On the Method for Proving the RH Using the Alcantara-Bode Equivalence

1. Introduction

For those who need an answer to RH

The Riemann Hypothesis (RH) claims (1859) that the Riemann Zeta function defined by the infinite sum $\zeta \left(s\right)={\sum }_{n=1}^{\infty }1/n^s$ has its non trivial zeros on the vertical line $\sigma =1/2$ ([7], [11])$.$ The Hilbert-Schmidt integral operator $T_{\rho }$ defined on $L^2(0,1)$ having the kernel function $\rho (y,x)=\{y/x\}$, the fractional part of the quantity between brackets,

$$\left(T_{\rho }u\right)\left(y\right)={\int }_0^1\rho (y,x)u\left(x\right)dx,u\in L^2(0,1)$$

(1)

has been used by Alcantara-Bode ([2], pg. 151) in his theorem of the equivalent formulation of the RH. Denoting by $N_{T_{\rho }}$ the null space of this operator, the equivalent formulation consists in:

The Riemann Hypothesis holds if and only if $N_{T_{\rho }}$ = {0}.

A connection between the Zeta function and the integral operator $T_{\rho }$ could be observed ([4], pg. 313) reformulating the left term in the equality:

$${\int }_0^1\rho (\theta /x)x^{s-1}dx=\theta /(s-1)-{\theta }^s\zeta \left(s\right)/s,\sigma >0,s=\sigma +it$$

as $\left(T_{\rho }{x}^{s-1}\right)\left(\theta \right)$. The equivalent formulation of RH given by Beurling in [4] has been used by Alcantara-Bode in his equivalence ([2]). Both outstanding equivalences reduced RH, a problem from the pure math still unsolved since 1859 and listed on Millennium Problems ([7], [16]), to a concrete problem: the injectivity of the Hilbert-Schmidt integral operator $T_{\rho }$ defined in (1). Its kernel function $\rho \in L^2{(0,1)}^2$ defined by the fractional part of the ratio $(y/x)$ is continue almost everywhere, the discontinuities in ${(0,1)}^2$ consisting in a set of numerable one dimensional lines of the form $y=kx,k\in N$ and so, being of the Lebesgue measure zero. The integral operator is Hilbert-Schmidt ([2]) and so compact, allowing us to consider its approximations on finite dimension subspaces ([3]).

2. Two Injectivity Theorems and Approximations on Subspaces

Let H be a separable Hilbert space and T be a linear bounded operator on H positive on a dense set $S\subset H$: $\langle Tv,v\rangle >0\forall v\in S$. Then it has no zeros in the dense set otherwise, if there exists $w\in S$ such that $Tw=0$ then $\langle Tw,w\rangle =0$ contradicting the positivity of T on S. Follows: its ’eligible’ zeros are all in the difference set $E:=H\setminus S$, i.e. $N_T\subset E$. In order to avoid any misunderstanding, we mention that on finite dimension subspaces the positivity and strict positivity of an operator coincide. On a infinite dimension (sub)space S, the strict positivity involves a constant $C>0$ for which $\langle Tv,v\rangle \ge {C\parallel v\parallel }^2,\forall v\in S$ and T is only positive when $\langle Tv,v\rangle >0,\forall v\in S$. The norm on H is the norm induced by the inner product.

The next theorem (introduced in [11]) proves that if T is strict positive definite on the dense set it has no zeros in the difference set either, obtaining $N_T=\left\{0\right\}$.

Theorem 1. 

A linear bounded operator T strict positive definite on a dense set of a separable Hilbert space is injective, equivalently $N_T=\left\{0\right\}$.

Proof. 

Let’s take in consideration only the set of eligible zeros that are on the unit sphere without restricting the generality, once for an element $w\in H,w\ne 0$ and for $w/\parallel w\parallel$ both are or both are not in $N_T$. The set $S\subset H$ is dense if its closure coincides with H. Then, if $w\in E$, for every $\epsilon >0$ there exists $u_{\epsilon ,w}\in S$ such that $\parallel w-u_{\epsilon ,w}\parallel <\epsilon$. If $\parallel w\parallel \ge \parallel u_{\epsilon ,w}\parallel$:

$0\le \parallel w\parallel -\parallel u_{\epsilon ,w}\parallel =\parallel w-u_{\epsilon ,w}+u_{\epsilon ,w}\parallel -\parallel u_{\epsilon ,w}\parallel \le \parallel w-u_{\epsilon ,w}\parallel +\parallel u_{\epsilon ,w}\parallel -\parallel u_{\epsilon ,w}\parallel <\epsilon$.

If $\parallel u_{\epsilon ,w}\parallel \ge \parallel w\parallel$ instead, then: $0\le \parallel u_{\epsilon ,w}-w+w\parallel -\parallel w\parallel <\epsilon$ to obtain:

given $w\in E$, for every $\epsilon >0$ there exists $u_{\epsilon ,w}\in S$ such that

$$|\parallel w\parallel -\parallel u_{\epsilon ,w}\parallel |<\epsilon$$

(2)

Let w be an eligible element from the unit sphere, $\parallel w\parallel =1$ and take ${\epsilon }_n=1/n$. Then there exists at least one element in the dense set $u_{{\epsilon }_n,w}\in S$ such that $\parallel u_{{\epsilon }_n,w}-w\parallel <{\epsilon }_n$ holds. Follows from (2), ∣ 1 - $\parallel u_{{\epsilon }_n,w}\parallel \mid$$<1/n$ showing that, for any choices of a sequence approximating w, $u_{{\epsilon }_n,w}\in S,n\ge 1$, it verifies $\parallel u_{{\epsilon }_n,w}\parallel \to 1$.

If T is a linear bounded operator on H strict positive on S, then there exists $\alpha >0$ such that $\forall u\in S$, $\langle Tu,u\rangle \ge {\alpha \parallel u\parallel }^2$.

Suppose that there exists $w\in E,\parallel w\parallel =1$ a zero of $T,\mathrm{i}.\mathrm{e}.w\in N_T$ and consider a sequence of approximations of w, $u_{{\epsilon }_n,w}\in S,n\ge 1$ that, as we showed, has its normed sequence converging in norm to 1. From the positivity of T on the dense set S, follows:

$$\alpha \parallel u_{{\epsilon }_n,w}{\parallel }^2\le \langle T{u}_{{\epsilon }_n,w},u_{{\epsilon }_n,w}\rangle =\langle T(u_{{\epsilon }_n,w}-w),u_{{\epsilon }_n,w}\rangle <{\epsilon }_n\parallel T\parallel \parallel u_{{\epsilon }_n,w}\parallel$$

(3)

With c=$\parallel T\parallel /\alpha$, we obtain $\parallel u_{{\epsilon }_n,w}\parallel \le c/n$. Then, $\parallel u_{{\epsilon }_n,w}\parallel \to 0$ with $n\to \infty$, in contradiction with its convergence $\parallel u_{{\epsilon }_n,w}\parallel \to 1$ with $n\to \infty$.

Or, this happen for any choice of the sequence of approximations of w, verifying $\parallel w-u_{{\epsilon }_n,w}\parallel <{\epsilon }_n,n\ge 1$, when $Tw=0$.

Thus $w\notin N_T$, valid for any $w\in E,\parallel w\parallel =1$, proving the theorem because no zeros of T there are in S either. □

From now on, we suppose that:

- The dense set S is the result of an union of finite dimension subspaces of a family F: $S={\cup }_{n\ge 1}{S}_n,\overline{S}=H$. It is not mandatory but will ease our proofs considering that the subspaces are including: $S_n\subset S_{n+1},n\ge 1$.

- The linear operator on H is positive definite on the dense set S. A such operator has no zeros in the dense set, otherwise if $u\in S$ then there exists a subspace containing it such that $Tu=0$ attracting the violation of its positivity: $\langle Tu,u\rangle =0$.

Observation 1. 

Let ${\beta }_n\left(u\right):=\parallel u-u_n\parallel$ be the normed residuum of the eligible element $u\in E$ after its orthogonal projection on $S_n$. Then, ${\beta }_n\left(u\right)\to 0$ with $n\to \infty$.

Proof. 

Given $ϵ>0$, from the density of the set S in H there exists $u_{ϵ}\in S$ verifying $\parallel u-u_{ϵ}\parallel <ϵ$, as per the observations made in the proof of the Theorem 1. Let $S_{n_{ϵ}}$ be the coarsest subspace, i.e. with the smallest dimension, from the family of subspaces containing $u_{ϵ}$. Because the best approximation of u in $S_{n_{ϵ}}$ is its orthogonal projection, we obtain

${\beta }_{n_{ϵ}}\left(u\right):=\parallel u-P_{n_{ϵ}}u\parallel \le \parallel u-u_{ϵ}\parallel <ϵ$,

inequality valid for every $ϵ>0$, proving our assertion. □

Theorem 2. 

Let T be a linear bounded operator on H, positive definite on a dense set S result of the union of including finite dimension subspaces of a family F. Consider $\{T_n,n\ge 1\}$ be a sequence of its approximations on the family F.If

i) $\langle T_{n}v,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2,\forall v\in S_n$, with ${\alpha }_n\ge \alpha >0,n\ge 1$;

ii) ${ϵ}_n:=\parallel T-T_n\parallel \to 0$ with $n\to \infty$,

then$N_T=\left\{0\right\}$.

Proof. 

Being positive on S, the operator has no zeros in the dense set. Showing now, that T has no zeros in the difference set. For $u\in E:=H\setminus S$, $\parallel u\parallel =1$ having the not null orthogonal projections $P_{n}u,n\ge n_0:=n_0\left(u\right)$, from Observation 1 we have its residuum sequence on the approximation subspaces ${\beta }_n:={\beta }_n\left(u\right)=\parallel u-P_{n}u\parallel \to 0$.

If there exists $u\in N_T\cap E$, then:

${\alpha }_n\parallel P_n{u\parallel }^2\le \langle T_n{P}_{n}u,P_{n}u\rangle \le \parallel T_n{P}_{n}u\parallel \parallel P_{n}u\parallel$

$\le (\parallel T_n{P}_{n}u-T{P}_{n}u+T{P}_{n}u-Tu\parallel )\parallel P_{n}u\parallel$

$\le (\parallel T-T_n\parallel \parallel P_{n}u\parallel +\parallel T\parallel \parallel u-P_{n}u\parallel )\parallel P_{n}u\parallel$,

obtaining

$\alpha \le \left({ϵ}_n+\parallel T\parallel {\beta }_n/\sqrt{1-{\beta }_n^2}\right)\to 0$.

The inequality is violated from an $n_1\ge n_0$, $n\ge n_1$ involving $u\notin N_T$, valid for any supposed zero of T in E. Once T has no zeros in the dense set, $N_T=\left\{0\right\}$. □

Lemma 1 

(Finite rank approximations). Let T be positive definite on the dense set S. If $\{P_n^r,n\ge 1\}$ is a sequence of finite rank operator orthogonal projections converging strong to identity operator such that:

$\langle P_n^r\left(T\right)v,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2$, ${\alpha }_n\ge \alpha >0$$\forall v\in S_n$, for every $S_n\in F$,

then T is injective.

Proof. 

Taking $T_n=P_n^r\left(T\right),n\ge 1$ then

$\parallel T-T_n\parallel =\parallel (I-P_n^r)T\parallel \le \parallel T\parallel \parallel I-P_n^r\parallel \to 0$. Follows: both requests of the Theorem 2 are fulfilled proving our assumption.

In Note 2(Appendix) we showed that a such operator satisfies the requests of the generic Theorem 1 being strict positive on the dense set. □

3. Dense Sets in $L^2(0,1)$

The intervals of equal lengths $h=2^{-m},m\in N$, nh = 1, ${\Delta }_{h,k}=((k-1)/2^m,k/2^m]$, $k=1,n-1$ together with ${\Delta }_{h,n}$ are defining for $m\ge 1$ a partition of (0,1), k=1,n, $n=2^m,nh=1$. Consider the interval indicator functions having the supports these intervals (k=1,n), nh=1:

$$I_{h,k}\left(t\right)=1\mathrm{for}t\in {\Delta }_{h,k}\mathrm{and}0\mathrm{otherwise}$$

(4)

The family F of finite dimensional subspaces $S_h$ that are the linear spans of interval indicator functions of the h-partitions defined by (4) with disjoint supports, $S_h=span\{I_{h,k};k=1,n,nh=1\}$, built on a multi-level structure, are including $S_h\subset S_{h/2}$ by halving the mesh h. In fact, due to (4) any $I_{h,k}\in S_h$ is contained in $S_{h/2}$ by $I_{h,k}\left(t\right)=I_{h/2,(2k-1)}\left(t\right)+I_{h/2,2k}\left(t\right)$ for any $t\in \left(0.1\right)$, k=1,n, nh=1 (or (2n)(h/2)=1). So, $S_h\ni u_h={\sum }_{k=1,n}{c}_k^h{I}_{h,k}={\sum }_{j=1,2n}{c}_j^{h/2}{I}_{h/2,j}\in S_{h/2}$ with $c_k^h:=c_{2k-1}^{h/2}=c_{2k}^{h/2},k=1,n$; equivalently, $c_{j-1}^{h/2}=c_j^{h/2}=c_{j/2}^h$, for $j/2=1,n,nh=1$.

We now consider the indicator open-interval functions $I_h^k,k=1,n,nh=1$ having the supports ${\Delta }_h^k=((k-1)/2^m,k/2^m),k=1,n$ that differ from the semi-open intervals ${\Delta }_{h,k}$ of the defined partitions only by the end-points $kh,k=1,(n-1)$. Denoting with $F^o$ the family of subspaces $S_n^o$ spanned by the open interval indicator functions $\{I_k^k,k=1,n,nh=1\}$, and denoting with $S^o=U_{n\ge 2}{S}_n^o$, then from ${\int }_0^1{I}_h^k\left(t\right)dt={\int }_0^1{I}_{h,k}\left(t\right)dt,k=1,n,n\ge 2,nh=1$, we obtain $\parallel I_h^k-I_{h,k}\parallel =0$ in $L^2(0,1)$; moreover:

Remark 2. 

If one of the sets S and $S^o$ is dense, then another is dense.

Proof. 

Let $S^o$ be dense, known in literature. If f is orthogonal on any $I_{h,k}\in S$ then:

$|\langle f,I_h^k\rangle |=|\langle f,I_h^k-I_{h,k}\rangle |\le \parallel f\parallel \parallel I_h^k-I_{h,k}\parallel =0$, k=1,n, $\forall n,nh=1$, proving that f is orthogonal to any $I_h^k$ and so f should be 0 because $S^0$ is dense. So, S is dense. Now, the reverse statement is also true, if S is dense then $S^o$ is dense using the same arguments. □

Note 1. Because on every level of discretization any pair of indicator functions in each dense set have disjoint supports, then on S (or on $S^o$):

i) for any $\phi \in L^2{(0,1)}^2$, $I_{h,k}\left(y\right)\phi (y,x)I_{h,j}\left(x\right)=0$, for every $k\ne j$;

ii) because ${\sum }_{i=1,n}{I}_{h,i}\left(t\right)=1$ for every $t\in (0,1)$, ${\int }_0^1{\int }_0^1{I}_{h,k}\left(y\right)\phi (y,x)I_{h,j}\left(x\right)=0$ for $k\ne j$, and

${\int }_0^1{\int }_0^1\phi (y,x)I_{h,j}\left(x\right)={\int }_{{\Delta }_{h,j}}{\int }_{{\Delta }_{h,j}}\phi (y,x)I_{h,j}\left(x\right)$, valid also for replacing the arguments of $\phi$.

iii) Let $M_h^o\left(T_{\phi }\right)$ and $M_h\left(T_{\phi }\right)$ be the matrix representations of the integral operator on the finite dimension subspaces in $F^o$ and F spanned by the corresponding orthogonal indicator interval functions.

Then both matrices $M_h^o$ and $M_h$ are identical and sparse diagonal matrices with the entries verifying

$d_{kj}^h:=\langle T_{\phi }{I}_{h,k},I_{h,j}\rangle ={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,j}}{I}_{h,k}\left(x\right)\phi (y,x)I_{h,j}\left(y\right)dxdy=0$ for $k\ne j$ and,

Remark 3. 

If $d_{kk}^h:={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\rho (y,x)dxdy>0,k=1,n,nh=1$ then

- on every level, the matrix representations of its restrictions on $S_h$ and $S_h^o$ are diagonal, identical and strict positive definite;

- the integral operator $T_{\rho }$ defined in (1) is positive definite on the dense sets S and $S^o$.

Proof. 

From Note 1-ii), both matrices are sparse diagonal on every level h, identical and strict positive. Now, we have for $v={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$, $n\ge 2,nh=1$ from $\langle T_{\rho }{I}_{h,k},I_{h,k}\rangle =\langle T_{\rho }{I}_h^k,I_h^k\rangle :=d_{kk}^h$:

$\langle T_{\rho }v,v\rangle ={\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h=(c_1,...,c_n)M_h\left(T_{\rho }\right){(\overline{c_1},...,\overline{c_n})}^T$

$=(c_1,...,c_n)\left[diag\left(d_{kk}^h\right)\right]{(\overline{c_1},...,\overline{c_n})}^T\ge h^{-1}{min}_{k=1,n}\left(d_{kk}^h\right){\parallel v\parallel }^2>0$

Then, $T_{\rho }$ is (strict) positive definite on $S_h$ with the positivity parameter:

$${\alpha }_h\left(T_{\rho }\right)=h^{-1}mi{n}_{k=1,n}{d}_{kk}^h;n\ge 2,nh=1.$$

(5)

and so, $\forall v\in S_h,\langle T_{\rho }v,v\rangle >0$, following: $\langle T_{\rho }v,v\rangle >0$, $\forall v\in S$.

Because on $S_h^o$$M_h^o\left(T_{\rho }\right)=M_h\left(T_{\rho }\right)$, $T_{\rho }$ has its restriction strict positive definite with the same parameter ${\alpha }_h\left(T_{\rho }\right)$. So, if $d_{kk}^h>0$, $k=1,n,nh=1$, then $T_{\rho }$ is positive definite on the dense sets S and $S^o$. □

4. The Proof of the RH Equivalent

The entries in the diagonal matrix representation $M_h\left(T_{\rho }\right)$ of the integral operator $T_{\rho }$ restriction to $S_h\in F$, $d_{kk}^h={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\rho (y,x)dxdy$, are valued (see [1]) as follows:

$$d_{11}^h=h^2(3-2\gamma )/4$$

$$d_{kk}^h={\int }_{{\Delta }_{h,k}}\left[{\int }_{{\Delta }_{h,k}}(y/x)dx-{\int }_{(k-1)h}^{y}dx\right]dy={\displaystyle \frac{h^2}{2}}(-1+{\displaystyle \frac{2k-1}{k-1}}ln{\left({\displaystyle \frac{k}{k-1}}\right)}^{k-1})$$

(6)

for $k\ge 2$, where $\gamma$ is the Euler-Mascheroni constant (≃ 0.5772156). The sequence

$f\left(k\right):=h^{-2}{d}_{kk}^h=(-1+{\displaystyle \frac{2k-1}{k-1}}ln{\left({\displaystyle \frac{k}{k-1}}\right)}^{k-1})/2$ is monotone decreasing for $k\ge 2$ and converges to 0.5 for k $\to \infty$. Then: $\forall h$, $d_{kk}^h>h^2/2>d_{11}^h,k\ge 2$.

So, the Remark 3 is fulfilled $T_{\rho }$ being positive definite on the dense set S with the positivity parameters of the integral operator restrictions on $S_h$ in (5) given by

${\alpha }_h\left(T_{\rho }\right)=h(3-2\gamma )/4>0$, $n\ge 2,nh=1$. Then we will take in consideration the finite rank approximations of the integral operator on the same dense set in order to apply Lemma 1.

Citing [5], (pg 986), the integral operator $P_h^r,n\ge 1$ having the kernel function:

$$r_h(y,x)=h^{-1}\sum _{k=1,n}{I}_{h,k}\left(y\right)I_{h,k}\left(x\right)$$

(7)

is an finite rank integral operator orthogonal projection having the spectrum ({0, 1}) with the eigenvalue 1 of the multiplicity n (nh=1) corresponding to the orthogonal eigenfunctions $I_{h,k},k=1,n$.

Thus, $\parallel I-P_h^r\parallel \to 0$ for $n\to \infty ,nh=1$. As result, the integral operator approximation of $T_{\rho }$ on $S_h$ is an finite rank operator approximation, $T_{{\rho }_h}$, having the kernel function ([5])

$${\rho }_h(y,x)=h^{-1}\sum _{k=1,n}{I}_{h,k}\left(y\right)\rho (y,x)I_{h,k}\left(x\right):=h^{-1}\sum _{k=1,n}{\rho }_h^k(y,x)$$

(8)

From $\langle T_{{\rho }_h}{I}_{h,k},I_{h.k}\rangle =h^{-1}{d}_{kk}^h$, we have $M_h\left(T_{{\rho }_h}\right)=h^{-1}{M}_h\left(T_{\rho }\right):=h^{-1}diag\left[d_{kk}^h\right]$; $k=1,n,nh=1$, and the positivity parameter of the finite rank approximation of $T_{{\rho }_h}$ is a result of: for $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$,

$$\begin{array}{cc}\hfill \langle T_{{\rho }_h}{v}_h,v_h\rangle & =h^{-1}\sum _{k=1,n}{c}_k\overline{c_k}{\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}{I}_{h,k}\rho (y,x)I_{h,k}dxdy=h^{-1}\sum _{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h\hfill \\ \hfill & \ge h^{-2}\left(mi{n}_{k=1,n}{d}_{kk}^h\right)\parallel v_h{\parallel }^2:={\alpha }_h\left(T_{{\rho }_h}\right){\parallel v_h\parallel }^2\hfill \end{array}$$

(9)

Follows:

$${\alpha }_h\left(T_{{\rho }_h}\right):=h^{-2}{d}_{11}^h=(3-2\gamma )/4,\mathrm{a}\mathrm{constant}\mathrm{mesh}\mathrm{independent}.$$

(10)

Theorem 3. 

(Finite Rank Approximations): The Alcantara-Bode equivalent holds and, the Riemann Hypothesis is true.

Proof. 

From (10), the sequence of the positivity parameters corresponding to the finite rank operator approximations $T_{{\rho }_h}$ on the dense family F is inferior bounded, obtaining from Lemma 1 that $N_{T_{\rho }}=\left\{0\right\}$. Now, obtaining $N_{T_{\rho }}=\left\{0\right\}$ half from equivalent formulation of the Alcantara-Bode is holds. Then, the other half should be true i.e.: the Riemann Hypothesis is true. □