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

1. Introduction

The result obtained (Theorem 1) shows that a linear bounded operator T strictly positive on a dense set S in a separable Hilbert space H, is injective. Equivalently, its null space does not contain non null elements: $N_T=\left\{0\right\}$.

The positivity of a linear bounded operator T on S, $\langle Tv,v\rangle >0$ $\forall v\in S$ not null, ensures that the null space of T contains from S only the element 0, i.e. $N_T\cap S=\left\{0\right\}$. Thus, a zero of T could be only in the difference set $E:=H\setminus S$ considering that a linear combination of $u\in S$ and $v\in E$ not null, is inside the difference set. Let observe that $\left(T^{*}T\right)$ is non negative on the entire space and, an integral operator having the kernel function non negative valued enters in this category, making the method useful for any linear bounded operator provided that the operator is strict positive on the finite dimension subspaces of the family whose union is dense.

We will deal with positive operators on a dense set and the norm used here is the norm induced by the inner product. The idea is to consider the dense set in H be the union of a family F of finite dimension including subspaces $S_n$, $S_n\subset S_{n+1},n\ge 1$. For obtaining the necessary criteria for injectivity, we will exploit the relationship between the orthogonal projections of the eligible elements onto the family subspaces built in a multi-level fashion and the positivity parameters of the operator or its operator approximations on these subspaces. This framework is similar to multigrid discretisation (multi-level) methods used in applied mathematics.

Now, a linear bounded operator T positive on a finite dimension subspace is in fact strictly positive on it, i.e. there exists ${\alpha }_n\left(T\right)>0$ such that $\langle Tv,v\rangle \ge {\alpha }_n\left(T\right){\parallel v\parallel }^2\forall v\in S_n$. Suppose T be positive in each subspace $S_n\in F$. If there exists $\alpha >0$ such that ${\alpha }_n\ge \alpha$ for any $n\ge 1$ then T is strict positive on the dense set S and, by Theorem 1 introduced below, $N_T=\left\{0\right\}$. In this case is no need for further investigations.

If the sequence of the positivity parameters of T is not bounded, ${\alpha }_n\left(T\right)\to 0$ with $n\to \infty$, we consider two directions for investigation:

· involving the adjoint operator restrictions on the subspaces of the family, improving in the new context with Lemma 2 below the criteria introduced in [1] or,

· considering a sequence of positive operator approximations on subspaces.

An inferior bound of the positivity parameters of operator approximations, ensures strict positivity of the operator on the dense set. Lemma 1 address this case.

Both cases are analysed in the next paragraph. The third paragraph is dedicated to analyse the dense set most appropriate for obtaining operator or operator approximations having sparse matrix representations on the finite dimension subspaces whose union is dense in $L^2(0,1)$. The last paragraph is used for showing that on these subspaces the operator considered by Alcantara-Bode in [2] has the sequence of positivity parameters bounded inferior and so, verifying the criteria introduced in the third paragraph.

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. $\langle Tv,v\rangle >0$ $\forall v\in S$ not null, then T has no zeros in the dense set. Otherwise, if there exists $w\in S$ such that $Tw=0$ then $\langle Tw,w\rangle =0$ contradicts its positivity.

Follows: its ’eligible’ zeros are all in the difference set $E:=H\setminus S$, i.e. $N_T\subset E$. In our analysis we will take in consideration only the collection of eligible zeros that are on the unit sphere, without restricting the generality once for an element that is not null $w\in H,$ both w and $w/\parallel w\parallel$ are or are not together in $N_T$.

Theorem 1.

If $T\in \mathcal{L}\left(H\right)$ is strictly positive on a dense set of a separable Hilbert space then T is injective, equivalently $N_T=\left\{0\right\}$.

Proof.

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, the (1) 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:

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

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

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

(1)

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. From (1), ∣ 1 - $\parallel u_{{\epsilon }_n,w}\parallel$ ∣ $<1/n$ 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 $w\in E\cap N_T,\parallel w\parallel =1$ 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 dense set S, follows:

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

(2)

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 $n\to \infty$, contradicting its convergence $\parallel u_{{\epsilon }_n,w}\parallel \to 1$ with $n\to \infty$.

This occurs 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. $\square$

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: $S_n\subset S_{n+1},n\ge 1$.

Observation 1.

Let ${\beta }_n\left(u\right):=\parallel u-u_n\parallel$ be the normed residuum of element $u\in E$ after its orthogonal projection onto $S_n$. 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

${\beta }_{n_{ϵ}}\left(u\right):=\parallel u-P_{n_{ϵ}}u\parallel \le \parallel u-u_{ϵ}\parallel <ϵ$, valid for every $ϵ>0$, proving our assertion. Rewriting this, ${\beta }_n\left(u\right):=\parallel u-P_{n}u\parallel =\parallel (I-P_n)u\parallel \le \parallel I-P_n\parallel \parallel u\parallel \to 0$ for $n\to \infty$ for any $u\in H$ with $P_n$ the orthogonal projection onto $S_n$. □

Theorem 2.

Suppose that $T\in \mathcal{L}\left(H\right)$ has a sequence of operator approximations on the dense set S, $\{S_n,n\ge 1\},$ having the following properties:

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

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

If T is positive on S and there exists$\alpha >0$ such that

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

then $N_T=\left\{0\right\}$.

Proof.

Being positive on S, the operator does not have zeros in the dense set.

For $u\in E:=H\setminus S$, $\parallel u\parallel =1$ denoting the not null orthogonal projection over $S_n$ by $u_n:=P_{n}u,n\ge n_0:=n_0\left(u\right)$, then on any subspace $S_n$, 1 = ${\parallel u\parallel }^2={\parallel u_n\parallel }^2+{\beta }_n^2\left(u\right)$. If there exists $u\in N_T\cap E$, $\parallel u\parallel =1$, for it denoting ${\beta }_n:={\beta }_n\left(u\right)$ we have from ii):

$0<{\alpha }_n\parallel u_n{\parallel }^2\le \langle T_n{u}_n,u_n\rangle \le \parallel T_n{u}_n\parallel \parallel u_n\parallel$.

Estimating $\parallel T_n{u}_n\parallel$,

$\parallel T_n{u}_n\parallel =\parallel T_n{u}_n-T{u}_n+T{u}_n-Tu\parallel$$\le (\parallel T-T_n\parallel \parallel u_n\parallel +\parallel T\parallel \parallel u-u_n\parallel )$

= $({ϵ}_n\parallel u_n\parallel +\parallel T\parallel {\beta }_n)$,

we observe that $\parallel T_n{u}_n\parallel \to 0$ because $\parallel u_n\parallel \to 1$, ${ϵ}_n\to 0$ (from i)) and, ${\beta }_n\to 0$ (from Observation 1). Now, from iii)

$\alpha \parallel u_n{\parallel }^2\le ({ϵ}_n+\parallel T\parallel {\beta }_n/\parallel u_n\parallel )\parallel u_n{\parallel }^2$.

From Observation 1 we have ${\beta }_n/\parallel u_n\parallel ={\beta }_n/\sqrt{1-{\beta }_n^2}\to 0$. So,

$\alpha \le \left({ϵ}_n+\parallel T\parallel {\beta }_n/\parallel u_n\parallel \right)\to 0$.

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

Let $T:=T_{\rho }$ be a Hilbert-Schmidt integral operator. A technique for obtaining approximations for $T:=T_{\rho }$ to verify i) was used in [5], [6]. When $T_n,n\ge 1$, $T_n:=P_n^r\left(T\right)$ are approximations of $T_{\rho }$ on the subspaces of family F obtained through a class of finite rank operators - that are orthogonal projection integral operators $\{P_n^r,n\ge 1\}$, then from $\parallel I-P_n^r\parallel \to 0$, we obtain the property i). In the next paragraph we show that $\{P_n^r,n\ge 1\}$ is a collection of finite rank projection operators on a family of finite dimension subspaces (see [5]) whose union is dense in $H:=L^2(0,1)$. Moreover, if the operator approximations $\{T_n,n\ge 1\}$ verifies ii), we can show that the operator T is strictly positive on the dense set S provided that their positivity parameters sequence is bounded.

Lemma 1.

(Criteria for operator approximations). If the finite rank approximations of a positive Hilbert-Schmidt integral operator $T_{\rho }$ verify the conditions ii) and iii) from Theorem 2, then $T_{\rho }$ is strictly positive on the dense set.

Proof.

From the convergence to zero of the sequence ${ϵ}_n,n\ge 1$ there exists ${ϵ}_0$ a parameter ${ϵ}_0$ such that ${ϵ}_0:=ma{x}_n\{{ϵ}_n;{ϵ}_n<\alpha \}$, corresponding to a subspace $S_{n_0},n_0<\infty$. This parameter is independent of any $v\in S$ and, because of the inclusion property, for any $n<n_0$ we have $S_n\subset S_{n_0}$. We could consider $S_{n_0}$ to be $S_1$ discarding a finite number of subspaces or, we could consider v to be inside of $S_{n_0}$. Then:

${\alpha }_n\ge \alpha >{ϵ}_0\ge {ϵ}_n$ for $n\ge 1$, resulting $({\alpha }_n-{ϵ}_n)>(\alpha -{ϵ}_0)>0$ $\forall n\ge 1$.

For an arbitrary $v\in S$ there exists a coarser subspace (i.e. with a smaller dimension) $S_n,n\ge n_1:=n_1\left(v\right)$, for which $v\in S_n$. For it, with $T:=T_{\rho }$ we have:

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

$\langle Tv,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2-\langle (T_n-T)v,v\rangle$.

Because T and $T_n$ are positive on $S_n$, the inner product in the right side of the inequality is real valued and, $|\langle (T_n-T)v,v\rangle |\le {ϵ}_n{\parallel v\parallel }^2$.

So, if $\langle (T_n-T)v,v\rangle \ge 0$, then $\langle (T_n-T)v,v\rangle \le {ϵ}_n{\parallel v\parallel }^2$. From ${ϵ}_n<{\alpha }_n$, follows:

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

Now, if $\langle (T_n-T)v,v\rangle <0$, then $\langle Tv,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2\ge {\alpha \parallel v\parallel }^2\ge (\alpha -{ϵ}_0){\parallel v\parallel }^2$.

Thus, taking $\alpha \left(T\right)=(\alpha -{ϵ}_0)$, for any $v\in S$ we obtain

$\langle Tv,v\rangle \ge {\alpha \left(T\right)\parallel v\parallel }^2$, i.e. $T_{\rho }$ is strict positive on the dense set S. □

Corollary. 

If $Q\in \mathcal{L}\left(H\right)$ is an Hermitian Hilbert-Schmidt operator verifying on a dense set S the properties ii) and iii) from Theorem 2, then Q is injective.

Proof.

Being Hermitian, the operator verifies $\langle Qv,v\rangle \ge 0$, for every $v\in H$. Being Hilbert-Schmidt it could be approximated on a dense family of finite dimension subspaces, its sequence of operator approximations verifying i). Then,

$\langle Qv,v\rangle =\langle Q_{n}v,v\rangle -\langle (Q_n-Q)v,v\rangle \ge 0$ for any $v\in S$. Following the steps from the proof of Lemma 1 we obtain that:

$\langle Qv,v\rangle \ge (\alpha -{ϵ}_0){\parallel v\parallel }^2$ meaning that Q is strictly positive on the dense set. Thus, due to Theorem 1/Lemma 1, we obtain $N_Q=\left\{0\right\}$. □

Now, reformulating the injectivity criteria introduced in [1], we have the following lemma, useful when a sequence of operator approximations could not be obtained.

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 parameters:

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

If $C>0$ exists 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, denoting with ${\beta }_n:={\beta }_n\left(u\right)=\parallel u-u_n\parallel$, we obtain from the (strict) positivity of T on each of the subspaces $S_n,n\ge 1$ (as in (2)),

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

Rewriting,

$C\le {\mu }_n\le {\beta }_n/\sqrt{1-{\beta }_n^2}\to 0$ that is a contradiction from a range $n_0\left(u\right)$. Thus, $u\notin N_T$ $\forall u\in H\setminus S$. Follows: $N_T=\left\{0\right\}$. □

3. Approximations on Subspaces

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 ${\Delta }_{h,n}$ define for $m\ge 1$ a partition of (0,1), k=1,n, $n=2^m,nh=1$. Consider the interval indicator functions that have as support these intervals (k=1,n), nh=1:

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

(3)

The family F of finite dimensional subspaces $\{S_h,nh=1,n\ge 2\}$ that are the linear spans of interval indicator functions of the h-partitions defined by (3) 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, this property is obtained from (3) observing that any $I_{h,i}\in S_h,i=1,n,nh=1$ can be rewritten as

$I_{h,i}=I_{h/2,2i-1}+I_{h/2,2i}\in S_{h/2}$.

With the observation that the set $S={\cup }_{n\ge 1}{S}_h,nh=1$ is dense in H well known in the literature, until now we have met the requests of previous lemmas needed to investigate the injectivity of an integral operator $T_{\rho }$.

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

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

(4)

is a finite rank integral operator orthogonal projection having the spectrum {0, 1} with eigenvalue 1 of 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, as a consequence, obviously ${\left(P_h^r\right)}^2=P_h^r$ for $n\ge 2,nh=1$. For any $u\in H$,

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

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

that is the standard orthogonal projection of u onto $S_h$. Now, for $I_{h,j}\in S_h$,

$P_h^r\left(I_{h,j}\right)=h^{-1}{\sum }_{k=1,n}{c}_k{I}_{h,k}$, where

$c_k={\int }_{{\Delta }_{h,k}}{I}_{h,j}\left(x\right)dx$, is valued as $c_k=h\mathrm{for}\mathrm{k}=\mathrm{j}$ and 0 for $k\ne j$.

$P_h^r\left(I_{h,j}\right)=I_{h,j}$ and therefore, $P_h^r\left(v_h\right)=v_h$ for every $v_h\in S_h$ involving ${\left(P_h^r\right)}^{2}u=P_h^{r}u$ for any $u\in H$.

Because $P_h^r$ is an orthogonal projection onto $S_h$ and due to the including properties of the finite dimension subspaces whose union is dense,

$\parallel I-P_h^r\parallel \to 0$ for $n\to \infty ,nh=1$. Therefore, from $(T_{\rho }-P_h^r\left(T_{\rho }\right))=(I-P_h^r)T_{\rho }$ the property i) in Theorem 2 holds for any integral operator $T_{\rho }\in \mathcal{L}\left(H\right)$ on the family of finite dimension subspaces spanned by indicator interval functions associated with partitions defined by (3). In fact, ${ϵ}_n:=\parallel T-T_n\parallel =su{p}_{u\in H,\parallel u\parallel =1}\parallel (T-T_n)\left(u\right)\parallel =su{p}_{u\in H,\parallel u\parallel =1}\parallel (I-P_h^r)\left(Tu\right)\parallel \le \parallel (I-P_h^r)\parallel \parallel T\parallel \to 0$.

Remark 1.

The matrix representation of $T_{\rho }$ restriction to $S_h$ is a sparse diagonal matrix: its elements outside the diagonal are zero valued.

Proof.

The inner product on the subspace $S_h$ between $u\notin S_h$ and $v_h\in S_h$ is the result of the orthogonal projection of u and $v_h$, similar to an inner product between two step functions: $\langle u,v_h\rangle :=\langle P_h^{r}u,v_h\rangle$. If $P_h^{r}u:=u_h={\sum }_{k=1,n}{a}_k{I}_{h,k}$ and $v_h={\sum }_{j=1,n}{c}_j{I}_{h,j}$, owing to the disjoint supports of the indicator interval functions, $\langle I_{h,k},I_{h,j}\rangle =0$ for $k\ne j$ and, their inner product is

$\langle u_h,v_h\rangle ={\sum }_{k=1,n}{a}_k\overline{c_k}\langle I_{h,k},I_{h,k}\rangle$.

Let $T_{\rho }$ be a Hilbert-Schmidt integral operator on H. Now,

$T_{\rho }{I}_{h,k}={\int }_0^1\rho (y,x)I_{h,k}\left(x\right)dx={\int }_{{\Delta }_{h.k}}\rho (y,x)I_{h,k}\left(x\right)dx$.

Follows:

$\langle T_{\rho }{I}_{h,k},I_{h,j}\rangle ={\int }_0^1\left[{\int }_{{\Delta }_{h,k}}\rho (y,x)I_{h,k}\left(x\right)dx\right]I_{h,j}\left(y\right)dy$

$={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,j}}{I}_{h,j}\left(y\right)\rho (y,x)I_{h,k}\left(x\right)dxdy=0$ for $k\ne j$ because $I_{h,k}$ and $I_{h,j}$ have disjoint supports for $k\ne j$.

Then, the matrix representation of $T_{\rho }$ restriction on $S_h$, $M_h\left(T_{\rho }\right)$ is a sparse diagonal matrix having the diagonal entries

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

$\langle T_{\rho }{v}_h,v_h\rangle ={\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h$ for any $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}$ from $S_h$. □

The integral operator approximation of $T_{\rho }$ on $S_h$ denoting it by $T_{{\rho }_h}$, is a finite rank operator approximation, with a kernel function ([5])

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

(5)

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

Remark 2.

The matrix representation of $T_{{\rho }_h}$ is a sparse diagonal matrix and,

$M_h^r\left(T_{\rho }\right)=h^{-1}{M}_h\left(T_{\rho }\right)$.

Evaluating $T_{{\rho }_h}v$ for $v=I_{h,i}$, we obtain

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

$\langle T_{{\rho }_h}{I}_{h.i},I_{h,j}\rangle =0$ for $i\ne j$ and the matrix representation of the finite rank operator $P_h^r\left(T_{\rho }\right):=T_{{\rho }_h}$, is: $M_h^r\left(T_{\rho }\right)=h^{-1}\mathrm{diag}{\left[d_{kk}^h\right]}_{k=1,n}$. It is a sparse diagonal matrix because $d_{ij}^h=0$ for $i\ne j$ having the diagonal entries

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

(6)

Follows: $M_h^r\left(T_{\rho }\right)=h^{-1}{M}_h\left(T_{\rho }\right)$, showing that both matrices are or are not together positive. More, $\forall v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$: $\langle T_{{\rho }_h}{v}_h,v_h\rangle =h^{-1}\langle T_{\rho }{v}_h,v_h\rangle$ □

Remark 3.

If the diagonal entries of the matrix representations are strict positive, $d_{kk}^h>0,\forall k=1,n,nh=1$, then $T_{\rho }$ is positive.

Proof.

From ${\parallel v_h\parallel }^2=h{\sum }_{k=1,n}{c}_k\overline{c_k}$ we obtain:

$\langle T_{{\rho }_h}{v}_h,v_h\rangle \ge {\alpha }_h\left(T_{{\rho }_h}\right){\parallel v_h\parallel }^2$ where ${\alpha }_h\left(T_{{\rho }_h}\right)$ is the positivity parameter of the finite rank operator approximation $T_{{\rho }_h}$ given by

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

(7)

and, $\langle T_{\rho }{v}_h,v_h\rangle ={\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h$ $=h\langle T_{{\rho }_h}{v}_h,v_h\rangle$ showing that $T_{\rho }$ is positive on $S_h$ if and only if $T_{{\rho }_h}$ is positive on $S_h$. Moreover, the following relationship holds

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

(8)

4. Proof of the Alcantara-Bode Equivalent

Alcantara-Bode ([2], pg. 151) in his theorem of the equivalent formulation obtained from the Beurling equivalent formulation ([4]) of RH, states:

$$\mathit{The}\mathit{Riemann}\mathit{Hypothesis}\mathit{holds}\mathit{if}\mathit{and}\mathit{only}\mathit{if} T_{T_{\rho }}=\left\{0\right\}$$

where $T_{\rho }$ is a Hilbert-Schmidt integral operator ([2]) whose kernel $\rho (y,x)=\{y/x\}$ is the fractional part function of the ratio ($y/x$). The kernel function $\rho \in L^2{(0,1)}^2$ is continue almost everywhere. Its discontinuities in ${(0,1)}^2$ consist of a set of numerable one dimensional lines of the form $y=kx,k\in N$, with Lebesgue measure zero.

The entries in the diagonal matrix representation $M_h\left(T_{{\rho }_h}\right)$ of the finite rank integral operator $T_{{\rho }_h}$ are given by: $d_{kk}^h={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\{y/x\}dxdy$, as valued in [1]:

$$d_{11}^h=h^2(3-2\gamma )/4;d_{kk}^h={\displaystyle \frac{h^2}{2}}(-1+{\displaystyle \frac{2k-1}{k-1}}ln{\left({\displaystyle \frac{k}{k-1}}\right)}^{k-1}),\mathrm{for}k\ge 2$$

(9)

where $\gamma$ is the Euler-Mascheroni constant (≃ 0.5772156...).

The formulae in (9) were computed using the suggestion found in [4] for the fractional part: for $0<a<b<2a$, $\{b/a\}=(b/a)-1$. Subsequently,

${\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\{y/x\}dxdy={\int }_{{\Delta }_{h,k}}\left[{\int }_{{\Delta }_{h,k}}(y/x)dx-{\int }_{(k-1)h}^{y}dx\right]dy.$ The sequence

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

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

(10)

showing that the positivity parameters of the sequence of operator approximations $\{T_{{\rho }_h},nh=1,n\ge 2\}$ verifies ii) and iii) properties in Lemma 1 (Theorem 2).

Theorem 3.

The Alcantara-Bode equivalent holds involving that RH is true.

Proof.

Having $d_{kk}^h>0$ for any $k=1,n$, $nh=1$ (see (9)), the operator is positive on the dense set S. Thus, we can consider both cases of the numerical method.

A) 

Finite rank approximations.

With (10), we obtain the bound of the positivity parameters of the operator approximations on the subspaces of the family F. From Lemma 1 follows the strict positivity of the operator on the dense set. Subsequently, from Theorem 1, $N_{T_{\rho }}=\left\{0\right\}$.

B) 

Injectivity criteria.

Using (8), (9) and (10) we obtained the positivity of the operator on the dense set, observing that ${\alpha }_h\left(T_{\rho }\right)\to 0$. Therefore, Lemma 2 must be used invoking the adjoint operator whose kernel function is

${\rho }^{*}(y,x)=\overline{\rho (x,y)}=\rho (x,y)$. For $v_h={\sum }_{k=1,n}{c}_k{I}_h^k\in S_h$

$T_{\rho }^{*}{v}_h={\sum }_{k=1,n}{c}_k{\int }_{{\Delta }_{h,k}}\rho (x,y)I_h^k\left(y\right)dy$$={\sum }_{k=1,n}{c}_k{\int }_{{\Delta }_{h,k}}{\rho }_{h,k}(x,y)dy$,

where ${\rho }_{h,k}=I_{h,k}\left(x\right)\rho (x,y)I_{h,k}\left(y\right)$. Follows:

${\parallel T_{\rho }^{*}{v}_h\parallel }^2=\langle {\sum }_{k=1,n}{c}_k{\int }_{{\Delta }_{h,k}}{\rho }_{h,k}(x,y)dy,{\sum }_{j=1,n}{c}_j{\int }_{{\Delta }_{h,j}}{\rho }_{h,j}(x,y)dy\rangle$

$={\sum }_{k=1,n}{c}_k\overline{c_k}\left({\int }_{{\Delta }_{h,k}}{\left[{\int }_{{\Delta }_{h,k}}\rho (x,y)I_{h,k}\left(y\right)dy\right]}^2{I}_{h,k}\left(x\right)dx\right)$.

Because $\rho (x,y)$ is valued in [0,1), $\rho (x,y)<1$ for every $x,y\in (0,1)$, obtaining:

${\parallel T_{\rho }^{*}{v}_h\parallel }^2\le {\sum }_{k=1,n}{c}_k\overline{c_k}{h}^3=h^2{\parallel v_h\parallel }^2$ and, $\parallel T_{\rho }^{*}{v}_h\parallel \le h\parallel v_h\parallel$ for every $v_h\in S_h$.

With ${\omega }_h\left(T_{\rho }^{*}\right)=h$, the injectivity parameter of T on $S_h$ given by ${\mu }_h={\alpha }_h\left(T_{\rho }\right)/{\omega }_h\left(T_{\rho }^{*}\right)$ is evaluated as

$${\mu }_h=(3-2\gamma )/4>0,\mathrm{for}\mathrm{any}h,nh=1.$$

(11)

a constant on every subspace. Then, applying Lemma 2 we obtain $N_{T_{\rho }}=\left\{0\right\}$.

In each case the result is $N_{T_{\rho }}=\left\{0\right\}$. Then in each case the conclusion is: half of Alcantara-Bode equivalent formulation of the Riemann Hypothesis holds involving the other half should hold. Therefore the Riemann Hypothesis is true. □

Observations.

· A connection between Zeta function $\zeta$ and the integral operator $T_{\rho }$ can be observed in [4] by reformulating the left term in the expression as $\left(T_{\rho }{x}^{s-1}\right)\left(\theta \right)$:

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

· Considering the indicator of semi-open intervals functions of a partition of the domain, the subspaces are including ($S_h\subset S_{h/2}$) ensuring the monotony of the positivity parameters. If we replace the indicator open-interval functions for generating the subspace $S_h^o$ as well as the indicator closed-interval functions generating the subspace $S_h^c$, $nh=1,n\ge 1$ then both sets $S^o$ and $S^c$ are still dense like S losing instead the including subspaces property. Information on the density of the set $S^o$ could be found in textbooks of functional analysis or on math.stackexchange.com. On the density of S, we showed in V4 of [11] that if one of the sets S, $S^c$ and $S^o$ is dense, then others are dense. A sketch of proof follows. Let $S^o$ be dense. 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, $\forall n,nh=1$, showing that f is orthogonal to any $I_h^k$ and so f should be 0 because $S^0$ is dense. So, S is dense.

· The dense sets S and $S^c$ have been used in [5] and [6] to obtain optimal evaluations of the decay rate of convergence to zero of the eigenvalues of Hermitian integral operators having a kernel function such as Mercer kernels ([9]).

The references [13-16] are related to other RH equivalents, [8] to exotic integrals and [12] to multi-level discretisations on separable Hilbert spaces.