A Proof of the Riemann Hypothesis via the Divisibility of Entire Functions

1. Introduction

The Riemann zeta function is originally defined in the half-plane $\Re \left(s\right)>1$ by the absolutely convergent series

$$\zeta \left(s\right)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n^s}},\Re \left(s\right)>1$$

(1)

The connection between the above-defined Riemann zeta function and prime numbers was discovered by Euler, i.e., the famous Euler product

$$\zeta \left(s\right)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n^s}}=\prod _p{(1-p^{-s})}^{-1},\Re \left(s\right)>1$$

(2)

where p runs over the prime numbers.

Riemann showed in his paper in 1859 how to extend the zeta function to the whole complex plane $\mathbb{C}$ by analytic continuation [1]

$$\zeta \left(s\right)={\displaystyle \frac{\mathrm{\Gamma }(1-s)}{2\pi j}}{\int }_{\infty }^{\infty }{\displaystyle \frac{{(-x)}^s}{e^x-1}}·{\displaystyle \frac{dx}{x}}$$

(3a)

where $“{\int }_{\infty }^{\infty }”$ is the symbol adopted by Riemann to represent the contour integral from $+\infty$ to $+\infty$ around a domain which includes the value 0 but no other point of discontinuity of the integrand in its interior, ’j’ is the imaginary unit ($j^2=-1$).

Or equivalently,

$$\zeta \left(s\right)={\displaystyle \frac{{\pi }^{s/2}}{\mathrm{\Gamma }(s/2)}}\{{\displaystyle \frac{1}{s(s-1)}}+{\int }_1^{\infty }(x^{{\displaystyle \frac{s}{2}}-1}+x^{-{\displaystyle \frac{s}{2}}-{\displaystyle \frac{1}{2}}})·\left({\displaystyle \frac{\theta \left(x\right)-1}{2}}\right)dx\}$$

(3b)

where $\theta \left(x\right)={\sum }_{-\infty }^{\infty }{e}^{-n^2\pi x}$ is the Jaccobi theta function, $\mathrm{\Gamma }\left(s\right)$ is the Gamma function in the following Weierstrass expression

$${\displaystyle \frac{1}{\mathrm{\Gamma }\left(s\right)}}=s·e^{\gamma s}{\displaystyle \prod _{n=1}^{\infty }}(1+{\displaystyle \frac{s}{n}})e^{-s/n}$$

(4)

where $\gamma$ is the Euler-Mascheroni constant.

As shown by Riemann, $\zeta \left(s\right)$ extends to $\mathbb{C}$ as a meromorphic function with only a simple pole at $s=1$, with residue 1, and satisfies the following functional equation

$${\pi }^{-{\displaystyle \frac{s}{2}}}\mathrm{\Gamma }\left({\displaystyle \frac{s}{2}}\right)\zeta \left(s\right)={\pi }^{-{\displaystyle \frac{1-s}{2}}}\mathrm{\Gamma }\left({\displaystyle \frac{1-s}{2}}\right)\zeta (1-s)$$

(5)

The Riemann zeta function $\zeta \left(s\right)$ has zeros at the negative even integers: $-2$, $-4$, $-6$, $-8$, … and one refers to them as the trivial zeros. The other zeros of $\zeta \left(s\right)$ are complex numbers, i.e., non-trivial zeros.

In 1896, Hadamard [2] and Poussin [3] independently proved that no zeros could lie on the line $\Re \left(s\right)=1$, together with the functional equation $\xi \left(s\right)=\xi (1-s)$ and the fact that there are no zeros with real part greater than 1, this showed that all non-trivial zeros must lie in the interior of the critical strip$0<\Re \left(s\right)<1$. Later on, Hardy (1914) [4], Hardy and Littlewood (1921) [5] showed that there are infinitely many zeros on the critical line $\Re \left(s\right)={\displaystyle \frac{1}{2}}$.

To give a summary of the related research works on the RH, we have the following results on the properties of the non-trivial zeros of $\zeta \left(s\right)$ [2,3,4,5,6].

Lemma 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.

As further study, the completed zeta function $\xi \left(s\right)$ is proposed, i.e.

$$\xi \left(s\right)={\displaystyle \frac{1}{2}}s(s-1){\pi }^{-{\displaystyle \frac{s}{2}}}\mathrm{\Gamma }\left({\displaystyle \frac{s}{2}}\right)\zeta \left(s\right)$$

(6)

It is well-known that $\xi \left(s\right)$ is an entire function of order 1. This implies $\xi \left(s\right)$ is analytic, and can be expressed as infinite product of polynomial factors, in the whole complex plane $\mathbb{C}$. In addition, replacing s with $1-s$ in Eq.(6), and combining Eq.(5), we obtain the following functional equation

$$\xi \left(s\right)=\xi (1-s)$$

(7)

According to the definition of $\xi \left(s\right)$, and recalling Eq.(4), the trivial zeros of $\zeta \left(s\right)$ are canceled by the poles of $\mathrm{\Gamma }\left({\displaystyle \frac{s}{2}}\right)$. The zero of $s-1$ and the pole of $\zeta \left(s\right)$ cancel; the zero $s=0$ and the pole of $\mathrm{\Gamma }\left({\displaystyle \frac{s}{2}}\right)$ cancel [7,8,9]. Thus, all the zeros of $\xi \left(s\right)$ are exactly the nontrivial zeros of $\zeta \left(s\right)$. Then we have the following Lemma 2.

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

Consequently, the following two statements are equivalent.

Statement 1: All the non-trivial zeros of $\zeta \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$.

Statement 2: All zeros of $\xi \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$.

To prove the RH, a natural thinking is to estimate the numbers of non-trivial zeros of $\zeta \left(s\right)$ inside or outside some certain areas according to Argument Principle. Along this train of thought, there are many research works. Let $N\left(T\right)$ denote the number of non-trivial zeros of $\zeta \left(s\right)$ inside the rectangle: $0<\alpha <1,0<\beta \le T$, and let $N_0\left(T\right)$ denote the number of non-trivial zeros of $\zeta \left(s\right)$ on the line $\alpha ={\displaystyle \frac{1}{2}},0<\beta \le T$. Selberg proved that there exist positive constants c and $T_0$, such that $N_0\left(T\right)>cN\left(T\right),(T>T_0)$ [10], later on, Levinson proved that $c\ge {\displaystyle \frac{1}{3}}$ [11], Lou and Yao proved that $c\ge 0.3484$ [12], Conrey proved that $c\ge {\displaystyle \frac{2}{5}}$ [13], Bui, Conrey and Young proved that $c\ge 0.41$ [14], Feng proved that $c\ge 0.4128$ [15], Wu proved that $c\ge 0.4172$ [16].

On the other hand, many non-trivial zeros have been calculated by hand or by computer programs. Among others, Riemann found the first three non-trivial zeros [17]. Gram found the first 15 zeros based on Euler-Maclaurin summation [18]. Titchmarsh calculated the 138^th to 195^th zeros using the Riemann-Siegel formula [19,20]. Here are the first three (pairs of) non-trivial zeros: ${\displaystyle \frac{1}{2}}\pm j14.1347251;{\displaystyle \frac{1}{2}}\pm j21.0220396;{\displaystyle \frac{1}{2}}\pm j25.0108575$.

The idea of this paper originates from Euler’s work on proving the famous equality

$$1+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{3^2}}+{\displaystyle \frac{1}{4^2}}+{\displaystyle \frac{1}{5^2}}+\dots ={\displaystyle \frac{{\pi }^2}{6}}$$

(8)

This result was deduced by comparing the coefficients of two infinite expressions of ${\displaystyle \frac{sinx}{x}}$: one as a power series and the other as an infinite product,

$$\begin{array}{cc}\hfill {\displaystyle \frac{sinx}{x}}& =1-{\displaystyle \frac{x^2}{3!}}+{\displaystyle \frac{x^4}{5!}}-{\displaystyle \frac{x^6}{7!}}+\dots =(1-{\displaystyle \frac{x^2}{{\pi }^2}})(1-{\displaystyle \frac{x^2}{4{\pi }^2}})(1-{\displaystyle \frac{x^2}{9{\pi }^2}})\dots \hfill \end{array}$$

(9)

Motivated by this approach, we infer that $\xi \left(s\right)$ can be factored into factors of the form $(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})$, which is achieved by pairing ${\rho }_i$ and ${\overline{\rho }}_i$ in the Hadamard product representation of $\xi \left(s\right)$, i.e. $(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})={\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})$.

The Hadamard product expansion of $\xi \left(s\right)$, first proposed by Riemann and later rigorously justified by Hadamard [21], is given by

$$\xi \left(s\right)=\xi \left(0\right)\prod _{\rho }(1-{\displaystyle \frac{s}{\rho }}),s\in \mathbb{C}$$

(10)

where $\xi \left(0\right)={\displaystyle \frac{1}{2}}$, $\rho$ runs over all zeros of $\xi \left(s\right)$.

Hadamard showed that to ensure the absolute convergence of this infinite product expansion, $\rho$ and $1-\rho$ must be paired. Later in Section 4, we will demonstrate that pairing $\rho$ with $\overline{\rho }$ (together with the boundedness of ${\alpha }_i$ guaranteed by the Euler product) also ensures the absolute and uniform convergence of the infinite product on compact subsets of $\mathbb{C}$.

2. Preliminary Lemmas

This section provides preliminary lemmas supporting the proof of the key lemma - Lemma 6 in the next section.

We begin with the ring of real polynomials $\mathbb{R}\left[x\right]$, defined as

$$\mathbb{R}\left[x\right]=\left\{{\displaystyle \sum _{i=0}^{\infty }}{a}_i{x}^i\right|a_i\in \mathbb{R},a_i\ne 0\mathrm{for}\mathrm{all}\mathrm{but}\mathrm{a}\mathrm{finite}\mathrm{number}\mathrm{of}i\}$$

and equipped with the operations + (addition) and · (multiplication).

The ring of real polynomials is a subset of the ring of entire functions, which is defined as the set of all holomorphic functions on the whole complex plane $\mathbb{C}$, together with the operations of addition and multiplication, denoted as $\mathbb{H}\left(\mathbb{C}\right)$ [22,23].

Both rings possess properties of divisibility, coprimality, and the greatest common divisor, denoted as "gcd". There are also differences between these two rings. Among others, the polynomial ring is a unique factorization domain (UFD), while the ring of entire functions is not a UFD. For entire functions, their divisibility, coprimality and common factors are determined by the relationships between their zero sets [23,24].

Lemma 3: Let $f\left(s\right)$ be an entire function on $\mathbb{C}$ with $f\left(0\right)\ne 0$. Suppose $f\left(s\right)={\prod }_{i=1}^{\infty }{g}_i\left(s\right)$ converges absolutely and uniformly on compact subsets of $\mathbb{C}$, where each $g_i\left(s\right)\in \mathbb{R}\left[s\right]$ is irreducible of degree $d\in \{1,2\}$. If $m\left(s\right)\in \mathbb{R}\left[s\right]$ is irreducible of degree d and divides $f\left(s\right)$, then $m\left(s\right)$ divides $g_i\left(s\right)$ for some $i\ge 1$.

Proof: Let $\alpha$ be a root of $m\left(s\right)$, i.e., $m\left(\alpha \right)=0$. Since $m\left(s\right)\mid f\left(s\right)$, we have $f\left(\alpha \right)=0$. By absolute convergence of ${\prod }_{i=1}^{\infty }{g}_i\left(s\right)$ at $s=\alpha$, 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 ${\prod }_{i=1}^{\infty }{g}_i\left(\alpha \right)$ converges to a non-zero limit, contradicting $f\left(\alpha \right)=0$.

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

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

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

That completes the proof of Lemma 3.

Remark: Although the uniform convergence is not used directly in the proof of Lemma 3, it guarantees (by Weierstrass’s theorem) that the infinite product defines an entire function.

The following Lemma 4 and Lemma 5 help to understand the finiteness and uniqueness of the multiplicities of zeros of $\xi \left(s\right)$.

Lemma 4: 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.

Proof: Let $f\left(s\right)\neg \equiv 0,s\in \mathbb{C}$, be an entire function, which means it is holomorphic on the whole complex plane. Suppose $f\left(s\right)$ has a zero at $s_0\in \mathbb{C}$ of multiplicity m, then $f\left(s\right)={(s-s_0)}^{m}g\left(s\right)$, where $g\left(s\right)$ is also an entire function and $g\left(s_0\right)\ne 0$.

Assume for contradiction that m is infinite, which implies there exists an accumulation point of zeros in the neighbor of $s_0$. Then, by Identity Theorem for holomorphic functions, and considering "0" is also an entire function, we have $f\left(s\right)\equiv 0,s\in \mathbb{C}$, which contradicts the given condition that $f\left(s\right)\neg \equiv 0,s\in \mathbb{C}$. Thus, the assumption is false, i.e., m must be a finite positive integer.

That completes the proof of Lemma 4.

Remark: Statements similar to Lemma 4 can be found in Ref.[25] and other related textbooks/monographs.

Lemma 5: 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.

Proof: Let $f\left(s\right)\neg \equiv 0,s\in \mathbb{C}$, be an entire function, which has a multiple zero at $s_0\in \mathbb{C}$ of multiplicity m. We can write: $f\left(s\right)={(s-s_0)}^{m}g\left(s\right)$, where $g\left(s\right)$ is also an entire function and $g\left(s_0\right)\ne 0$.

Assume for contradiction that there exists another integer $n\ne m$ such that n is also a multiplicity of the zero $s_0$. This means we can also write: $f\left(s\right)={(s-s_0)}^{n}h\left(s\right)$, where $h\left(s\right)$ is an entire function and $h\left(s_0\right)\ne 0$.

Since both expressions for $f\left(s\right)$ must be equal, we then obtain ${(s-s_0)}^{m}g\left(s\right)={(s-s_0)}^{n}h\left(s\right)$. Without loss of generality, consider $m>n$, then we have: ${(s-s_0)}^{m-n}g\left(s\right)=h\left(s\right)\Rightarrow h\left(s_0\right)=0$, which is a contradiction to $h\left(s_0\right)\ne 0$. Thus, the assumption is false, i.e., the multiplicity of a zero of any non-zero entire function is unique.

That completes the proof of Lemma 5.

Remark: Based on Lemma 4 and Lemma 5, the multiplicity $m_i$ of the i-th quadruplet of zeros $({\alpha }_i+j{\beta }_i,{\alpha }_i-j{\beta }_i,1-{\alpha }_i-j{\beta }_i,1-{\alpha }_i+j{\beta }_i)$ of $\xi \left(s\right)$ is finite and uniquely determined.

3. Key Lemma

In this section, we prove the key lemma - Lemma 6, which is substantial for the proof of the RH. The entire function $f\left(s\right)$ considered in Lemma 6, has the same zero set as $\xi \left(s\right)$. In fact, they differ only by a constant (see Eq.(29) in Section 4 for more details).

Lemma 6: Given a symmetric entire function $f\left(s\right)=f(1-s)$ with the infinite product representation absolutely and uniformly convergent on compact subsets of $\mathbb{C}$:

$$f\left(s\right)={\displaystyle \prod _{i=1}^{\infty }}{\left(g_i\left(s\right)\right)}^{m_i},s\in \mathbb{C}$$

(11)

$$f(1-s)={\displaystyle \prod _{i=1}^{\infty }}{\left(g_i(1-s)\right)}^{m_i},s\in \mathbb{C}$$

(12)

where $g_i\left(s\right)=1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}$ is an irreducible quadratic polynomial in $\mathbb{R}\left[s\right]$ with zeros ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ ranging over all conjugate zero pairs, $0<{\alpha }_i<1$ and ${\beta }_i$ are real numbers with $0<|{\beta }_1|\le |{\beta }_2|\le \dots$, ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$, and $m_i\ge 1$ is the uniquely determined multiplicity of the i-th quadruplet of zeros of $f\left(s\right)$: $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)=({\alpha }_i+j{\beta }_i,{\alpha }_i-j{\beta }_i,1-{\alpha }_i-j{\beta }_i,1-{\alpha }_i+j{\beta }_i)$.

Then we have

$$f\left(s\right)=f(1-s)⟺{\alpha }_i={\displaystyle \frac{1}{2}}(i=1,2,3,\dots );0<|{\beta }_1|<|{\beta }_2|<\dots$$

(13)

Remark: It is worth pointing out that the absolute and uniform convergence on compact subsets of $\mathbb{C}$ serves two purposes here: absolute convergence justifies the rearrangement of factors in the infinite product, while uniform convergence guarantees that the infinite product defines an entire function.

Proof: Considering

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

(14)

we have from $f\left(s\right)=f(1-s)$ and Eqs.(11)-(12) that

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

(15)

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

Based on the conditions $0<{\alpha }_i<1$, ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$ and the Weierstrass M-Test, both sides of Eq.(15) are infinite products absolutely and uniformly convergent on compact subsets of $\mathbb{C}$, see the next section for relevant details. Hence each side of Eq.(15) can be rearranged into any quadratic polynomial times the remaining product, which also converges absolutely and uniformly on compact subsets of $\mathbb{C}$ (This reduces to the fact that omitting finitely many terms does not affect the convergence of an infinite product.), thus defines an entire function. Then by the divisibility property of entire functions [23,24], each quadratic factor on one side divides the product on the other side, i.e.,

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

(16)

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

(16a)

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 $1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}$ have discriminant $\Delta =-4·{\displaystyle \frac{1}{{\beta }_i^2}}<0$. Hence they are irreducible in $\mathbb{R}\left[s\right]$. By Lemma 3, Eq.(16) yields (Eq.(16a) plays the same role as Eq.(16) and can be ignored):

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

(17)

For the special kind of polynomials in Eq.(17), "divisible" means "equal", which can be verified by comparing the like terms in the following equation to get $k=1$.

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

(18)

Further, due to the uniqueness of the multiplicity $m_i$, the only solution to Eq.(17) is

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

(19)

i.e.,

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

(19a)

since otherwise we have

$$(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})\mid (1+{\displaystyle \frac{{(1-s-{\alpha }_l)}^2}{{\beta }_l^2}}),i,l\in \mathbb{N},l\ne i$$

(19b)

i.e.,

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

(19c)

Eq.(19c) leads to ${\alpha }_i+{\alpha }_l=1,{\beta }_i^2={\beta }_l^2,l\ne i$, which means the i-th and l-th quadruplets of zeros coincide on the complex plane (see Figure 1 for illustration), violating the uniqueness of $m_i$.

Therefore, by comparing the like terms in Eq.(19a), we obtain ${\alpha }_i={\displaystyle \frac{1}{2}},i\in \mathbb{N}$. Further, to ensure the uniqueness of $m_i$ while ${\alpha }_i={\displaystyle \frac{1}{2}}$, we need to restrict the ${\beta }_i$ values to be distinct, i.e., $0<|{\beta }_1|<|{\beta }_2|<|{\beta }_3|<\dots$. Thus, we obtain

$$f\left(s\right)=f(1-s)⟹{\alpha }_i={\displaystyle \frac{1}{2}}(i=1,2,3,\dots );0<|{\beta }_1|<|{\beta }_2|<\dots$$

(20)

On the other hand, we have the following fact.

$$\begin{array}{cc}\hfill & {\alpha }_i={\displaystyle \frac{1}{2}}(i=1,2,3,\dots );0<|{\beta }_1|<|{\beta }_2|<\dots \hfill \\ \hfill & \Rightarrow (\sin \mathrm{ce}{\beta }_i\ne 0)\hfill \\ \hfill & (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})=(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}})\hfill \\ \hfill & \Rightarrow (\sin \mathrm{ce}{m}_i\ge 1)\hfill \\ \hfill & (1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}=(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \\ \hfill & \Rightarrow \left(\mathrm{taking}\mathrm{infinite}\mathrm{products}\mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equations}\right)\hfill \\ \hfill & {\displaystyle \prod _{i=1}^{\infty }}(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}={\displaystyle \prod _{i=1}^{\infty }}(1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}{)}^{m_i}\hfill \end{array}$$

(21)

From Eqs.(20) and (21), we finally get Eq.(13).

That completes the proof of Lemma 6.

Remark: In Lemma 6 and similar situations, the condition ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$ with $0<|{\beta }_1|\le |{\beta }_2|\le \dots$ (where each $|{\beta }_i|$ appears $m_i$ times) is equivalent to ${\sum }_{i=1}^{\infty }{\displaystyle \frac{m_i}{{\beta }_i^2}}<\infty$ with $0<|{\beta }_1|<|{\beta }_2|<\dots$, where $m_i$ is the multiplicity of the i-th quadruplet of zeros $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$.

Remark: Lemma 6 is not applicable to the Davenport-Heilbronn counterexample. That is because, in the proof of Lemma 6, we need the conditions ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\beta }_i{|}^2}}<\infty$ and $0<{\alpha }_i<1$ to guarantee the divisibility of the special kind of entire functions expressed as ${\prod }_{i=1}^{\infty }{\underbrace{(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})\dots }}_{m_i\mathrm{times}}$. Unfortunately, the Davenport-Heilbronn counterexample does not have the property that all its zeros lie inside the critical strip since it possesses no Euler product; therefore, one cannot deduce that ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\beta }_i{|}^2}}<\infty$ from ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty$ without $0<{\alpha }_i<1$.

In addition, Lemma 7 will also be used in the proof of the RH in the next section.

Lemma 7: 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: $0<{\alpha }_i<1,0<|{\beta }_1|\le |{\beta }_2|\le |{\beta }_3|,\dots ,{\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$, and $m_i\ge 1$ is the multiplicity of the i-th quadruplet of zeros $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$ of $\xi \left(s\right)$.

Proof: Let $a_i={\displaystyle \frac{{\alpha }_i^2}{{\alpha }_i^2+{\beta }_i^2}}$, then ${\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}=1-a_i$. Since $0<{\alpha }_i<1$ and ${\beta }_i\ne 0$, we have $0<a_i<{\displaystyle \frac{1}{{\beta }_i^2}}$. By the given condition ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$ (with each $|{\beta }_i|$ appearing $m_i$ times), we have equivalently ${\sum }_{i=1}^{\infty }{\displaystyle \frac{m_i}{{\beta }_i^2}}<\infty$ (with $|{\beta }_1|<|{\beta }_2|<\dots$). Hence ${\sum }_{i=1}^{\infty }{m}_i{a}_i<{\sum }_{i=1}^{\infty }{\displaystyle \frac{m_i}{{\beta }_i^2}}<\infty$, which guarantees that the product ${\prod }_{i=1}^{\infty }{(1-a_i)}^{m_i}={\prod }_{i=1}^{\infty }{\left({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}\right)}^{m_i}$ converges to a non-zero constant.

That completes the proof of Lemma 7.

4. A Proof of the RH

This section presents a proof of the Riemann Hypothesis. We first prove that Statement 2 of the RH is true, and then by Lemma 2, Statement 1 of the RH is also true. To be brief, to prove the Riemann Hypothesis, it suffices to show that ${\alpha }_i={\displaystyle \frac{1}{2}},i=1,2,3,\dots$ in the new expression of $\xi \left(s\right)$ as shown in Eq.(25).

Proof of the RH: The proof proceeds in three steps.

Step 1:

It is well-known that zeros of $\xi \left(s\right)$ always come in complex conjugate pairs. Then by pairing ${\rho }_i={\alpha }_i+j{\beta }_i$ and ${\overline{\rho }}_i={\alpha }_i-j{\beta }_i$ in the Hadamard product as shown in Eq.(10), we have

$$\begin{array}{cc}\hfill \xi \left(s\right)& =\xi \left(0\right)\prod _{\rho }(1-{\displaystyle \frac{s}{\rho }})=\xi \left(0\right){\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\rho }_i}})(1-{\displaystyle \frac{s}{{\overline{\rho }}_i}})\hfill \\ \hfill & =\xi \left(0\right){\displaystyle \prod _{i=1}^{\infty }}(1-{\displaystyle \frac{s}{{\alpha }_i+j{\beta }_i}})(1-{\displaystyle \frac{s}{{\alpha }_i-j{\beta }_i}})=\xi \left(0\right){\displaystyle \prod _{i=1}^{\infty }}({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}})\hfill \end{array}$$

(22)

where $0<{\alpha }_i<1,{\beta }_i\ne 0$ (according to Lemma 1) are real numbers with $|{\beta }_1|\le |{\beta }_2|\le |{\beta }_3|\le \dots$, $\rho$ ranges over all zeros of $\xi \left(s\right)$, $({\rho }_i,{\overline{\rho }}_i)$ range over all conjugate zero pairs of $\xi \left(s\right)$.

The absolute convergence of the infinite product in Eq.(22) in the form

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

(23)

depends on the convergence of infinite series ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}$ (since $\left|s\right|<\infty ,0<{\alpha }_i<1\Rightarrow \left|s(2{\alpha }_i-s)\right|<\infty$), which is a well-known fact according to Theorem 2 in Section 2, Chapter IV of Ref.[9] (also listed in Appendix for convenience). Thus, the infinite products as shown in Eq.(23) and Eq.(22) are absolutely convergent on the whole complex plane $\mathbb{C}$.

According to the Weierstrass M-test, these two infinite products as shown in Eq.(23) and Eq.(22) are also uniformly convergent on any compact subset of $\mathbb{C}$. Indeed, let $K\subset \mathbb{C}$ be an arbitrary compact set and let $R_K={sup}_{s\in K}\left|s\right|$. Then for all $s\in K$ and all i, since $0<{\alpha }_i<1$, we have

$$|s(2{\alpha }_i-s)|\le |s\left|\right(2{\alpha }_i+\left|s\right|)<R_K(2+R_K)=M_K.$$

Hence

$${\displaystyle \frac{\left|s\right(2{\alpha }_i-s\left)\right|}{|{\rho }_i{|}^2}}<{\displaystyle \frac{M_K}{|{\rho }_i{|}^2}},\mathrm{and}{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{M_K}{|{\rho }_i{|}^2}}=M_K{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty .$$

Therefore, both infinite products as shown in Eq.(23) and Eq.(22) converge absolutely and uniformly on K.

Further, considering the absolute convergence of

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

(24)

we have the following new expression of $\xi \left(s\right)$ by putting all the possible multiple factors (zeros) together:

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

(25)

where $m_i\ge 1$ is the multiplicity of the i-th quadruplet of zeros $({\rho }_i,{\overline{\rho }}_i,1-{\rho }_i,1-{\overline{\rho }}_i)$, $i=1,2,3,\dots$.

Step 2: Replacing s with $1-s$ in Eq.(25), we obtain the infinite product expression of $\xi (1-s)$, i.e.,

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

(26)

Step 3: According to the functional equation $\xi \left(s\right)=\xi (1-s)$, and considering Eq.(25) and Eq.(26), we have

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

(27)

which is equivalent to

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

(28)

Since $0<{\alpha }_i<1$, then ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}<\infty \Rightarrow {\displaystyle \frac{1}{|{\rho }_i{|}^2}}\to 0\Rightarrow |{\rho }_i{|}^2\to \infty \Rightarrow {\left|{\beta }_i\right|}^2\to \infty$.

Further we have ${lim}_{i\to \infty }{\displaystyle \frac{{\beta }_i^2}{|{\rho }_i{|}^2}}={lim}_{i\to \infty }{\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}=1$, that means ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}$ and ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{|{\rho }_i{|}^2}}$ have the same convergence, i.e., ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{{\beta }_i^2}}<\infty$. Thus, the infinite products on both sides of Eq.(28) are absolutely and uniformly convergent on compact subsets of $\mathbb{C}$, with reference to the analysis of Eqs. (22) and (23) in Step 1.

Furthermore, both sides of Eq.(28) converge to entire functions, because they differ with the entire functions $\xi \left(s\right)$ and $\xi (1-s)$ respectively by a non-zero multiplicative constant, i.e.

$$\begin{array}{cc}\hfill \xi \left(s\right)=& \xi \left(0\right){\displaystyle \prod _{i=1}^{\infty }}{({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\alpha }_i^2+{\beta }_i^2}})}^{m_i}\hfill \\ \hfill =& \xi \left(0\right){\displaystyle \prod _{i=1}^{\infty }}{\left({\displaystyle \frac{{\beta }_i^2}{{\alpha }_i^2+{\beta }_i^2}}\right)}^{m_i}{\displaystyle \prod _{i=1}^{\infty }}{(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})}^{m_i}\hfill \\ \hfill =& c·{\displaystyle \prod _{i=1}^{\infty }}{(1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}})}^{m_i}\hfill \end{array}$$

(29)

where c is a non-zero constant according to Lemma 7.

In addition, the polynomial factors in both sides of Eq.(28), i.e., $1+{\displaystyle \frac{{(s-{\alpha }_i)}^2}{{\beta }_i^2}}$ and $1+{\displaystyle \frac{{(1-s-{\alpha }_i)}^2}{{\beta }_i^2}}$ have discriminant $\Delta =-4·{\displaystyle \frac{1}{{\beta }_i^2}}<0$. Hence they are irreducible real polynomials of degree 2 in $\mathbb{R}\left[s\right]$.

Finally, according to Lemma 6, Eq.(28) is equivalent to

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

(30)

Thus, we conclude that all the zeros of the completed zeta function $\xi \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$, i.e., Statement 2 of the RH is true. According to Lemma 2, Statement 1 of the RH is also true, i.e., all the non-trivial zeros of the Riemann zeta function $\zeta \left(s\right)$ have real part equal to ${\displaystyle \frac{1}{2}}$.

That completes the proof of the RH.