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

1. Introduction

Let H be a separable Hilbert space. A result obtained (Theorem 1 below) shows that the null space of a linear bounded operator strict positive on a dense set in H, does not contain non null elements. Consider F be a family of finite dimension including subspaces $S_n,n\ge 1$ such that their union S is a dense set in the separable Hilbert space H. Dense sets having such properties there exist, for example when $H:=L^2(0,1)$ then S could be built in a multi-level fashion using indicator functions of the disjoint intervals of the domain partitions (see the paragraph 3). Such families could be obtained also from a basis in H, a subspace $S_n$ being spanned by first n elements from the basis for example.

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. $N_T\cap S=\left\{0\right\}$.

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 $\langle Tv,v\rangle \ge {\alpha }_n\left(T\right){\parallel v\parallel }^2$ for every $v\in S_n$.

Suppose T positive on each subspace from the family 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 below $N_T=\left\{0\right\}$.

If instead the sequence of the positivity parameters of T is converging to zero, ${\alpha }_n\left(T\right)\to 0$ with $n\to \infty$, we consider two directions for the investigation of its injectivity providing the theory and the methods needed for:

- involving the adjoint operator restrictions on the subspaces of the family, having as support Lemma 2 below;

- considering a sequence of positive operator approximations on subspaces, if there exists one such that the sequence is convergent in norm to the operator

$\parallel T-T_n\parallel \to 0\mathrm{with}n\to \infty$ and, whose corresponding sequence of positivity parameters is inferior bounded: there exists $\alpha >0$ such that

$\langle T_{n}v,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2\mathrm{for}\mathrm{any}n\ge 1$ where ${\alpha }_n:={\alpha }_n\left(T_n\right)\ge \alpha$. The Theorem 2 and Lemma 1 below are dealing with this method.

While the criteria involving the adjoint operator (introduced in [1 ]) could be applied to any positive, linear bounded operator, for involving the operator approximations we have to find the means for obtaining a proper approximation schema in terms of the convergence of the approximations to the original operator and, such operator approximations should be positive on the subspaces in F having an inferior bound for the sequence of the positivity parameters.

However, note that the positivity of the operator could be solved by replacing it by its associated Hermitian ($T^{*}T$) that has the same null space with T and it is non negative definite on H.

Let observe that there is a connection between the two kind of positivity parameters on each subspace: if h is the length of the intervals in a partition of (0,1), $nh=1$, then for the Hilbert-Schmidt integral operator on $L^2(0,1)$ of our interest, we obtained ${\alpha }_n\left(T\right)=n^{-1}{\alpha }_n\left(T_n\right),n\ge 1$ with ${\alpha }_n\left(T_n\right)$ a constant mesh independent.

2. Two Theorems on Injectivity and Associated Methods

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

Follows: its ’eligible’ zeros are all in the difference set $E:=H\setminus S$, i.e. $N_T\subset E$.

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 $w\in H,w\ne 0$ both w and $w/\parallel w\parallel$ are or 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, (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:

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

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

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

(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. Follows from (1), ∣ 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:

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

(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 $n\to \infty$, 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. □

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 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. □

Rewriting it, ${\beta }_n\left(u\right):=\parallel u-P_{n}u\parallel =\parallel (I-P_n)u\parallel \le \parallel I-P_n\parallel \parallel u\parallel \to 0$ for $n\to \infty$ for any $u\in H$ with $P_n$ the orthogonal projection onto $S_n$.

For $T\in \mathcal{L}\left(H\right)$, let $T_n,n\ge 1$ be a sequence of operator approximations on $S_n,n\ge 1$ having the property ${ϵ}_n:=\parallel T-T_n\parallel \to 0$ and, 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 $T\in \mathcal{L}\left(H\right)$ be positive on the dense set S. If the sequence $\{T_n,n\ge 1\}$ of its approximations on the family F verifies:

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

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

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.

For $u\in E:=H\setminus S$, $\parallel u\parallel =1$ denoting the not null orthogonal projection over $S_n$ by $u_n:=P_{n}u,n\ge n_0:=n_0\left(u\right)$, we have on any subspace $S_n$, ${\parallel u\parallel }^2={\parallel u_n\parallel }^2+{\beta }_n^2\left(u\right)$ where ${\beta }_n\left(u\right)=\parallel u-P_{n}u\parallel :={\beta }_n$ is its (normed) residuum. We have: $\parallel u_n\parallel \uparrow 1$ and ${\beta }_n\left(u\right)\to 0$.

If there exists $u\in N_T\cap E$, $\parallel u\parallel =1$ then for it:

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

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

$\le (\parallel T-T_n\parallel \parallel u_n\parallel +\parallel T\parallel \parallel u-u_n\parallel )\parallel u_n\parallel$

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

evaluation obtained because $\parallel u_n\parallel <1$. Then from Observation 1 and iii) we have:

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

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

We will deal now, with the special case of the approximations of the Hilbert-Schmidt integral operators that, being compact operators could be approximated in a proper manner on finite dimension subspaces so, the condition i) is satisfied ([3]).

Let $T:=T_{\phi }$ be a Hilbert-Schmidt integral operator. A technique for obtaining the approximations for an integral operator is used in [5]. Thus, the condition i) 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 $T_n:=P_n^r\left(T\right)$. Then, for every $u\in H$ not null:

$\parallel Tu-T_{n}u\parallel =\parallel (I-P_n^r)Tu\parallel \le \parallel I-P_n^r\parallel \parallel Tu\parallel \to 0$

Lemma 1

((Criteria for finite rank approximations)). If the finite rank approximations of a positive linear Hilbert-Schmidt integral operator $T_{\phi }$ on a dense set S are positive on the family of approximation subspaces F and the sequence of the positivity parameters is inferior bounded,

then $T_{\phi }$ is strict positive on the dense set so, it is injective.

Proof. 

The requests i), ii) in the Theorem 2 hold 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, due to the including property, for any $n<n_0$ we have $S_n\subset S_{n_0}$. We could consider $S_{n_0}$ be $S_1$ or, your choice, we could consider v as being 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, 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 $S_n$,

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

Now, T and $T_n$ are positive on $S_n$. Then the inner product in the right side of the equality 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 $\langle (T_n-T)v,v\rangle \le {ϵ}_n{\parallel v\parallel }^2$. Because ${ϵ}_n<{\alpha }_n$, follows:

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

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

Thus, taking $\alpha \left(T\right)=(\alpha -{ϵ}_0)$,

$\langle Tv,v\rangle \ge {\alpha \left(T\right)\parallel v\parallel }^2$ for any $v\in S$ meaning that T is strict positive on the dense set S and from Theorem 1, $N_T=\left\{0\right\}$. □

Corollary 1.

If $Q\in \mathcal{L}\left(H\right)$ is a Hermitian compact operator verifying on a dense set the properties i) and ii) from Theorem 2, then it is injective.

Proof. 

Being Hermitian, the operator verifies $\langle Qv,v\rangle \ge 0$, for every $v\in H$. Being compact it admits on a dense family of finite dimension subspaces a sequence of approximations. Then, for any $v\in S$,

$\langle Qv,v\rangle =\langle Q_{n}v,v\rangle -\langle (Q_n-Q)v,v\rangle \ge 0$ obtaining following the steps from the proof of Lemma 1 that in the hypotheses i) and ii) holds:

$\langle Qv,v\rangle \ge (\alpha -{ϵ}_0){\parallel v\parallel }^2$ meaning that Q is strict positive on the dense set. Thus, $N_Q=\left\{0\right\}$ due to the Theorem 1. Let observe that if $Q=T^{*}T$ then $N_T=N_Q=\left\{0\right\}$ result obtained without requesting the positivity of Q or T on the dense set. □

The following lemma is dealing with the cases in which a proper sequence of operator approximations could not be defined (see the Injectivity Criteria in [1]).

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 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 each of the subspaces $S_n,n\ge 1$ (see (2)):

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

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, $u\notin N_T$ affirmation valid 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:

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

(3)

The family F of finite dimensional subspaces $S_h$ 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, the property is obtained from (3) 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:

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

(4)

is a 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, 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 f = $I_{h,j}\in S_h$, $j=1,n,nh=1$:

$P_h^r\left(f\right)=h^{-1}{\sum }_{k=1,n}{c}_k{I}_{h,k}$, where $c_k={\int }_{{\Delta }_{h,k}}{I}_{h,j}\left(x\right)dx$, with $c_k=h\mathrm{for}\mathrm{k}=\mathrm{j}$ and 0 for $k\ne j.$ Then,

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

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

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

Remark 1.

The matrix representation of $T_{\rho }$ on $S_h$ 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 a result between the orthogonal projection of u and $v_h$, like 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 $v_h={\sum }_{j=1,n}{c}_j{I}_{h,j}$, due 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 $T_{\rho }$ 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, with $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}$,

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

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

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

(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 ${\Delta }_{h,k}\times {\Delta }_{h,k},k=1,n,nh=1$.

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 the previous relationship 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:

$M_h^r\left(T_{\rho }\right)=h^{-1}\mathrm{diag}{\left[d_{kk}^h\right]}_{k=1,n}$, a sparse diagonal because $d_{ij}^h=0$ for $i\ne j$ and with the diagonal entries given by:

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

(6)

Follows: $M_h^r\left(T_{\rho }\right)=h^{-1}{M}_h\left(T_{\rho }\right)$, i.e. both matrices are or are not simultaneous positive. □

Pointing out:

$\langle T_{{\rho }_h}{v}_h,v_h\rangle =h^{-1}\langle T_{\rho }{v}_h,v_h\rangle =h^{-1}{\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h$, for any $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h,v_h\ne 0$.

Remark 3.

If $d_{kk}^h>0,\forall k=1,n,nh=1$, because $\parallel v_h{\parallel }^2=h{\sum }_{k=1,n}{c}_k\overline{c_k}$ 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

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

(7)

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

From $\langle T_{\rho }{v}_h,v_h\rangle ={\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h$ $=h\langle T_{{\rho }_h}{v}_h,v_h\rangle$ results that $T_{\rho }$ is positive on $S_h$ if and only if $T_{{\rho }_h}$ is positive on $S_h$. Then, if on every subspace $S_h\in F$, $d_{kk}^h>0$$k=1,n,nh=1$, 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)

Remark 4.

Thus, the positivity of the linear bounded integral operator $T_{\rho }$ on the dense set S is determined by the diagonal entries in its matrix representations.

4. Proof of the Alcantara-Bode Equivalent

We are now in position to prove RH showing that the integral operator

$\left(T_{\rho }u\right)\left(y\right)={\int }_0^1\rho (y,x)u\left(x\right)dx,u\in L^2(0,1),$ where $\rho (y,x)=\{y/x\}$ is the fractional function of the ratio $y/x$, has its null space $N_{T_{\rho }}=\left\{0\right\}$.

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

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

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:

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

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

(9)

for $k\ge 2$, where $\gamma$ is the Euler-Mascheroni constant (≃ 0.5772156...). The formula (9) has been computed using for the fractional part the suggestion found in [4]: for $0<a<b<2a$, $\{b/a\}=(b/a)-1$. Then,

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

The sequence from (9)

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

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

(10)

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 of RH holds.

Proof. 

From (10) results that the sequence of the positivity parameters of the finite rank operator approximations $T_{{\rho }_h}$ on the dense family F is inferior bounded, ${\alpha }_h\left(T_{{\rho }_h}\right)$ being a constant mesh independent. Then, $N_{T_{\rho }}=\left\{0\right\}$ is obtained from Lemma 1 (or Theorem 2). □

We use now the method covered by Lemma 2 (see also [1], Injectivity Criteria). The integral operator $T_{\rho }$ is (strict) positive on $S_h\in F,n\ge 2,nh=1$ with the parameter valued from (8) and (10),

${\alpha }_h\left(T_{\rho }\right)=h{\alpha }_h\left(T_{{\rho }_h}\right)=h(3-2\gamma )/4\to 0$ with $n\to \infty$.

In order to apply Lemma 2, we should invoke the adjoint operator of $T_{\rho }$ 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), $\rho (y,x)<1$ for every $x,y\in (0,1)$, we 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, $\parallel T_{\rho }^{*}{v}_h\parallel \le h\parallel v_h\parallel$.

Taking ${\omega }_h\left(T_{\rho }^{*}\right)=h$, the injectivity parameter of T on $S_h$ is given by:

$${\mu }_h=(3-2\gamma )/4,\mathrm{a}\mathrm{mesh}\mathrm{independent}\mathrm{constant}\forall n,nh=1$$

(11)

Theorem 4

((Injectivity Criteria):). The Alcantara-Bode equivalent of RH holds.

Proof. 

Because ${\mu }_h$ is a constant (see (11)) for any $h,nh=1$, applying Lemma 2 we obtain $N_{T_{\rho }}=\left\{0\right\}$. □

Proposition 1.

The Riemann Hypothesis is true.

Proof. 

From Theorem 3 or Theorem 4 we obtained $N_{T_{\rho }}=\left\{0\right\}$, meaning that half from Alcantara-Bode equivalent of RH holds. Then, the other half should hold, i.e.: the Riemann Hypothesis is true. □

Observations.

We considered the subspaces $S_h$ spanned by indicator of semi-open intervals functions of a partition of the domain and so, the subspaces are including ($S_h\subset S_{h/2}$) providing the monotony of the positivity parameters. If we take instead the indicator open-intervals functions for generating the subspace $S_h^o$ as well of the indicator closed-intervals functions generating the subspace $S_h^c$, $nh=1,n\ge 1$ then both sets $S^o$ and $S^c$ are dense like S, easy to show ([11]).

The dense sets S and $S^c$ have been used in [5] and respectively [6] for obtaining optimal evaluations of the decay rate of convergence to zero of the eigenvalues of Hermitian integral operators having the kernel like Mercer kernels ([9]).