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

1. Introduction.

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])$.$ 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 obtained from Beurling equivalent formulation ([4]). 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_Tρ = {0}.

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 $y=kx,k\in N$, being of Lebesgue measure zero. The integral operator is Hilbert-Schmidt ([2]) and so compact, allowing us to consider its approximations on finite dimension subspaces ([3]). From the strict positivity of an integral operator like in (1) on such subspaces whose union is dense, we will obtain that it is injective, equivalently $N_{T_{\rho }}=\left\{0\right\}$.

2. Two Theorems on Injectivity

Let H be a separable Hilbert space and denote with $\mathcal{L}\left(H\right)$ the class of the linear bounded operators on H. If $T\in \mathcal{L}\left(H\right)$ is positive on a dense set $S\subset H$, i.e. $\langle Tv,v\rangle >0\forall v\in S$ then T 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. $N_T\subset E$.

An operator linear T and its associated Hermitian $T^{*}T$ have the same null space. We could switch T with its Hermitian when no information on its positivity there exists. So, our method could be applied to any linear bounded operator, that is the reason we will consider the linear operator be positive definite $\langle Tu,u\rangle >0$ on a dense set. The next theorem ([11]) is the starting point of our solution.

Theorem 1. 

If $T\in \mathcal{L}\left(H\right)$ is strict positive on a dense set of a separable Hilbert space then T 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 $w\in E:=H\setminus S$, for every $\epsilon >0$ there exists $u_{\epsilon ,w}\in S$ such that $\parallel w-u_{\epsilon ,w}\parallel <\epsilon$. Now, (2) results as follows. 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:

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

So, 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 $u_{{\epsilon }_n,w}\in S$ 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 $T\in \mathcal{L}\left(H\right)$ is 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. $\square$

From now on, we will 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: $S_n\subset S_{n+1},n\ge 1$. There are such approximation subspaces having the including property. For example one in which the subspaces are spanned by functions indicator of semi-open disjoint intervals of the same length $h,nh=1$. In this case, by halving the intervals we obtain the including property.

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. □

Pointing out. Let $T\in \mathcal{L}\left(H\right)$ be positive on a dense set S. Consider the operator restricted to the subspaces $S_n\in F,n\ge 1$ and define its positivity parameters by ${\alpha }_n\left(T\right)$ verifying $\langle Tv,v\rangle \ge {\alpha }_n\left(T\right){\parallel v\parallel }^2$, $\forall v\in S_n,n\ge 1$. We distinguish three cases:

1.) ${\alpha }_n\left(T\right)\ge \alpha >0,\forall n\ge 1$. Then T is strict positive on the dense set and by Theorem 1, $N_T=\left\{0\right\}$.

2.) ${\alpha }_n\left(T\right)\to 0$ then we have to provide a criteria for the investigation of the operator injectivity.

3.) If T is not positive on the dense set, we could replace it with its associated Hermitian operator $T^{*}T$. We could consider this case to fit in one of the previous cases provided that the Hermitian is positive definite on the dense set (thus it has not zeros in the dense set). Thus, we have to provide a criteria only for the case when the positivity parameters sequence of the operator restrictions on the approximation subspaces $S_n$ converges to 0. A such criteria, used in Appendix involving the adjoint operator is given in [1]. The following result shows that, if there exists a dense set on which the operator has a sequence of operator approximations positive with the corresponding positivity parameters bounded inferior then it is injective. More, the operator is strict positive on the dense set.

Given $T\in \mathcal{L}\left(H\right)$, let $T_n,n\ge 1$ be a sequence of operator approximations on $S_n,n\ge 1$ having the property ${ϵ}_n:=\parallel T-T_n\parallel \to 0$.

Suppose that for every $n\ge 1$, the operator approximation $T_n$ is positive on $S_n$ and denote with ${\alpha }_n:={\alpha }_n\left(T_n\right)$ its positivity parameter.

Theorem 2. 

Let T be a linear bounded operator on H, positive on a dense set S, $\langle Tv,v\rangle >0$ for every $v\in S$.

If S is a result of the union of including finite dimension subspaces of a family F on which the sequence $\{T_n,n\ge 1\}$ of its approximations on the family F verifies:

i) $\langle T_{n}v,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2,\forall v\in S_n$, $S_n\in F$,

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

iii) ${\alpha }_n\ge \alpha >0,n\ge 1$

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. Next observation is a consequence of the fact that if a not null $u\in N_T$ then $u/\parallel u\parallel \in N_T$. For $u\in E:=H\setminus S$, $\parallel u\parallel =1$ having the not null orthogonal projections $P_{n}u,n\ge n_0:=n_0\left(u\right)$, we have on any subspace $S_n$, ${\parallel u\parallel }^2={\parallel P_{n}u\parallel }^2+{\beta }_n^2\left(u\right)$ where ${\beta }_n\left(u\right)=\parallel u-P_{n}u\parallel$ is its residuum.

If there exists $u\in N_T\cap E$, i.e. $Tu=0$, then for it:

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

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

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

obtaining from Observation 1 and iii):

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

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

When $T_n,n\ge 1$ are the restrictions of T onto the subspaces $S_n\in F$, both i) and ii) are fulfilled and, the condition iii) could be avoided by using the Injectivity Criteria. We will deal with the special case of the approximations of the Hilbert-Schmidt integral operators that, being compact operators could be approximated on finite dimension subspaces such that the condition ii) is satisfied ([3]).

Let $T:=T_{\phi }$ be an Hilbert-Schmidt integral operator. A method for obtaining the approximations for an integral operator is explained and used in [5]. Thus, ii) in the Theorem 2 is fulfilled when $T_n,n\ge 1$ are finite rank approximations on the subspaces of the family F obtained by orthogonal projection integral operators like in [5]: $T_n:=P_n^r\left(T\right)$. Then:

$\parallel Tu-T_{n}u\parallel =\parallel (I-P_n^r)Tu\parallel \le \parallel I-P_n^r\parallel \parallel Tu\parallel \to 0$, for every $u\in H$.

Moreover, the bounding of the positivity parameter sequence of the finite rank operator approximations involves the strict positivity of the operator on the dense set.

Lemma 1. (Criteria for finite rank approximations). 

If the finite rank approximations of a positive Hilbert-Schmidt operator on a dense set S are positive on the family of approximation subspaces F and the sequence of the positivity parameters are inferior bounded, then it is injective.

Proof. 

The requests i), ii) in the Theorem 2 are fulfilled by the previous observations. From the convergence to zero of the sequence ${ϵ}_n,n\ge 1$ there exists ${ϵ}_0$ a ’compactness’ parameter verifying ${ϵ}_0:=ma{x}_n\{{ϵ}_n;{ϵ}_n<\alpha \}$ corresponding to a subspace $S_{n_0},n_0<\infty$. The parameter ${ϵ}_0$ in independent from any $v\in S$ and,

${\alpha }_n\ge \alpha >{ϵ}_0\ge {ϵ}_n$, for $n\ge 1$.

For an arbitrary $v\in S$ there exists a coarser subspace (i.e. with a smaller dimension) $S_n,n\ge n_0:=n_0\left(v\right)$, for which $v\in S_n$. For it, we have:

$\langle Tv,v\rangle =\langle T_{n}v,v\rangle -\langle (T_n-T)v,v\rangle >0$.

Because T and $T_n$ are positive on S both inner products in the right side of the equality being real valued. So, if $\langle (T_n-T)v,v\rangle \ge 0$, then $\langle (T_n-T)v,v\rangle \le {ϵ}_n{\parallel v\parallel }^2\le {\alpha }_n{\parallel v\parallel }^2$. Follows,

$\langle Tv,v\rangle \ge ({\alpha }_n-{ϵ}_n){\parallel v\parallel }^2\ge \alpha -\left({ϵ}_0\right){\parallel v\parallel }^2$.

Now, if $\langle (T_n-T)v,v\rangle <0$, $\langle Tv,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2\ge {\alpha \parallel v\parallel }^2\ge (\alpha -{ϵ}_0){\parallel v\parallel }^2$ proving the condition iii) in the Theorem 2 with $\alpha \left(T\right)=\alpha -{ϵ}_0>0$ and so, $N_T=\left\{0\right\}$. □

The following lemma is dealing with the cases in which no finite rank approximations could be defined (see [1], [11]).

Lemma 2. (Criteria for operator restrictions.) 

Let $T\in \mathcal{L}\left(H\right)$ positive on the subspaces $S_n,n\ge 1$ whose union S is a dense set S, verifying: $\langle Tv,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2$ for every $v\in S_n$, where ${\alpha }_n\to 0$ with $n\to \infty$. Consider now the parameter:

${\mu }_n:={\alpha }_n\left(T\right)/{\omega }_n$where${\omega }_n$verifies$\parallel T^{*}v\parallel \le {\omega }_n\parallel v\parallel ,\forall v\in S_n,n\ge 1$.

If there exists $C>0$such that${\mu }_n\ge C$for every$n\ge 1$, then$N_T=\left\{0\right\}$.

Proof. 

Suppose that there exists $u\in (H\setminus S)\cap N_T$, $\parallel u\parallel =1$ and let $u_n$ its orthogonal projection on $S_n,n\ge 1$. Then, from the (strict) positivity of T on the subspaces $S_n,n\ge 1$ (see (3)):

${\alpha }_n\left(T\right)\parallel u_n{\parallel }^2\le \langle T{u}_n,u_n\rangle =\langle T(u_n-u),u_n\rangle =\langle (u_n-u),T^{*}{u}_n\rangle \le {\beta }_n{\omega }_n\parallel u_n\parallel$

Then, from

$C\le {\mu }_n\le {\beta }_n/\sqrt{1-{\beta }_n^2}\to 0$ where ${\beta }_n:={\beta }_n\left(u\right)=\parallel u-u_n\parallel$, we obtain a contradiction. Thus, $u\notin N_T$ for any $u\in H\setminus S$. Follows: $N_T=\left\{0\right\}$. □

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

Let $H:=L^2(0,1)$. The semi-open 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 the open ${\Delta }_{h,n}$ 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, the property is obtained from (4) observing that $S_h\ni I_{h,i}=I_{h/2,2i-1}+I_{h/2,2i}\in S_{h/2},i=1,n$.

The set $S={\cup }_{n\ge 1}{S}_h,nh=1$ is dense in H well known in literature.

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

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

(5)

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 $I_{h,k},k=1,n$. We will show it, by proving that $\forall u\in H$, $P_h^{r}u\in S_h$ and ${\left(P_h^r\right)}^2=P_h^r$ for $n\ge 2,nh=1$. For any $u\in H$,

$P_h^{r}u={\int }_{x\in (0,1)}\left(h^{-1}{\sum }_{k=1.n}{I}_{h,k}\left(y\right)I_{h,k}\left(x\right)\right)u\left(x\right)dx$

$=h^{-1}{\sum }_{k=1,n}{c}_k{I}_{h,k}\left(y\right)$, $c_k:={\int }_{{\Delta }_{h,k}}u\left(x\right)dx$,

that is the standard orthogonal projection onto $S_h$ up to the variable switches that come from the definition of an integral operator ($\left(T_{\phi }u\right)\left(y\right)={\int }_{x\in (0,1)}\phi (y,x)u\left(x\right)dx$).

Now, if f = ${\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$,

$P_h^r\left(f\right)=h^{-1}{\sum }_{j=1,n}{\int }_{{\Delta }_{h,j}}{I}_{h,j}\left(y\right)\left({\sum }_{k=1,n}{c}_k{I}_{h,k}\left(y\right)\right)I_{h,j}\left(x\right)dy$

$=h^{-1}{\sum }_{j=1,n}{I}_{h,j}\left(x\right){\int }_{{\Delta }_{h,j}}{c}_j{I}_{h,j}\left(y\right)dy$

$=h^{-1}{\sum }_{j=1,n}{c}_j{I}_{h,j}\left(x\right){\int }_{{\Delta }_{h,j}}{I}_{h,j}\left(y\right)dy$$={\sum }_{j=1,n}{c}_j{I}_{h,j}=f$,

i.e. $P_h^{r}f=f$ and so, ${\left(P_h^r\right)}^{2}u=P_h^{r}u$ for any $u\in H$. Thus,

$\parallel I-P_h^r\parallel \to 0$ for $n\to \infty ,nh=1$. So, ii) in Theorem 2 is fulfilled.

Remark 1. 

Because $P_h^r$ is an orthogonal projection, the inner product on the subspace $S_h$ between $u\notin S_h$ and $v\in S_h$ is a result between the orthogonal projection of u and v is like between two step functions in $S_h$: $\langle u,v\rangle :=\langle P_h^{r}u,v\rangle$.

Let $T_{\rho }$ be an Hilbert-Schmidt integral operator on H. Its integral operator approximation of $T_{\rho }$ on $S_h$ is an finite rank operator approximation, $T_{{\rho }_h}$, having the kernel function ([5])

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

(6)

where the pieces of the kernel function ${\rho }_h$ in the sum have disjoint supports in $L^2{(0,1)}^2$ namely ${\Delta }_{h,k}\times {\Delta }_{h,k},k=1,n,nh=1$.

This observation is crucial in obtaining diagonal matrix representations of the integral operators onto the finite dimension subspaces $S_h,n\ge 2,nh=1$.

So, for $v\in S_h,v=\sum c_k{I}_{h,k}$,

$T_{{\rho }_h}v=h^{-1}{\sum }_{k=1,n}{c}_k{I}_{h,k}\left(y\right){\sum }_{j=1,n}{\int }_{{\Delta }_{h,j}}{I}_{h,j}\left(y\right)\rho (y,x)I_{h,j}\left(x\right)dx$

$=h^{-1}{\sum }_{k=1,n}{c}_k\left[{\int }_{{\Delta }_{h,k}}\rho (y,x)I_{h,k}\left(x\right)dx\right]I_{h,k}\left(y\right)$.

Then, for $v=I_{h,i}$, $\left(T_{{\rho }_h}{I}_{h,i}\right)\left(y\right)=h^{-1}\left[{\int }_{{\Delta }_{h,i}}\rho (y,x)I_{h,i}\left(x\right)dx\right]I_{h,i}\left(y\right)$.

So, $\langle T_{{\rho }_h}{I}_{h.i},I_{h,j}\rangle =0$ for $i\ne j$ and the matrix representation of the finite rank operator $P_h^r\left(T_{\rho }\right):=T_{{\rho }_h}$, $M_h^r\left(T_{\rho }\right)=h^{-1}{\left[d_{ij}^h\right]}_{i,j=1,n}$ is sparse diagonal i.e. $d_{ij}^h=0,i\ne j$ with the diagonal entries given by:

$$d_{kk}^h={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}{I}_{h,k}\left(y\right)\rho (y,x)I_{h,k}\left(x\right)dxdy:={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\rho dxdy,k=1,n,nh=1$$

(7)

For finding the positivity parameter of $T_{{\rho }_h}$, let $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$. Because ${\parallel v\parallel }^2=h{\sum }_{k=1,n}{c}_k\overline{c_k}$, the inner product becomes

$\langle T_{{\rho }_h}{v}_h,v_h\rangle =h^{-1}{\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h$.

Thus, if $d_{kk}^h>0,\forall k=1,n,nh=1$, $\langle T_{{\rho }_h}{v}_h,v_h\rangle \ge {\alpha }_h\left(T_{{\rho }_h}\right){\parallel v_h\parallel }^2$ where

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

(8)

is the positivity parameter of the finite rank operator approximation $T_{{\rho }_h}$.

In order to meet the requests of the Lemma 1 (or Theorem 2), we need:

i.)- to show that the integral operator $T_{\rho }$ is positive on the dense set and,

ii.)- the sequence of the positivity parameters ${\alpha }_h\left(T_{{\rho }_h}\right)$ is inferior bounded.

i.) The kernel function $\rho$ is positive valued almost everywhere in $\left(\right(0,1)$, then:

${\int }_{(0,1)}{\int }_{(0,1)}\rho (y,x)dxdy>0$ and on any of the subintervals on a h-partition, $\langle T_{\rho }{I}_{h,k},I_{h,j}\rangle =0$ for $k\ne j$ because the functions of the indicator of intervals have disjoint supports. Now, for $v={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$:

$\langle T_{\rho }v,v\rangle ={\sum }_{k=1,n}{c}_k\overline{c_k}{\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\rho (y,x)dxdy={\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h>0$ proving the positivity of $T_{\rho }$ on every subspace in F and so, on the dense set S.

Let remark that $d_{kk}^h$ has the same formula like in (7) and so, we could define the positivity parameter of the restriction of $T_{\rho }$ onto $S_h$:

$${\alpha }_h\left(T_{\rho }\right)=h^{-1}mi{n}_{(k=1,n)}\left(d_{kk}^h\right):=h{\alpha }_h\left(T_{{\rho }_h}\right),nh=1$$

(9)

ii.) At this point, for proving the Alcantara-Bode equivalent we have only to show that the sequence of the positivity parameters of the finite rank approximations ${\alpha }_h\left(T_{{\rho }_h}\right)$ is inferior bounded.

4. The Proof of the RH Equivalent

The entries in the diagonal matrix representation $M_h\left(T_{{\rho }_h}\right)$ of the finite rank integral operator $T_{{\rho }_h}$ where the kernel function is $\rho (y.x)=\{y/x\}$ the fractional part of the ratio ($y/x$), $x,y\in (0,1)$ are given by:

$d_{kk}^h={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\rho (y,x)dxdy$, and valued (see also [1]) as follows:

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

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

(10)

for $k\ge 2$, where $\gamma$ is the Euler-Mascheroni constant (≃ 0.5772156). The expression in (10) for computing the fractional part of the ratio $\{y/x\}$ is suggested in [4]: for $0<a<b<2a$, $\{b/a\}=(b/a)-1$. 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 $\forall h$, we have with $k\ge 2$ : $d_{kk}^h>0.5h^2>d_{11}^h$. So,

$${\alpha }_h\left(T_{{\rho }_h}\right)=h^{-2}{d}_{11}^h=(3-2\gamma )/4>0,n\ge 2,nh=1.$$

(11)

showing that the finite rank approximations of the integral operator have the sequence of the positivity parameters inferior bounded.

Theorem 3. 

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

Proof. 

From (8), 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 holds. Then, the other half should be true i.e.: the Riemann Hypothesis is true. □