadamard product:

$$\Lambda (\chi ,s)=e^{A\left(\chi \right)+B\left(\chi \right)s}\prod _{\rho }\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }$$

(35)

where the product is over all zeros $\rho$ of $\Lambda (\chi ,s)$, and $A\left(\chi \right)$ and $B\left(\chi \right)$ are constants depending on $\chi$.

Next we prove the Generalized Riemann Hypothesis.

Theorem 5: The nontrivial zeros of the above-described Dirichlet L-functions lie on the critical line.

Remark: We only need to prove that all the zeros of the completed Dirichlet L-function $\Lambda (\chi ,s)$ have real part ${\displaystyle \frac{1}{2}}$, i.e., all the zeros of $\Lambda (\chi ,s)$ lie on the critical line.

Proof. 

We conduct the proof in two cases.

CASE 1: $\chi =\overline{\chi }$ (self-dual)

It suffices to verify that the properties of $\Lambda (\chi ,s)$ with $\chi =\overline{\chi }$ match the conditions of Theorem 4 with $\lambda =\chi$, $\epsilon \left(\lambda \right)=W\left(\chi \right)$, $k=1$. Eq.(35) is equivalent to Eq.(21) by separating all zeros into two sets ${\mathcal{Z}}_{\mathrm{real}}$ and ${\mathcal{Z}}_{\mathrm{complex}}$. Actually, to restrict $\chi =\overline{\chi }$ is to guarantee that the conjugate zeros of $\Lambda (\chi ,s)$ appear in pairs, and then the quadruplets of non-trivial zeros $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$ with their multiplicities appear together according to Eq.(33). The condition ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ can be assured by Lemma 5.5, considering that ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}$ is a subseries of ${\sum }_{\rho \ne 0,1}{\displaystyle \frac{1}{{\left|\rho \right|}^2}}$; The condition ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀$ holds because ${\mathcal{Z}}_{\mathrm{real}}$ and ${\mathcal{Z}}_{\mathrm{complex}}$ are mutually exclusive sets, i.e., if $\rho \in {\mathcal{Z}}_{\mathrm{real}}$, then $\Im \left(\rho \right)=0$; if $\rho \in {\mathcal{Z}}_{\mathrm{complex}}$, then $\Im \left(\rho \right)\ne 0$.

Therefore, by Theorem 4 with $\lambda =\chi$, $\epsilon \left(\lambda \right)=W\left(\chi \right)$, $k=1$, we know that both the real (if exists) and the complex zeros of $\Lambda (\chi ,s)$ with $\chi =\overline{\chi }$ lie on the critical line.

CASE 2: $\chi \ne \overline{\chi }$

In this case, the conjugate non-trivial zeros do not appear together in Eq.(35), because if $\rho$ is a zero of $\Lambda (\chi ,s)$, then $\overline{\rho }$ is a zero of $\Lambda (\overline{\chi },s)$.

Thus, we need to extend Eq.(33) to another form, i.e.,

$$\Lambda (\overline{\chi },s)=W\left(\overline{\chi }\right)\Lambda (\chi ,1-s)$$

(36)

Combining (36) with (33), we get a new functional equation

$$\Lambda (\overline{\chi },s)\Lambda (\chi ,s)=W\left(\overline{\chi }\right)W\left(\chi \right)\Lambda (\chi ,1-s)\Lambda (\overline{\chi },1-s)$$

(37)

Both sides of Eq.(37) are the products of entire functions, thus they are still entire functions. And we know that the conjugate zeros of $\Lambda (\overline{\chi },s)\Lambda (\chi ,s)$ appear in pairs, and then the quadruplets of non-trivial zeros $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$ with their multiplicities appear together according to Eq.(37). Further, based on Eq.(35), we have

$$\begin{array}{cc}\hfill \Lambda (\chi ,s)\Lambda (\overline{\chi },s)& =e^{A\left(\chi \right)+A\left(\overline{\chi }\right)+[B\left(\chi \right)+B\left(\overline{\chi }\right)+c]s}\prod _{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}{\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})\hfill \end{array}$$

(38)

where $c={\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{2{\alpha }_i}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}$.

The condition ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ and condition ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀$ hold for the same reasons as in CASE 1.

Therefore, by Theorem 4, we know that both the real (if exists) and the complex zeros of $\Lambda (\chi ,s)\Lambda (\overline{\chi },s)$ (thus of $\Lambda (\chi ,s)$) with $\chi \ne \overline{\chi }$ lie on the critical line.

Combining CASE 1 and CASE 2, we conclude that Theorem 5 holds as a specific case of Theorem 4 with $k=1$ and $\lambda =\chi$. □

3.2. Dedekind Zeta Function

Definition: For a number field K with ring of integers ${\mathcal{O}}_K$, the Dedekind zeta function is defined for $\Re \left(s\right)>1$ by:

$${\zeta }_K\left(s\right)=\sum _{\mathfrak{a}}{\displaystyle \frac{1}{N{\left(\mathfrak{a}\right)}^s}}$$

(39)

where the sum is over all non-zero ideals $\mathfrak{a}$ of ${\mathcal{O}}_K$, and $N\left(\mathfrak{a}\right)$ is the norm of the ideal.

Completed Zeta Function: The completed Dedekind zeta function is defined as:

$${\Lambda }_K\left(s\right)={\left|D_K\right|}^{s/2}{\left({\pi }^{-s/2}\Gamma \left({\displaystyle \frac{s}{2}}\right)\right)}^{r_1}{\left({\left(2\pi \right)}^{-s}\Gamma \left(s\right)\right)}^{r_2}{\zeta }_K\left(s\right)$$

(40)

where $D_K$ is the discriminant of K, $r_1$ is the number of real embeddings of K, $r_2$ is the number of pairs of complex embeddings of K.

Functional Equation: The completed Dedekind zeta function satisfies:

$${\Lambda }_K\left(s\right)=\epsilon \left(K\right){\Lambda }_K(1-s)$$

(41)

where $\epsilon \left(K\right)=1$ for all number fields K, showing the symmetry of the functional equation.

Hadamard Product: The completed Dedekind zeta function has a simple pole at $s=1$ with residue ${\displaystyle \frac{2^{r_1}{\left(2\pi \right)}^{r_2}{h}_K{R}_K}{w_K\sqrt{|D_K|}}}$, where $h_K$ is the class number, $R_K$ is the regulator, and $w_K$ is the number of roots of unity in K. The function $s(s-1){\Lambda }_K\left(s\right)$ is entire and has the Hadamard product:

$$s(s-1){\Lambda }_K\left(s\right)=e^{A_K+B_{K}s}\prod _{\rho \ne 0,1}\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }$$

(42)

where the product is over all zeros $\rho$ of ${\Lambda }_K\left(s\right)$ except $\rho =0$ and $\rho =1$, and $A_K$ and $B_K$ are constants depending on K.

For more details of the completed Dedekind zeta function, please be referred to Ref.[5] (Chapter 5.10) and Ref.[6] (Section 10.5.1).

Theorem 6: The nontrivial zeros of the above-described Dedekind zeta function lie on the critical line.

Remark: We only need to prove that all the zeros of ${\Lambda }_K\left(s\right)$ have real part ${\displaystyle \frac{1}{2}}$, i.e., all the zeros of ${\Lambda }_K\left(s\right)$ lie on the critical line.

Proof. 

It suffices to show that the properties of ${\Lambda }_K\left(s\right)$ match the conditions of Theorem 4 with $\lambda =K$, $\epsilon \left(\lambda \right)=1$, $k=1$.

Actually, Eq.(41) and Eq.(42) guarantee that

$$e^{A_K+B_{K}s}\prod _{\rho \ne 0,1}\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }=e^{A_K+B_K(1-s)}\prod _{\rho \ne 0,1}\left(1-{\displaystyle \frac{1-s}{\rho }}\right)e^{1-s/\rho }$$

(43)

where $\prod _{\rho \ne 0,1}\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }=\prod _{\rho \in {\mathcal{Z}}_{\mathrm{real}}\setminus \{0,1\}}\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }\prod _{\rho \in {\mathcal{Z}}_{\mathrm{complex}}}\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }$.

And we know that the conjugate zeros of ${\Lambda }_k\left(s\right)$ appear in pairs, and then the quadruplets of non-trivial zeros $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$ with their multiplicities appear together according to Eq.(41). The condition ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ and condition ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀$ hold for the same reasons as in CASE 1 in the proof of Theorem 5.

Therefore, by Theorem 4 we know that both the real (if exists) and the complex zeros of ${\Lambda }_K\left(s\right)$ lie on the critical line, i.e., Theorem 6 holds as a specific case of Theorem 4 with $\lambda =K$, $\epsilon \left(\lambda \right)=1$, $k=1$. □

In the following contents, we prove the Grand Riemann Hypothesis for modular form L-Functions and automorphic L-functions, respectively. After that we will make a summarization, and conclude that the Grand Riemann Hypothesis holds for all L-functions satisfying some general properties.

3.3. Modular Form L-Function

Definition: For a modular form $f\left(z\right)={\displaystyle {\sum }_{n=1}^{\infty }}{a}_n{e}^{2\pi inz}$ of weight k for a congruence subgroup $\Gamma$, the associated L-function is defined for $\Re \left(s\right)>{\displaystyle \frac{k+1}{2}}$ by:

$$L(f,s)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{a_n}{n^s}}$$

(44)

Completed L-function: For a cusp form f of weight k for ${\Gamma }_0\left(N\right)$ (level N is any positive integer) with Nebentypus character $\chi$, the completed modular form L-function is defined as:

$$\Lambda (f,s)=N^{s/2}{\left(2\pi \right)}^{-s}\Gamma \left(s\right)L(f,s)$$

(45)

Functional Equation: For a normalized Hecke eigenform f of weight k for ${\Gamma }_0\left(N\right)$ with Nebentypus character $\chi$, the completed modular form L-function satisfies:

$$\Lambda (f,s)=\epsilon \left(f\right)\Lambda (\overline{f},k-s)$$

(46)

where $\epsilon \left(f\right)=\pm 1$ is the epsilon factor, which is the eigenvalue of f under the Atkin-Lehner involution, and $\overline{f}$ is the modular form with Fourier coefficients ${\overline{a}}_n$.

Hadamard Product: For a cusp form f, the completed L-function $\Lambda (f,s)$ is an entire function of order 1 and has the Hadamard product:

$$\Lambda (f,s)=e^{A\left(f\right)+B\left(f\right)s}\prod _{\rho }\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }$$

(47)

where the product is over all zeros $\rho$ of $\Lambda (f,s)$, and $A\left(f\right)$ and $B\left(f\right)$ are constants depending on f.

For more details of the completed modular form L-function, please be referred to Refs.[5][7].

We have the following result about the non-trivial zero distribution of modular form L-Functions.

Theorem 7: The non-trivial zeros of the above-described modular form L-Functions lie on the critical line.

Remark: We only need to prove that all the zeros of $\Lambda (f,s)$ have real part ${\displaystyle \frac{k}{2}}$, i.e., all the zeros of $\Lambda (f,s)$ lie on the critical line.

Proof. 

We conduct the proof in two cases.

CASE 1: $f=\overline{f}$ (self-dual)

It suffices to show that the properties of $\Lambda (f,s)$ with $f=\overline{f}$ match the conditions of Theorem 4 with $\lambda =f$. Eq.(47) is equivalent to Eq.(21) by separating all zeros into two sets ${\mathcal{Z}}_{\mathrm{real}}$ and ${\mathcal{Z}}_{\mathrm{complex}}$. Actually, to restrict $f=\overline{f}$ is to guarantee that the conjugate zeros of $\Lambda (f,s))$ appear in pairs. Then the quadruplets of non-trivial zeros $({\rho }_i,{\overline{\rho }}_i,k-{\rho }_i,k-{\overline{\rho }}_i)$ with their multiplicities appear together according to Eq.(46). The condition ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ and condition ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀$ hold for the same reasons as in CASE 1 in the proof of Theorem 5.

Therefore, by Theorem 4, we know that both the real (if exists) and the complex zeros of $\Lambda (f,s)$ with $f=\overline{f}$ lie on the critical line.

CASE 2: $f\ne \overline{f}$

To deal with this case $f\ne \overline{f}$, we need first to extend Eq.(46) to another form, i.e.,

$$\Lambda (\overline{f},s)=\epsilon \left(\overline{f}\right)\Lambda (f,k-s)$$

(48)

Combining (48) with (46), we get a new functional equation

$$\Lambda (\overline{f},s)\Lambda (f,s)=\epsilon \left(f\right)\epsilon \left(\overline{f}\right)\Lambda (f,k-s)\Lambda (\overline{f},k-s)$$

(49)

Obviously, both sides of Eq.(49) are the products of entire functions, thus they are still entire functions. And we know that the conjugate zeros of $\Lambda (\overline{f},s)\Lambda (f,s)$ appear in pairs, and then the quadruplets of non-trivial zeros $({\rho }_i,{\overline{\rho }}_i,k-{\rho }_i,k-{\overline{\rho }}_i)$ with their multiplicities appear together according to Eq.(49). Further, based on Eq.(47), we have

$$\Lambda (f,s)\Lambda (\overline{f},s)=e^{A\left(f\right)+A\left(\overline{f}\right)+[B\left(f\right)+B\left(\overline{f}\right)+c]s}\prod _{\rho \in {\mathcal{Z}}_{\mathrm{real}}}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}{\displaystyle \prod _{i=1}^{\infty }}\left(1-{\displaystyle \frac{s}{{\rho }_i}}\right)\left(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}}\right)$$

(50)

where $c={\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{2{\alpha }_i}{{\displaystyle {\alpha }_i^2}+{\displaystyle {\beta }_i^2}}}$.

The condition ${\displaystyle {\sum }_{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ and condition ${\mathcal{Z}}_{\mathrm{real}}\cap {\mathcal{Z}}_{\mathrm{complex}}=⌀$ hold for the same reasons as in CASE 1 in the proof of Theorem 5.

Therefore, by Theorem 4, we know that both the real (if exists) and the complex zeros of $\Lambda (f,s)\Lambda (\overline{f},s)$ (thus of $\Lambda (f,s)$) with $f\ne \overline{f}$ lie on the critical line.

Combining CASE 1 and CASE 2, we conclude that Theorem 7 holds as a specific case of Theorem 4 with $\lambda =f$. □

3.4. Automorphic L-Function

Definition: For an automorphic representation $\pi$ of ${\mathrm{GL}}_n\left({\mathbb{A}}_{\mathbb{Q}}\right)$, the associated L-function is defined for $\Re \left(s\right)>1$ by:

$$L(\pi ,s)=\prod _p{L}_p({\pi }_p,s)$$

(51)

where $L_p({\pi }_p,s)$ is the local L-factor at the prime p. For unramified ${\pi }_p$ with Satake parameters $\{{\alpha }_{1,p},\dots ,{\alpha }_{n,p}\}$,

$$L_p({\pi }_p,s)={\displaystyle \prod _{i=1}^n}{\left(1-{\displaystyle \frac{{\alpha }_{i,p}}{p^s}}\right)}^{-1}$$

(52)

Completed L-function: The completed automorphic L-function is defined as:

$$\Lambda (\pi ,s)={\displaystyle Q_{\pi }^{s/2}}{\displaystyle \prod _{i=1}^n}{\displaystyle {\Gamma }_{*}^{\left(i\right)}}(s+{\mu }_{i,\pi })·L(\pi ,s)$$

(53)

where $Q_{\pi }$ is the conductor of $\pi$, ${\mu }_{i,\pi }$ are complex numbers determined by the i-th local component of ${\pi }_{\infty }$, and

$${\displaystyle {\Gamma }_{*}^{\left(i\right)}}\left(s\right)=\left\{\begin{array}{cc}& {\Gamma }_{\mathbb{R}}\left(s\right)={\pi }^{-{\displaystyle \frac{s}{2}}}\Gamma \left({\displaystyle \frac{s}{2}}\right),\mathrm{for}\mathrm{real}\mathrm{representations}.\hfill \\ \hfill & {\Gamma }_{\mathbb{C}}\left(s\right)={\left(2\pi \right)}^{-s}\Gamma \left(s\right),\mathrm{for}\mathrm{complex}\mathrm{representations}.\hfill \end{array}\right.$$

(54)

Functional Equation: The completed automorphic L-function satisfies:

$$\Lambda (\pi ,s)=\epsilon \left(\pi \right)\Lambda (\tilde{\pi },1-s)$$

(55)

where $\tilde{\pi }$ is the contragredient representation of $\pi$ and $\epsilon \left(\pi \right)$ is the epsilon factor, a complex number of absolute value 1.

Hadamard Product: For a cuspidal automorphic representation $\pi$, the completed L-function $\Lambda (\pi ,s)$ is an entire function of order 1 and has the Hadamard product:

$$\Lambda (\pi ,s)=e^{A\left(\pi \right)+B\left(\pi \right)s}\prod _{\rho }\left(1-{\displaystyle \frac{s}{\rho }}\right)e^{s/\rho }$$

(56)

where the product is over all zeros $\rho$ of $\Lambda (\pi ,s)$, and $A\left(\pi \right)$ and $B\left(\pi \right)$ are constants depending on $\pi$.

For more details of the completed automorphic L-function, please be referred to Refs.[5,7].

We have the following result about the non-trivial zero distribution of automorphic L-Functions.

Theorem 8: The non-trivial zeros of the above-described automorphic L-Functions lie on the critical line.

Remark: We only need to prove that all the zeros of $\Lambda (\pi ,s)$ have real part ${\displaystyle \frac{1}{2}}$, i.e., all the zeros of $\Lambda (\pi ,s)$ lie on the critical line.

Proof. 

The proof procedures of Theorem 8 is similar to that of Theorem 7 with $k=1$ and f replaced by $\pi$, $\overline{f}$ replaced by $\tilde{\pi }$. Thus the proof details are omitted for simplicity. □

Actually, from the above proofs of Theorem 5, Theorem 6, and Theorem 7, we can note that each proof does not depend on the specific definition of the L-function $L(\lambda ,s)$, but rather relies on the following general properties of the corresponding completed L-function $\Lambda (\lambda ,s)$:

P1: 

Symmetric functional equation between $\Lambda (\lambda ,s)$ and $\Lambda (\overline{\lambda },k-s)$: $\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$;

P2: 

Hadamard product expression of entire function $\Lambda (\lambda ,s)$ or $s^{m_0}{(k-s)}^{m_1}\Lambda (\lambda ,s)$, $m_0\ge 1,m_1\ge 1$ are the multiplicities (orders) of poles $s=0,s=k$, respectively;

P3: 

The zeros of $\Lambda (\lambda ,s)$ are precisely the non-trivial zeros of $L(\lambda ,s)$;

P4: 

Zero distribution related items: 1) the concurrence of quadruplets of complex non-trivial zeros with their multiplicities; 2) the property stated in Lemma 5.5; 3) the disjointness of real and complex non-trivial zero sets.

Therefore we conclude that the Grand Riemann Hypothesis holds for all kinds of L-functions satisfying properties P1, P2, P3, and P4, i.e., If only the completed L-function $\Lambda (\lambda ,s)$ satisfies the requirements of P1, P2, P3, and P4, the non-trivial zeros of the corresponding L-Function $L(\lambda ,s)$ lie on the critical line.