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

1. Preliminary Lemmas

In the following contents, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{R}\left[x\right]$ denote the field of real numbers, the field of complex numbers (the complex plane), the ring of real polynomials, respectively.

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

It is also worth noting that the following fact plays an important role in this paper: for any given non-zero entire function, the multiplicities of its zeros are finite and uniquely determined, hence invariant, even though their specific values may be unknown.

Lemma 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 $g_1\left(x\right),...,g_n\left(x\right)$.

Remark 1. 

The contents of Lemma 1 can be found in many textbooks of linear algebra, modern algebra, or abstract algebra.

Lemma 2. 

Let $f\left(x\right)$ be an entire function on $\mathbb{C}$. Suppose $f\left(x\right)={\prod }_{i=1}^{\infty }{g}_i\left(x\right)$ converges absolutely on $\mathbb{C}$, where each $g_i\left(x\right)\in \mathbb{R}\left[x\right]$ is irreducible in $\mathbb{R}\left[x\right]$ of degree $d\in \{1,2\}$. If $m\left(x\right)\in \mathbb{R}\left[x\right]$ is irreducible in $\mathbb{R}\left[x\right]$ of degree d and divides $f\left(x\right)$, then $m\left(x\right)$ divides $g_i\left(x\right)$ for some $i\ge 1$.

Proof. 

Let $\alpha$ be a root of $m\left(x\right)$, i.e., $m\left(\alpha \right)=0$. Since $m\left(x\right)\mid f\left(x\right)$, we have $f\left(\alpha \right)=0$. By absolute convergence of ${\prod }_{i=1}^{\infty }{g}_i\left(x\right)$, there exists at least one index $i\in \mathbb{N}$ such that $g_i\left(\alpha \right)=0$, otherwise $g_i\left(\alpha \right)\ne 0$ for all i, then ${\prod }_{i=1}^{\infty }{g}_i\left(x\right)$ converges to a non-zero limit, contradicting $f\left(\alpha \right)=0$.

As $g_i\left(x\right)$ and $m\left(x\right)$ are irreducible over $\mathbb{R}$ with $deg\left(g_i\left(x\right)\right)=deg\left(m\left(x\right)\right)=d$, they share the root $\alpha$. Thus:

• If $d=1$, then $g_i\left(x\right)=a(x-\alpha )$ and $m\left(x\right)=b(x-\alpha )$ for $a,b\ne 0$, so $m\left(x\right)\mid g_i\left(x\right)$.
• If $d=2$, then both $g_i\left(x\right)$ and $m\left(x\right)$ have roots $\{\alpha ,\overline{\alpha }\}$, so $g_i\left(x\right)=c·m\left(x\right)$ for $c\ne 0$, hence $m\left(x\right)\mid g_i\left(x\right)$.

In both cases, $m\left(x\right)$ divides $g_i\left(x\right)$.

That completes the proof of Lemma 2. □

Remark 2. 

Lemma 2 is actually Lemma 5 in Ref. [14], which is a preprint paper by this author.

Lemma 3. 

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

Proof. 

First of all, we know that

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

where in the right side expression, $i^{th}$ factor appears $m_i$ times.

Let $a_i={\displaystyle \frac{{\alpha }_i^2}{{\alpha }_i^2+{\beta }_i^2}}$, then ${\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}=1-{\displaystyle \frac{{\alpha }_i^2}{{\alpha }_i^2+{\beta }_i^2}}=1-a_i$.

Since $0\le {\alpha }_i\le k$ and ${\beta }_i\ne 0$, we have: $0\le a_i<{\displaystyle \frac{k^2}{{\beta }_i^2}}$. Then ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$ (given condition) implies ${\sum }_{i=1}^{\infty }\left|a_i\right|={\sum }_{i=1}^{\infty }{a}_i<k^2{\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$.

Further, the absolute convergence of ${\sum }_{i=1}^{\infty }{a}_i$ guarantees that the product ${\prod }_{i=1}^{\infty }(1-a_i)={\prod }_{i=1}^{\infty }{\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}={\prod }_{i=1}^{\infty }({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}{)}^{m_i}$ converges to a non-zero constant.

That completes the proof of Lemma 3. □

2. Foundational Theorems

As pointed out in Ref. [2] (p.57), the non-trivial zeros of the zeta function can be enumerated in order of increasing absolute value of their imaginary parts; where zeros whose imaginary parts have the same absolute value are arranged arbitrarily. Consequently, the same enumeration rule applies to the zeros of entire functions in the following contents. We therefore drop the default assumption $|{\beta }_1|\le |{\beta }_2|\le \dots$ as a condition hereafter for simplicity.

Theorem 1. 

Given an entire function

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

(1)

which converges absolutely on $\mathbb{C}$.

Here

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

Then

$$F\left(s\right)=F(k-s)⟹{\alpha }_i={\displaystyle \frac{k}{2}}(\forall i)\mathrm{and}0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots .$$

(2)

Remark 3. 

It should be noted that $m_i$ is actually the multiplicity of quadruplets of zeros $({\rho }_i,{\overline{\rho }}_i,k-{\rho }_i,k-{\overline{\rho }}_i)$ under the assumption $F\left(s\right)=F(k-s)$.

Proof. 

Considering

$$(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=\underset{m_i\mathrm{times}}{\underbrace{(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})\dots }}$$

we have from $F\left(s\right)=F(k-s)$ and Equation(1)

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

(3)

where the $i^{th}$ factor appears $m_i$ times in any order.

Due to absolute convergence, both products ${\prod }_{i=1}^{\infty }\left(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}\right)$ and ${\prod }_{i=1}^{\infty }\left(1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\beta }_i^2}}\right)$ can be rearranged into any single quadratic polynomial factor times the remaining product, which stays absolutely convergent and hence defines an entire function. Then by the definition of divisibility of entire functions $\left[3\right]\left[4\right]$, Equation(3) implies that each quadratic polynomial factor on either side divides the infinite product on the opposite side, i.e.,

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

(4)

where "∣" is the divisible sign, $i\in \mathbb{N}=\{1,2,3,\dots \}$.

Both polynomials $1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}$ and $1+{\displaystyle \frac{{(k-s-{\alpha }_i)}^2}{{\beta }_i^2}}$ have discriminant $\Delta =-4·{\displaystyle \frac{1}{{\beta }_i^2}}<0$. Hence they are irreducible in $\mathbb{R}\left[x\right]$. By Lemma 2, Equation(4) yields:

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

(5)

For the special kind of polynomials in Equation(5), "divisible" means "equal", which can be verified by comparing the like terms in the following equation to get $k^{\prime }=1$.

$$(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})=k^{\prime }(1+{\displaystyle \frac{{(k-s-{\alpha }_l)}^2}{{\beta }_l^2}}),k^{\prime }\ne 0\in \mathbb{R}$$

(6)

Further, due to the uniqueness of the multiplicity $m_i$, the only solution to Equation(5) is $i=l$, otherwise, duplicated zeros (in quadruplets) with ${\alpha }_i+{\alpha }_l=k,{\beta }_i^2={\beta }_l^2,l\ne i$ would be generated to change $m_i$. Therefore we have from Equation(5):

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

(7)

By comparing the like terms in Equation(7), we obtain ${\alpha }_i={\displaystyle \frac{k}{2}}(\forall i)$. 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., $0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots$.

That completes the proof of Theorem 1. □

Theorem 2. 

Given an entire function

$$G\left(s\right)=\prod _{i=1}^{\infty }(1-{\displaystyle \frac{s}{{\rho }_i}}),s\in \mathbb{C}$$

(8)

which converges absolutely on $\mathbb{C}$ in the form of

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

(9)

Here

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

Then

$$G\left(s\right)=G(k-s)⟹{\alpha }_i={\displaystyle \frac{k}{2}}(\forall i)\mathrm{and}0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots .$$

(10)

Proof. 

According to Theorem 1 and Lemma 3, we have

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

(11)

Then we know that

$$G\left(s\right)=G(k-s)⟹{\alpha }_i={\displaystyle \frac{k}{2}}(\forall i)\mathrm{and}0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots .$$

(12)

That completes the proof of Theorem 2. □

Theorem 3. 

Given an entire function represented by its Hadamard product:

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

(13)

and that

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

(14)

Here λ denotes a mathematical object, $\overline{\lambda }$ is the dual of λ, $k>0$ is a real constant.

• ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ (${\beta }_i\ne 0$) are the complex-conjugate zeros of $\Lambda (\lambda ,s)$,
• $0\le {\alpha }_i\le k$, ${\displaystyle \sum _{i=1}^{\infty }\frac{1}{|{\rho }_i{|}^2}<\infty }$,

Then

$$\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)⟹{\alpha }_i={\displaystyle \frac{k}{2}}(\forall i)\mathrm{and}0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots .$$

(15)

where $\epsilon \left(\lambda \right)$ is a complex number of absolute value 1.

Remark 4. 

For more details about $\epsilon \left(\lambda \right)$, see Ref. [2] (p.94).

Proof. 

First, we have

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

(16)

Noticing that ${\sum }_{i=1}^{\infty }{\displaystyle \frac{2{\alpha }_i}{{\alpha }_i^2+{\beta }_i^2}}\le 2k·{\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$, then we have ${\sum }_{i=1}^{\infty }{\displaystyle \frac{2{\alpha }_i}{{\alpha }_i^2+{\beta }_i^2}}=c,c\in \mathbb{R},c\ne 0$. Further, taking into account the multiplicity $m_i\ge 1$ of each zero, we rewrite the product as

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

(17)

Accordingly

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

(18)

Because $e^{A\left(\lambda \right)+\left[B\right(\lambda )+c]s}$ and $\epsilon \left(\lambda \right)e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c](k-s)}$ are units in the ring of entire functions, they have no zeros and therefore do not affect the complex zeros related divisibility in the functional equation $\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$. We thus conclude that Theorem 3 holds by Theorem 2, whose detailed conditions match those of Theorem 3.

In addition to Equation(15), we still have (by canceling the complex zeros related polynomial factors on both sides of functional equation: $\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$, since we already have ${\alpha }_i={\displaystyle \frac{k}{2}}$)

$$\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)⟹e^{A\left(\lambda \right)+\left[B\right(\lambda )+c]s}=e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c](k-s)}$$

That completes the proof of Theorem 3.

□

In the following Theorem 4, we make further efforts to lay a foundation for the study of completed L-functions that possess both real and complex zeros, denoted by $\rho \in {\mathcal{Z}}_r$ ($\Im \left(\rho \right)=0$) and $\rho \in {\mathcal{Z}}_c$ ($\Im \left(\rho \right)\ne 0$), respectively. When these two zero sets have no common elements, we express their disjointness by: ${\mathcal{Z}}_r\cap {\mathcal{Z}}_c=⌀$, which is automatically satisfied by our definitions of ${\mathcal{Z}}_r$ and ${\mathcal{Z}}_c$ because ${\mathcal{Z}}_r$ and ${\mathcal{Z}}_c$ are mutually exclusive sets, i.e., if $\rho \in {\mathcal{Z}}_r$, then $\Im \left(\rho \right)=0$; if $\rho \in {\mathcal{Z}}_c$, then $\Im \left(\rho \right)\ne 0$.

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.

Denote the set of real zeros as

$${\mathcal{Z}}_r=\left\{a_n\in \mathbb{R}\mid 0\le a_n\le k,n=1,2,\dots ,N\right\}$$

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.

Theorem 4. 

Given an entire function represented by its Hadamard product:

$$\begin{array}{cc}\hfill \Lambda (\lambda ,s)& =e^{A\left(\lambda \right)+B\left(\lambda \right)s}\prod _{\rho }(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}},s\in \mathbb{C}\hfill \\ \hfill & =e^{A\left(\lambda \right)+B\left(\lambda \right)s}\prod _{\rho \in {\mathcal{Z}}_r}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}\prod _{\rho \in {\mathcal{Z}}_c}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}\hfill \end{array}$$

(19)

and that

$$\begin{array}{cc}\hfill \Lambda (\overline{\lambda },k-s)& =e^{A\left(\overline{\lambda }\right)+B\left(\overline{\lambda }\right)(k-s)}\prod _{\rho }(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}},s\in \mathbb{C}\hfill \\ \hfill & =e^{A\left(\overline{\lambda }\right)+B\left(\overline{\lambda }\right)(k-s)}\prod _{\rho \in {\mathcal{Z}}_r}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}\prod _{\rho \in {\mathcal{Z}}_c}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}\hfill \end{array}$$

(20)

Here λ denotes a mathematical object, $\overline{\lambda }$ is the dual of λ, $k>0$ is a real constant.

• ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ (${\beta }_i\ne 0$) are the complex-conjugate zeros of $\Lambda (\lambda ,s)$,
• $0\le {\alpha }_i\le k$, ${\displaystyle \sum _{i=1}^{\infty }\frac{1}{|{\rho }_i{|}^2}<\infty }$,
• ${\mathcal{Z}}_r\cap {\mathcal{Z}}_c=⌀.$

Then

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

(21)

where $\epsilon \left(\lambda \right)$ is a complex number of absolute value 1.

i.e., all the zeros (both real and complex) of $\Lambda (\lambda ,s)$ in the critical strip $0\le \Re \left(s\right)\le k$ lie on the critical line $\Re \left(s\right)={\displaystyle \frac{k}{2}}$.

Proof. 

By Theorem 3, to determine the distribution of the complex zeros of $\Lambda (\lambda ,s)$, it suffices to show that the newly introduced factors ${\prod }_{\rho \in {\mathcal{Z}}_r}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}$ and ${\prod }_{\rho \in {\mathcal{Z}}_r}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}$ do not affect the complex zeros related divisibility in the functional equation $\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$. This follows from the given condition ${\mathcal{Z}}_r\cap {\mathcal{Z}}_c=⌀$, which implies that ${\prod }_{\rho \in {\mathcal{Z}}_r}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}$ and ${\prod }_{\rho \in {\mathcal{Z}}_c}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}$ are co-prime, ${\prod }_{\rho \in {\mathcal{Z}}_r}(1-{\displaystyle \frac{k-s}{\rho }})e^{{\displaystyle \frac{k-s}{\rho }}}$ and ${\prod }_{\rho \in {\mathcal{Z}}_c}(1-{\displaystyle \frac{s}{\rho }})e^{{\displaystyle \frac{s}{\rho }}}$ are co-prime, according to Ref. [2] (p.174, p.208) or Ref. [2] (see its THEOREM 4).

Thus, we conclude by Theorem 3 that

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

(22)

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

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

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

(23)

where constant c is the same as in the proof of Theorem 3.

Further Equation(23) is equivalent to

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

(24)

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

Suppose the multiplicity of zero $s=a_n$ is $m_n$ ($m_n\ge 1$) that is finite and unique although unknown. Then Equation(24) becomes

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

(25)

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

Considering $(s-a_t)$ and $(s-(k-a_t))$ are irreducible in $\mathbb{R}\left[x\right]$, then by Lemma 1, Equation(25) means

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

(26)

The only solution to Equation(26) is $t=l,a_t={\displaystyle \frac{k}{2}},t=1,\dots ,T$, otherwise the uniqueness of $m_t$ would be violated with $a_t+a_l=k,t\ne l$. To avoid changing the multiplicity of $m_t$ while $a_t={\displaystyle \frac{k}{2}}$, we need to limit $T=1$. Thus we get

$$\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)\Rightarrow a_t={\displaystyle \frac{k}{2}},t=1,m_1=N$$

(27)

Putting Equation(22) and Equation(27) together, we proved Equation(21).

In addition to Equation(21), we still have (by canceling the real zeros related polynomial factors on both sides of Equation(25))

$$\Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)⟹{(-1)}^N{e}^{A\left(\lambda \right)+[B\left(\lambda \right)+c^{\prime }]s}=\epsilon \left(\lambda \right)e^{A\left(\overline{\lambda }\right)+[B\left(\overline{\lambda }\right)+c^{\prime }](k-s)}$$

That completes the proof of Theorem 4. □

Regarding Theorem 4, we have the following supplementary remarks.

1): Theorem 4 actually implies that if there exists any real zero for the discussed entire function, there can be only one (with multiplicity $\ge 1$), and it must lie on the critical line.

2): If the completed L-function has poles at $s=0$ or $s=k$ with multiplicity $m\ge 1$, then the Hadamard product expression can only be applied to $s^m{(k-s)}^m\Lambda (\lambda ,s)$, not to $\Lambda (\lambda ,s)$ itself. This adjustment does not affect the results of Theorem 4, since we have $s^m{(k-s)}^m\Lambda (\lambda ,s)=\epsilon \left(\lambda \right){(k-s)}^m{s}^m\Lambda (\overline{\lambda },k-s)\iff \Lambda (\lambda ,s)=\epsilon \left(\lambda \right)\Lambda (\overline{\lambda },k-s)$.

3): As pointed out in Ref. [2] (p.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 in the related Hadamard products. For more details, see the proofs of Theorem 5 and Theorem 7.

3. Main Results

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 non-trivial zeros of Dirichlet L-functions in the critical strip $0<\Re \left(s\right)<1$ lie on the critical line.

The Extended Riemann Hypothesis: The non-trivial zeros of Dedekind zeta functions in the critical strip $0<\Re \left(s\right)<1$ lie on the critical line.

The Grand Riemann Hypothesis: The non-trivial zeros of all L-functions in the critical strip $0<\Re \left(s\right)<k$ lie on the critical line.

Another version of Grand Riemann Hypothesis: The non-trivial zeros of all automorphic L-functions in the critical strip $0<\Re \left(s\right)<k$ lie on the critical line.

To begin with, we provide a general property of L-functions, which was labeled Lemma 5.5 in Ref. [2] (p.101).

Lemma 5.5$\left[5\right]$: 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).

To adapt to the wider situation in this manuscript, i.e., $0\le \Re \left(s\right)\le k$, ${\sum }_{\rho \ne 0,k}{\left|\rho \right|}^{-1-ϵ}<+\infty$, Lemma 5.5 has been extended to Theorem 10 (positioned after Theorem 9).

Another general property of L-functions is as follows.

The zeros of $\Lambda (\lambda ,s)$ are precisely the non-trivial zeros of $L(\lambda ,s)$, as the trivial zeros of $L(\lambda ,s)$ are canceled by the poles of the Gamma factors in the completion process. See Ref. [2] (p.96) for more details, with f therein replaced by $\lambda$.

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:

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

(28)

For the principal (or trivial) 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:

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

(29)

Remark: The Riemann Zeta-function $\zeta \left(s\right)$ is a special case of $L(\chi ,s)$ with $\chi \left(n\right)\equiv 1$.

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

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

(30)

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:

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

(31)

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

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

(32)

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

Hadamard Product: For non-principal character $\chi$, the completed L-function $\Lambda (\chi ,s)$ is an entire function of order 1 and has the Hadamard product:

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

(33)

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 $\chi$.

For principal (trivial) character ${\chi }_0$, $\Lambda ({\chi }_0,s)$ carries a simple pole at $s=1$ (and also at $s=0$ precisely when $q=1$). Then the Hadamard product is applied to $s(1-s)\Lambda ({\chi }_0,s)$. But this (trivial) situation makes no difference to our concerned result under the symmetric functional equation, i.e.,

$$\Lambda ({\chi }_0,s)=W\left({\chi }_0\right)\Lambda ({\overline{\chi }}_0,1-s)⟹s(1-s)\Lambda ({\chi }_0,s)=(1-s)sW\left({\chi }_0\right)\Lambda ({\overline{\chi }}_0,1-s)$$

Thus this (trivial) situation is omitted in the proof of Theorem 5.

Next we prove the Generalized Riemann Hypothesis.

Theorem 5: The non-trivial zeros of Dirichlet L-functions in the critical strip $0<\Re \left(s\right)<1$ lie on the critical line.

Remark: We only need to prove that all the zeros of the completed Dirichlet L-function $\Lambda (\chi ,s)$ in the critical strip $0\le \Re \left(s\right)\le 1$ 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 $\lambda =\chi$, $\chi =\overline{\chi }$, $\epsilon \left(\lambda \right)=W\left(\chi \right)$, $k=1$ match the conditions of Theorem 4: 1) Functional equation; 2) Hadamard product; 3) The complex zeros appear in pairs $(\rho ,\overline{\rho })$; 4) $0\le {\alpha }_i\le 1$, ${\displaystyle \sum _{i=1}^{\infty }\frac{1}{|{\rho }_i{|}^2}<\infty }$; 5) ${\mathcal{Z}}_r\cap {\mathcal{Z}}_c=⌀.$

Equation(31) shows the functional equation; Hadamard product Equation(33) is equivalent to Equation(19) in Theorem 4 by separating all zeros into two sets ${\mathcal{Z}}_r$ and ${\mathcal{Z}}_c$; To restrict $\chi =\overline{\chi }$ is to guarantee that the complex conjugate zeros of $\Lambda (\chi ,s)$ appear in pairs; The condition $0\le {\alpha }_i\le 1$ and ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ can be assured by Lemma 5.5, considering that ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}},{\rho }_i\in {\mathcal{Z}}_c$ is a subseries of ${\sum }_{\rho \ne 0,1}{\displaystyle \frac{1}{{\left|\rho \right|}^2}},\rho \in {\mathcal{Z}}_r\cup {\mathcal{Z}}_c$; The condition ${\mathcal{Z}}_r\cap {\mathcal{Z}}_c=⌀$ holds because ${\mathcal{Z}}_r$ and ${\mathcal{Z}}_c$ are mutually exclusive sets, i.e., if $\rho \in {\mathcal{Z}}_r$, then $\Im \left(\rho \right)=0$; if $\rho \in {\mathcal{Z}}_c$, 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 all zeros (real, if any, and complex) of $\Lambda (\chi ,s)$ in the critical strip $0\le \Re \left(s\right)\le 1$ lie on the critical line for $\chi =\overline{\chi }$ .

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

In this case, the complex conjugate zeros do not appear together in Equation(33), 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 Equation(31) to another form, i.e.,

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

(34)

Combining (34) with (31), we get a new functional equation

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

(35)

The product $\Lambda (\overline{\chi },s)\Lambda (\chi ,s)$ is an entire function of finite order $\le 1$, as this holds for the product of any two entire functions of order 1. This, by Lemma 5.5, implies that its zeros $\rho$ satisfy the conditions: $0\le \alpha \le 1$ and ${\sum }_{\rho \ne 0,1}{\left|\rho \right|}^{-2}<\infty$. Moreover, its zero set—being the union of the zeros of the individual factors—possesses the conjugate zeros paring property.

Further, based on Equation(33), we have

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

(36)

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

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

Therefore, by Theorem 4, all zeros (real, if any, and complex) of $\Lambda (\chi ,s)\Lambda (\overline{\chi },s)$, and consequently of $\Lambda (\chi ,s)$, in the critical strip $0\le \Re \left(s\right)\le 1$, lie on the critical line for $\chi \ne \overline{\chi }$.

Combining CASE 1 and CASE 2, we conclude that Theorem 5 holds as a specific case of Theorem 4 with $k=1$, $\lambda =\chi$, and $0<\Re \left(s\right)<1$ is a subset of $0\le \Re \left(s\right)\le 1$. □

Remark: Based on Theorem 4, we actually proved: The non-trivial zeros of Dirichlet L-functions in the critical strip $0\le \Re \left(s\right)\le 1$ lie on the critical line ${\displaystyle \frac{1}{2}}$, which implies the non-existence of Landau-Siegel zeros. Hence, the Landau-Siegel zeros conjecture is justified.

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:

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

(37)

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.

Remark: The Riemann Zeta-function $\zeta \left(s\right)$ is a special case of ${\zeta }_K\left(s\right)$ with $K=\mathbb{Q}$, where $\mathbb{Q}$ denotes the field of rational numbers.

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

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

(38)

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:

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

(39)

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 an entire function of order 1 and has the Hadamard product:

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

(40)

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. [2] (Chapter 5.10) and Ref. [2] (Section 10.5.1).

Theorem 6: The non-trivial zeros of Dedekind zeta functions in the critical strip $0<\Re \left(s\right)<1$ lie on the critical line.

Remark: We only need to prove that all the zeros of ${\Lambda }_K\left(s\right)$ in the critical strip $0\le \Re \left(s\right)\le 1$ 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, Functional equation Equation(39) and Hadamard product Equation(40) guarantee that

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

(41)

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

And that the complex conjugate zeros of ${\Lambda }_k\left(s\right)$ appear in pairs since it is self-dual. The condition $0\le {\alpha }_i\le 1$, ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ and condition ${\mathcal{Z}}_r\cap {\mathcal{Z}}_c=⌀$ hold for the same reasons as in CASE 1 in the proof of Theorem 5.

Therefore, by Theorem 4 we know that all zeros (real, if any, and complex) of ${\Lambda }_K\left(s\right)$ in the critical strip $0\le \Re \left(s\right)\le 1$ lie on the critical line. Thus, Theorem 6 holds as a specific case of Theorem 4 with $\lambda =K$, $\epsilon \left(\lambda \right)=1$, $k=1$, and $0<\Re \left(s\right)<1$ is a subset of $0\le \Re \left(s\right)\le 1$. □

3.3. Modular Form L-Function

Definition: For a cusp form $f\left(z\right)={\sum }_{n=1}^{\infty }{a}_n{e}^{2\pi inz}$ of weight k for ${\Gamma }_0\left(N\right)$, the associated L-function is defined for $\Re \left(s\right)>{\displaystyle \frac{k+1}{2}}$ by:

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

(42)

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 L-function is defined as:

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

(43)

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

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

(44)

where $\epsilon \left(f\right)=\pm 1$ is the epsilon factor, 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:

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

(45)

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 modular form L-Functions in the critical strip $0<\Re \left(s\right)<k$ lie on the critical line.

Remark: We only need to prove that all the zeros of $\Lambda (f,s)$ in the critical strip $0\le \Re \left(s\right)\le k$ 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 $\lambda =f$. Equation(44) shows the functional equation; Hadamard product Equation(45) is equivalent to Equation(19) in Theorem 4 by separating all zeros into two sets ${\mathcal{Z}}_r$ and ${\mathcal{Z}}_c$; Actually, to restrict $f=\overline{f}$ is to guarantee that the complex conjugate zeros of $\Lambda (f,s))$ appear in pairs; The condition $0\le {\alpha }_i\le k$, ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ is satisfied by Theorem 10. The condition ${\mathcal{Z}}_r\cap {\mathcal{Z}}_c=⌀$ hold for the same reasons as in CASE 1 in the proof of Theorem 5.

Therefore, by Theorem 4, we know that all zeros (real, if any, and complex) of $\Lambda (f,s)$ in the critical strip $0\le \Re \left(s\right)\le k$ lie on the critical line for $f=\overline{f}$.

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

To deal with this case $f\ne \overline{f}$, we need first to extend Equation(44) to another form, i.e.,

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

(46)

Combining Equation(46) with Equation(44), we get a new functional equation

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

(47)

Both sides of Equation(47) are the products of entire functions of order 1, thus they are still entire functions of order $\le 1$, with $0\le \alpha \le k$, ${\sum }_{\rho \ne 0,k}{\displaystyle \frac{1}{{\left|\rho \right|}^2}}<\infty$, according to Theorem 10. And that the complex conjugate zeros of $\Lambda (\overline{f},s)\Lambda (f,s)$ appear in pairs.

Further, based on Equation(45), we have

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

(48)

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

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

Therefore, by Theorem 4, all zeros (real, if any, and complex) of $\Lambda (f,s)\Lambda (\overline{f},s)$, and consequently of $\Lambda (f,s)$, in the critical strip $0\le \Re \left(s\right)\le k$, lie on the critical line for $f\ne \overline{f}$.

Combining CASE 1 and CASE 2, we conclude that Theorem 7 holds as a specific case of Theorem 4 with $\lambda =f$, and $0<\Re \left(s\right)<k$ is a subset of $0\le \Re \left(s\right)\le k$. □

3.4. Automorphic L-Function

Definition: For an automorphic representation $\pi$ of ${\mathrm{GL}}_n\left({\mathbb{A}}_{\mathbb{Q}}\right)$ (${\mathbb{A}}_{\mathbb{Q}}$ is the adele ring over the field of rational numbers $\mathbb{Q}$, see Ref. [2] (p.5) for more details), the associated L-function is defined for $\Re \left(s\right)>1$ by:

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

(49)

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

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

(50)

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

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

(51)

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

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

(52)

Functional Equation: The completed automorphic L-function satisfies:

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

(53)

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:

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

(54)

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 automorphic L-Functions in the critical strip $0<\Re \left(s\right)<1$ lie on the critical line.

Remark: We only need to prove that all the zeros of $\Lambda (\pi ,s)$ in the critical strip $0\le \Re \left(s\right)\le 1$ have real part ${\displaystyle \frac{1}{2}}$, i.e., all the zeros of $\Lambda (\pi ,s)$ lie on the critical line.

Proof. 

The proof procedure of Theorem 8 is similar to that of Theorem 7 with $k=1$, f replaced by $\pi$, and $\overline{f}$ 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 $\Lambda (\lambda ,s)$ and $L(\lambda ,s)$:

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 $s^m{(k-s)}^m\Lambda (\lambda ,s),m\ge 1$;

P3: The complex zeros in the relevant Hadamard products appear in pairs $(\rho ,\overline{\rho })$;

P4: All zeros in the relevant Hadamard products lie in the critical strip $0\le \Re \left(\rho \right)\le k$ and satisfy ${\displaystyle \sum _{\rho \ne 0,k}\frac{1}{{\left|\rho \right|}^2}<\infty }$;

P5: The disjointness of real and complex non-trivial zero sets;

P6: The zeros of $\Lambda (\lambda ,s)$ are precisely the non-trivial zeros of $L(\lambda ,s)$.

Therefore, we have the following result on the Grand Riemann Hypothesis.

Theorem 9: The non-trivial zeros of all L-functions in the critical strip $0<\Re \left(s\right)<k$ lie on the critical line if only the properties P1, P2, P3, P4, P5, and P6 are satisfied.

Proof. 

It is not difficult to see that P1, P2, P3, P4, and P5 cover all the conditions in Theorem 4. Thus, we know by Theorem 4 that all the zeros, both real (if any) and complex, of $\Lambda (\lambda ,s)$ in the critical strip $0\le \Re \left(s\right)\le k$ lie on the critical line. Further, according to P6, we conclude that all the non-trivial zeros, both real (if any) and complex, of $L(\lambda ,s)$ in the critical strip $0<\Re \left(s\right)<k$ (as subset of $0\le \Re \left(s\right)\le k$) lie on the critical line.

That completes the proof of Theorem 9. □

Remark: Conditions P1-P3 and P6 are established properties of the relevant L-functions, see Chapter 5 of Ref. [2]. Condition P4 follows from Theorem 10, which is an extended result of Lemma 5.5. Condition P5 is automatically satisfied by our definitions of ${\mathcal{Z}}_r$ (the set of zeros with $\Im \left(\rho \right)=0$) and ${\mathcal{Z}}_c$ (the set of zeros with $\Im \left(\rho \right)\ne 0$). The sole purpose of P6 is to link the zeros of $\Lambda (\lambda ,s)$ with the non-trivial zeros of $L(\lambda ,s)$; further details can be found in Ref. [2] (p.96), with f therein replaced by $\lambda$.

Theorem 10: Let $L(\lambda ,s)$ be an L-function, $\Lambda (\lambda ,s)$, the corresponding completed L-function satisfying a functional equation of the form $\Lambda (\lambda ,s)=ϵ\left(\lambda \right)\Lambda (\overline{\lambda },k-s)$, with $k>0,k\in \mathbb{R}$. Then all zeros $\rho$ of $\Lambda (\lambda ,s)$ lie in the critical strip $0\le \Re \left(\rho \right)\le k$. Moreover, for any $ϵ>0$, we have

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

where $\lambda$ denotes a mathematical object, $\overline{\lambda }$ is the dual of $\lambda$; $\epsilon \left(\lambda \right)$ is a complex number of absolute value 1, called the "root number" of L-function $L(\lambda ,s)$.

Proof. 

Defining $s=kt$ and $F(\lambda ,t)=\Lambda (\lambda ,kt)$, the functional equation transforms as follows:

$$F(\lambda ,t)=\Lambda (\lambda ,kt)=ϵ\left(\lambda \right)\Lambda (\overline{\lambda },k-kt)=ϵ\left(\lambda \right)\Lambda (\overline{\lambda },k(1-t))=ϵ\left(\lambda \right)F(\overline{\lambda },1-t).$$

We thus obtain a new entire function $F(\lambda ,t)$ with the standard functional equation $F(\lambda ,t)=ϵ\left(\lambda \right)F(\overline{\lambda },1-t)$, to which the classical Lemma 5.5 applies.

Let $\tau$ be a zero of $F(\lambda ,t)$. By definition, $\rho =k\tau$ is then a zero of $\Lambda (\lambda ,s)$, and this correspondence is bijective. The classical Lemma 5.5 guarantees that $0\le \Re \left(\tau \right)\le 1$, ${\sum }_{\tau \ne 0,1}{\left|\tau \right|}^{-1-ϵ}<\infty$ (for any $ϵ>0$), which immediately implies $0\le \Re \left(\rho \right)\le k$, ${\sum }_{\rho \ne 0,k}{\left|\rho \right|}^{-1-ϵ}<\infty$ (for any $ϵ>0$).

That completes the proof of Theorem 10. □