On the Method for Proving RH Using the Alcantara-Bode Equivalence (II)

1. Introduction

The functional-numerical method is built on the result obtained with Theorem 1, a generic ’theoretical’ criterion: if a linear bounded operator T on a separable Hilbert space H is strict positive on a dense set S then T is injective - equivalently, its null space contains only element 0: $N_T=\left\{0\right\}$.

The author of [2] proved that the integral operator of interest involved in the equivalent formulation of RH is Hilbert-Schmidt and so, it is compact. T is a compact operator if and only if there exist a sequence $\left\{T_n\right\},n\ge 1$ of finite rank operators such that $\parallel T-T_n\parallel \to 0$. We showed the convergence of approximations of a linear bounded integral operator on $L^2(0,1)$ on a family of subspaces built on indicator of interval functions associated to partitions of the domain of uniform mesh. Like in discretization schema used in [5], the operator approximations on the finite dimension subspaces whose union is dense, are finite rank and convergent in norm to the original integral operator. If the approximations are positive definite and the sequence of positivity parameters is bounded by a strict positive constant, then the operator is injective. One could replace the operator with its associated Hermitian since both have the same null space and, it is at least semi-positive definite allowing the investigation of the associated operator approximations positivity.

2. Injectivity Criteria

Let H be a separable Hilbert space and denote with $\mathcal{L}\left(H\right)$ the class of linear bounded operators on H. We denote $E:=H\setminus S$ be the difference set or the set of eligible zeros. We will consider only eligible elements from the unit sphere ($\parallel u\parallel =1$) because $\forall u\in H$ not null, u and $u/\parallel u\parallel$ both are or are not in the null space of T.

Theorem 1.

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

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, (1) below results as follows. If $\parallel w\parallel \ge \parallel u_{\epsilon ,w}\parallel$:

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

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 \mid$$<1/n$ showing that, for any choices of a sequence approximating w, $u_{{\epsilon }_n,w}\in S,n\ge 1$, it verifies $\parallel u_{{\epsilon }_n,w}\parallel \to 1$.

Let $T\in \mathcal{L}\left(H\right)$ strict positive on S, i.e. $\exists \alpha >0$ such that $\forall u\in S$, $\langle Tu,u\rangle \ge {\alpha \parallel u\parallel }^2$. If $\exists w\in E\cap N_T,\parallel w\parallel =1$, 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 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 $\parallel u_{{\epsilon }_n,w}\parallel \le c/n$. Subsequently, $\parallel u_{{\epsilon }_n,w}\parallel \to 0$ 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 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. □

2.1 Finite rank operator approximations.

Suppose that there exists a dense set S as a result of an union of finite dimension of including subspaces of a family F: $S_n\subset S_{n+1},n\ge 1$, $S={\cup }_{n\ge 1}{S}_n,\overline{S}=H$.

For $T\in \mathcal{L}\left(H\right)$, let $\{T_n;n\ge 1\}$ be a sequence of its operator approximations on the family F. We will say that T satisfies the approximation properties if:

i) ${ϵ}_n:=\parallel T-T_n\parallel \to 0$ for $n\to \infty$; (convergence of approximations)

ii) $\langle T_{n}v,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2$, ${\alpha }_n>0$$\forall v\in S_n,S_n\in F$ (positivity of approximations).

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

${\beta }_n\left(u\right):=\parallel u-P_{n}u\parallel =\parallel (I-P_n)u\parallel \le \parallel I-P_n\parallel \parallel u\parallel \to 0$ for $n\to \infty$ for any $u\in H$ with $P_n$ being the orthogonal projection onto $S_n,\forall n\ge 1$.

Theorem 2.

If T satisfies the approximation properties and the set $\{{\alpha }_n,n\ge 1\}$ of positivity parameters is bounded, i.e. there exists $\alpha >0$ such that

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

then $N_T=\left\{0\right\}$, equivalently T is injective.

Proof. 

If there exists $u\in S\cap N_T$ then there exists a coarser subspace $S_{n_0}$ such that for $n\ge n_0$$u\in S_n$ and $Tu=0$. Then, on $S_n$ we have:

${0<\alpha \parallel u\parallel }^2\le {\alpha }_n{\parallel u\parallel }^2\le \langle T_{n}u,u\rangle =\langle (T_n-T)u,u\rangle \le {ϵ}_n{\parallel u\parallel }^2$ so, ${ϵ}_n\ge \alpha$ where ${ϵ}_n\to 0$. This is a contradiction showing that $S\cap N_T=\left\{0\right\}$.

We will show now, that the operator does not have any eligible zeros too.

For a not null $u\in E:=H\setminus S$, $\parallel u\parallel =1$ and $u_n:=P_{n}u$ its non null orthogonal projection over a coarser subspace $S_n$, $n\ge n_0:=n_0\left(u\right)$, from (ii),

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

(3)

If $u\in N_T$, then the estimation of $\parallel T_n{u}_n\parallel$ becomes:

$\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 )$

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

Because $\parallel u_n\parallel \to 1$, ${ϵ}_n\to 0$ (from i)) and ${\beta }_n:={\beta }_n\left(u\right)\to 0$ (see the observation above), then $\parallel T_n{u}_n\parallel \to 0$. From iii) and (3),

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

From 1 = ${\parallel u\parallel }^2={\parallel u_n\parallel }^2+{\beta }_n^2$, results ${\beta }_n/\parallel u_n\parallel ={\beta }_n/\sqrt{1-{\beta }_n^2}\to 0$. So,

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

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

Theorem 3.

If T is Hermitian verifying the approximation properties and the set $\{{\alpha }_n,n\ge 1\}$ of the positivity parameters is bounded, i.e. there exists $\alpha >0$ such that

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

then T is strictly positive on the dense set.

Proof. 

From the convergence to zero of the sequence ${ϵ}_n,n\ge 1$ there exists 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, amending the inequality in (3) to the semi-positivity property of the Hermitian operator yields:

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

Because $T_n$ is positive on $S_n$ and T is semi-positive, $T_n-T$ or $T-T_n$ is positive for that v and, the second inner product in the right side of the equality is real valued: $\langle (T_n-T)v,v\rangle \ge 0$ or, $\langle (T_n-T)v,v\rangle <0$.

If $\langle (T_n-T)v,v\rangle \ge 0$, then $0\le \langle (T_n-T)v,v\rangle \le {ϵ}_n{\parallel v\parallel }^2$ and,

$\langle T_{n}v,v\rangle -\langle (T_n-T)v,v\rangle \ge \langle T_{n}v,v\rangle -{ϵ}_n{\parallel v\parallel }^2\ge {\alpha }_n{\parallel v\parallel }^2-{ϵ}_n{\parallel v\parallel }^2>(\alpha -{ϵ}_0){\parallel v\parallel }^2$.

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

If instead $\langle (T_n-T)v,v\rangle <0$, then $\langle Tv,v\rangle$ is a sum of two positive quantities so, greater than any of them:

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

Thus, taking $\alpha \left(T\right)=(\alpha -{ϵ}_0)>0$, ${\langle Tv,v\rangle >\alpha \left(T\right)\parallel v\parallel }^2$ holds $\forall v\in S$.

So, T is strict positive on the dense set. From Theorem 1, $N_T=\left\{0\right\}$. □

Corollary. 

Let $T\in \mathcal{L}\left(H\right)$. If the sequence of operator approximations of its associated Hermitian operator $Q:=T^{*}T$ verifies:

$\langle Q_{n}v,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2\ge \alpha {\parallel v\parallel }^2,\forall v\in S_n$, on any $S_n\in F$, where $\alpha >0$ is a constant,

then Q is strictly positive on S and, $N_T=N_Q=\left\{0\right\}$.

Proof. 

Suppose that T and Q operator approximations are convergent, verifying both i) property. The associated Hermitian operator is also semi-positive on the subspaces of F. We are under the hypotheses of the Theorem 3, both positivity and iii) properties hold, showing that the operator is strictly positive on the dense set. According to Theorem 1, $N_Q=\left\{0\right\}$. From $N_T=N_Q$, we have $N_T=\left\{0\right\}$. □

3. Integral Operator Approximations

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, $n=2^m,nh=1$. As in [5], to build the subspaces with their union a dense set, consider the interval indicator functions with the supports of these intervals (k=1,n):

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

(4)

Consider now, 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, i.e. pairwise disjoint, $S_h=span\{I_{h,k};k=1,n,nh=1\}$, built on a multi-level structure and denote $S:={\cup }_{n\ge 2}{S}_h,nh=1$, their union that is a dense set in H well known in literature.

As an exercise in [9], the author shows the density of step functions built on functions of indicators of intervals pairwise disjoint in $L^p(\Omega ),1\le p<\infty$ where $\Omega$ is a domain in $\mathcal{R}$. In [5] and [6] the authors used the dense set built on (4) with the open subintervals respectively with the closed subintervals partitions, to obtain the best rate of convergence to zero of the eigenvalues of integral operators with Mercer kernels ([8]). Choosing open subintervals or closed subintervals for partitioning the domain, the unions of the subspaces generated in both cases are also dense ([1]): if one from three sets is dense, then the other two sets are dense.

Note 1.

a) The family F is an including family of subspaces.

b) S is dense and, $\forall u\in E$${\beta }_h\left(u\right):=\parallel u-P_{h}u\parallel \to 0$ with $n\to \infty ,nh=1$.

c) For any $u\in L^2(0,1)$ the functions $\left(u{I}_{h,k}\right)$, $\left(I_{h,k}u\right)$ not necessarily in S have the support ${\Delta }_{h,k}$: $\left(u{I}_{h,k}\right)\left(t\right)=\left(I_{h,k}u\right)\left(t\right)=0$ for $t\notin {\Delta }_{h,k}$).

Proof. 

a). The including property $S_h\subset S_{h/2},n\ge 2,nh=1$ is obtained from (4) by halving the mesh h, observing that any $I_{h,i}\in S_h,i=1,n$ could be embed into $S_{h/2}$ as per: $I_{h,i}=I_{h/2,2i-1}+I_{h/2,2i}\in S_{h/2}$.

b). For any $u\in H\setminus S,u\ne 0$, from an index $n_0,n_0{h}_0=1$ its orthogonal projection $P_{h}u=h^{-1}{\sum }_{k=1,n}{I}_{h,k}{c}_{h,k}\in S_h$, $P_{h}u\ne 0,nh=1$. Here, $c_{h,k}=\langle u,I_{h,k}\rangle$. From a) we have $P_{h}u\in S_{h/2}$. So, ${\beta }_{h/2}\le {\beta }_h$ because the best approximation of u in $S_{h/2}$ is given by $P_{h/2}u$. Therefore, ${\beta }_h$ is decreasing with h decreasing. We could remark that $\parallel P_{h}u-u\parallel \to 0$ for any $u\in E$ (equivalently with: given $ϵ>0$ there exists h such that ${\beta }_h\left(u\right)<ϵ$), holds iff S is dense.

c). For $t\notin {\Delta }_{h,k}$, k= 1,n, $I_{h,k}\left(t\right)=0$ (4). Then $\left(u{I}_{h,k}\right)\left(t\right)=\left(I_{h,k}u\right)\left(t\right)=0$. □

3.1 Convergent operator approximations. The option for semi-open subintervals partitioning ensures the subspaces including property and, every pair of indicator subinterval functions has disjoint supports obtaining as result sparse diagonal matrix representations.

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

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

(5)

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 $I_{h,k},k=1,n$. Then, $\forall u\in H$, $P_h^{r}u\in S_h$ and, ${\left(P_h^r\right)}^2=P_h^r$ for $n\ge 2,nh=1$. For any $u\in H$,

$P_h^{r}u={\int }_0^1{r}_{h}u\left(x\right)dx=h^{-1}{\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$, the constants $c_k$, k=1,n being given by

$c_k={\int }_0^{1}u\left(x\right)I_{h,k}\left(x\right)dx:=\langle u,I_{h,k}\rangle$.

Thus, $P_h^r$ is an orthogonal projection onto $S_h$ like is mentioned in [5].

Let $T_{\rho }\in \mathcal{L}\left(H\right)$. Its integral operator approximation on $S_h$ denoted by $T_{{\rho }_h}$ is a finite rank operator approximation, with the kernel function (citing again [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 ${\rho }_h^k,k=1,n$ of the kernel function ${\rho }_h$ in the sum have disjoint supports in $L^2{(0,1)}^2$, namely ${\Delta }_{h,k}^2,k=1,n,nh=1$.

The operator approximations $T_{{\rho }_h},n\ge 2,nh=1$ of $T_{\rho }$ are obtained as follows:

$P_h^r\left(\rho u\right)={\int }_0^{1}r(y,x)\rho (y,x)u\left(x\right)dx$

$=h^{-1}{\int }_0^1{\sum }_{k=1,n}{I}_{h,k}\left(y\right)I_{h,k}\left(x\right)\rho (y,x)u\left(x\right)dx$

$={\int }_0^1\left[h^{-1}{\sum }_{k=1,n}{I}_{h,k}\left(y\right)\rho (y,x)I_{h,k}\left(x\right)\right]u\left(x\right)dx$$={\int }_0^1{\rho }_h(y,x)u\left(x\right)dx$. So,

$T_{{\rho }_h}u:={\int }_0^1{\rho }_h(y,x)u\left(x\right)dx=P_h^r\left(\rho u\right)$, $T_{{\rho }_h}$ being a finite rank operator approximation on $S_h$ obtained through schema (4)-(6).

Lemma 1.

For any integral operator $T_{\rho }\in \mathcal{L}\left(H\right)$, the discretization schema (4)-(6) has the properties:

a.) the sequence of operator approximations $\{T_{{\rho }_h};n\ge 2,nh=1\}$ is convergent, that is: the convergence property i) holds;

b.) the matrix representations of the operator approximations over the subspaces in F are one-diagonal sparse matrices;

c.) if the diagonal entries of the matrix representations are strictly positive valued, $d_{kk}^h>0,\forall k=1,n,nh=1$, then the operator approximations $T_{{\rho }_h}$ are positive i.e. the positivity property ii) holds.

Proof. 

a.) From $T_{\rho }u-T_{{\rho }_h}u={\int }_0^1(\rho u-{\rho }_{h}u)dx={\int }_0^1(I-P_h^r)\left(\rho u\right)dx$. Then,

$\parallel T_{\rho }u-T_{{\rho }_h}u\parallel \le \parallel I-P_h^r\parallel \parallel \rho u\parallel$. Because $\{P_h^r;n\ge 2,nh=1\}$ is a sequence of orthogonal projections onto a family of including subspaces whose union is dense, $\parallel I-P_h^r\parallel \to 0$ with $n\to \infty ,nh=1$ holds; then the convergence of operator approximations property i) is satisfied:

${ϵ}_n:=\parallel T_{\rho }-T_{{\rho }_h}\parallel \to 0$ for $n\to 0$, nh=1.

b.) Now, evaluating $T_{{\rho }_h}v$ for $v=I_{h,i}$, we obtain

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

= $h^{-1}\left[{\int }_{{\Delta }_{h,i}}\rho (y,x)I_{h,i}\left(x\right)dx\right]I_{h,i}\left(y\right):=h^{-1}w\left(y\right)I_{h,i}\left(y\right)$

where $w=\left[{\int }_{{\Delta }_{h,i}}\rho (y,x)I_{h,i}\left(x\right)dx\right]\notin S_h$.

Because $\left(w{I}_{h,i}\right)$ has the support ${\Delta }_{h,i}$ (Note 2.c), $I_{h,j}$ has the support ${\Delta }_{h,j}$ and ${\Delta }_{h,i}\cap {\Delta }_{h,j}=\varnothing$ for $i\ne j$, we obtain

$\langle T_{{\rho }_h}{I}_{h,i},I_{h,j}\rangle =\langle w\left(y\right)I_{h,i}\left(y\right),I_{h,j}\left(y\right)\rangle =0$. Thus, the matrix representation of the finite rank operator $T_{{\rho }_h}$ on the basis $\{I_{h,k},k=1,n\}$ of $S_h$, is:

$M_h^r\left(T_{\rho }\right)=\mathrm{diag}{\left[h^{-1}{d}_{kk}^h\right]}_{k=1,n}$ where

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

(7)

and $d_{ij}^h=0$ for any pair (i,j), $i\ne j$ showing that the matrix is 1-diagonal.

c.) Given $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$, and from $\parallel v_h{\parallel }^2=h{\sum }_{k=1,n}{c}_k\overline{c_k}$, we obtain:

$\langle T_{{\rho }_h}{v}_h,v_h\rangle =h^{-1}{\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h\ge {\alpha }_h\left(T_{{\rho }_h}\right){\parallel v_h\parallel }^2$

where ${\alpha }_h\left(T_{{\rho }_h}\right)$ is the positivity parameter of the operator approximation $T_{{\rho }_h}$,

$${\alpha }_h\left(T_{{\rho }_h}\right)=h^{-2}mi{n}_{(k=1,n)}{d}_{kk}^h>0$$

(8)

that is strict positive for any $S_h$ because the minimum from a finite number of strict positive quantities $d_{kk}^h$, $k=1,n,nh=1$ can not be zero. □

Lemma 2. (Main Criterion)

If the operator approximations of $T_{\rho }\in \mathcal{L}\left(H\right)$ on schema (4)-(6) are positive and its sequence of positivity parameters is bounded,

then $N_{T_{\rho }}=\left\{0\right\}$.

Proof. 

The sequence $T_{{\rho }_h}$ converges using the approximation schema for $T_{\rho }$ (i.e. i) holds). Suppose that this sequence is also positive, that is $\langle T_{{\rho }_h}v,v\rangle \ge {\alpha }_h\left(T_{{\rho }_h}\right){\parallel v\parallel }^2>0$$\forall v\in S_h$, $n\ge 2,nh=1$, i.e. ii) holds. Let the positivity parameters in (8) be bounded, that is $\alpha >0$ exists such that ${\alpha }_h\left(T_{{\rho }_h}\right)\ge \alpha$, $n\ge 2,nh=1$, i.e. iii) holds. Then applying Theorem 2, $N_{T_{\rho }}=\left\{0\right\}$. □

Lemma 3. (Criterion for positive valued kernels)

If the kernel of the integral operator $T_{\rho }\in \mathcal{L}\left(H\right)$ is strictly positive valued on (0,1)² except for a set of measure Lebesgue 0 and the operator approximations sequence of the positivity parameters verifies for some$\alpha >0$

iii) ${\alpha }_h\ge \alpha$ for any $n\ge 2$, nh=1

then $N_{T_{\rho }}=\left\{0\right\}$.

Proof. 

The sequence $\{T_{{\rho }_h},nh=1,n\ge 2\}$ converges using the approximation schema for $T_{\rho }$. Now, let $\rho (y,x)\ge 0$ for any pair (x,y)$\in {(0,1)}^2$. Because the set of the points (x,y) in the domain of $\rho$ for which the kernel is null are of measure Lebesgue zero, $d_{kk}^h>0$ in (7) for any k=1,n, nh=1. Moreover, on a finite dimension subspace $S_h$, $mi{n}_{k=1,n}{d}_{kk}^h>0$ and, from ${\alpha }_h\left(T_{{\rho }_h}\right)>0$, ii) holds i.e. we have the positivity of the operator approximations $T_{{\rho }_h}$, $n\ge 2$, nh=1. Then from iii), invoking Theorem 2, $N_{T_{\rho }}=\left\{0\right\}$. □

4. Alcantara-Bode’s Equivalence of RH

The Hilbert-Schmidt integral operator $T_{\rho }$, with the kernel function $\rho (y,x)=\{y/x\}$ the fractional part of the quantity between brackets,

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

was used by Alcantara-Bode ([2]) for the equivalent formulation of the Riemann Hypothesis obtained from the Beurling equivalent formulation ([4]). Riemann conjectured (see [7]) that the Riemann Zeta function defined by the infinite sum:

$\zeta \left(s\right)=1+1/2^s+1/3^s+1/4^s+...$

has non trivial zeros s = $\sigma$ + it on the vertical line $\sigma =1/2$.

A connection between the Riemann Zeta function $\zeta$ and the integral operator $T_{\rho }$ can be found in [4] after reformulating as: $\left(T_{\rho }{x}^{s-1}\right)\left(\theta \right):={\int }_0^1\rho (\theta /x)x^{s-1}dx$, the left term in the equality below

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

The kernel function $\rho \in L^2{(0,1)}^2$ is continue and positive valued almost everywhere, and the discontinuities in ${(0,1)}^2$ consists of a set of numerable one dimensional lines of the form $y=kx,k\in N$, being of Lebesgue measure zero and thus, Riemann integrable. The integral operator $T_{\rho }$ is a Hilbert-Schmidt operator ([2]), therefore $T_{\rho }\in \mathcal{L}\left(H\right)$ together with its associated Hermitian. Then, for it will consider a finite rank operator approximations sequence built on a multi-level structure by (4)-(6).

Because the operator approximations are convergent (Lemma 1), following the algorithm we should compute the diagonal entries in the matrix representation, $M_h^r\left(T_{\rho }\right)$ which are given by:

$d_{kk}^h={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\rho (y,x)dxdy$, computed using the observation: for $x,y\in {\Delta }_{h,k}$$\{y/x\}=y/x$ for $y/x\le 1$ and $\{y/x\}=y/x-1$ for $y/x>1$, valued as:

$d_{11}^h=h^2(3-2\gamma )/4>0;d_{kk}^h={\displaystyle \frac{h^2}{2}}(-1+{\displaystyle \frac{2k-1}{k-1}}ln{\left({\displaystyle \frac{k}{k-1}}\right)}^{k-1})>0,\mathrm{for}k\ge 2$

where $\gamma$ is the Euler-Mascheroni constant. The sequence $\{f_k;k=2,n,nh=1\}$

$f_k:=h^{-2}{d}_{kk}^h=(-1+{\displaystyle \frac{2k-1}{k-1}}ln{\left({\displaystyle \frac{k}{k-1}}\right)}^{k-1})/2$ is decreasing for $k\ge 2$ and converges to 0.5 for k $\to \infty$. For $k\ge 3$, we have: $d_{22}^h>d_{kk}^h>0.5h^2>d_{11}^h$. Then from (8), ${\alpha }_h\left(T_{{\rho }_h}\right)=h^{-2}{d}_{11}^h$ we have the following positivity parameters of the operator approximations:

$${\alpha }_h\left(T_{{\rho }_h}\right)\ge \alpha :=(3-2\gamma )/4>0,\forall h,nh=1.$$

(9)

Theorem 4.

The Alcantara-Bode equivalent of Riemann Hypothesis holds, that is $N_{T_{\rho }}=\left\{0\right\}$.

Proof. 

Applying Lemma 2 (the operator approximations are built by schema (4)-(6)) or Lemma 3 (the kernel $\rho$ is positive valued a.e. on H) from (9) the bound of the positivity parameters is $\alpha =(3-2\gamma )/4>0$ a constant for any n, $nh=1$ obtaining $N_{T_{\rho }}=\left\{0\right\}$ and so, proving that the Alcantara-Bode equivalent holds. □

Proof Using Corollary.

The associated Hermitian integral operator has the kernel function $\widehat{\rho }(y,x):={\int }_0^1\rho (y,t)\overline{\rho (t,x)}dt={\int }_0^1\rho (y,t)\rho (t,x)dt={\int }_0^1\phi (y,x,t)dt$. We have to compute the corresponding entries in the matrix representations given by (7). For $y,x\in {\Delta }_{h,k},t\in {\Delta }_{h,j}$ and with $\lfloor \rfloor$ the floor function, the function $\phi$ to be integrated in variable t has the form

$\phi (y,x,t)=(y/t-\lfloor y/t\rfloor )(t/x-\lfloor t/x\rfloor )$

= $y/x-(t/x)\lfloor y/t\rfloor =(1/x)(y-t\lfloor y/t\rfloor )$ for $j<k,(t<kh)$:

= $y/x-(y/t)\lfloor t/x\rfloor =y(1/x-1/t\lfloor t/x\rfloor )$ for $j>k,(t>kh)$

Then with $d_{kk}^{h,j}:={\int }_{t\in {\Delta }_{h,j}}{\int }_{x\in {\Delta }_{h,k}}{\int }_{y\in {\Delta }_{h,k}}\phi (y,x,t)dtdxdy$,

${\widehat{d}}_{kk}^h={\sum }_{j=1,n}{d}_{kk}^{h,j}\ge n·mi{n}_{j=1,n}{d}_{kk}^{h,j}$. To compute ${\widehat{d}}_{kk}^h$, we will separate the terms in the sum as follows: $d_{kk}^{h,j=k}$, $d_{kk}^{h,j<k}$ and $d_{kk}^{h,j>k}$. The sequences $d_{kk}^{h,1<j<k}$ and $d_{kk}^{h,j>k}$ are increasing monotonically therefore, their components with the minimum values are $d_{kk}^{h,2}$ and $d_{kk}^{h,k}$. The term $d_{kk}^{h,1}\ge d_{kk}^{h,2}/8$. Therefore, for any $n\ge 2,nh=1$, the positivity parameter on $S_h$ is given by ${\alpha }_h\left(T_{{\widehat{\rho }}_h}\right)=h^{-2}mi{n}_{k=1,n}{\widehat{d}}_{kk}^h\ge {\displaystyle \frac{ln2}{16}}>0$. From Corollary, $N_{T_{\rho }}=\left\{0\right\}$. □

Note 2. (A Criterion for operator restrictions.)

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

${\mu }_n:={\alpha }_n\left(T{|}_{S_n}\right)/{\omega }_n$ 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 a constant $C>0$ such that ${\mu }_n\ge C$ for every $n\ge 1$ , then $N_T=\left\{0\right\}$.

If there exist $\alpha >0$ such that $\alpha \le {\alpha }_n,\forall n\ge 1$ then we are in the Theorem 1 hypothesis: T is strict positive on the dense set, so $N_T=\left\{0\right\}$.

Suppose that ${\alpha }_n\to 0$. If there exists $u\in (H\setminus S)\cap N_T$, $\parallel u\parallel =1$, let $u_n$ its orthogonal projection on $S_n,n\ge 1$. From the positivity on $S_n$ mentioning that for $u\in E$ its orthogonal projection verifies $P_{n}u:=u_n=P_n{u}_n$, if $Tu=0$ then:

${\alpha }_n\left(T\right){\parallel u_n\parallel }^2\le \langle T{|}_{S_n}{u}_n,u_n\rangle =\langle P_{n}T{|}_{S_n}{P}_n{u}_n,u_n\rangle$ = $\langle P_{n}T(P_n{u}_n-u),u_n\rangle$

= $\langle (u_n-u),T^{*}{u}_n\rangle \le {\beta }_n{\omega }_n\parallel u_n\parallel$. Or, ${\mu }_n:={\alpha }_n/{\omega }_n\le {\beta }_n\parallel u_n\parallel$. Let closest ${\mu }_n$ verifying $\forall v\in S_n$$\parallel T^{*}v\parallel \le {\mu }_n\parallel v\parallel$. If $\forall n$${\mu }_n\ge C$, from $C\le {\mu }_n\le {\beta }_n/\sqrt{1-{\beta }_n^2}\to 0$, we obtain a contradiction. Thus, $u\notin N_T$$\forall u\in H\setminus S$.

Now, if $\exists u\in S\cap N_T$, $\parallel u\parallel =1$ and $u\in S_n$ for some n, then $\langle T{|}_{S_n}u,u\rangle \ge {\alpha }_n>0$. Because ${Tu=T|}_{S_n}u,\forall u\in S_n$, the positivity is violated, so $u\notin N_T$ and $N_T=\left\{0\right\}$.

When T is Hermitian, ${\mu }_n={\lambda }_{min}\left(T{|}_{S_n}\right)/{\lambda }_{max}\left(T{|}_{S_n}\right)$, with $\lambda$’s are the eigenvalues of the operator restrictions, ${\mu }_n$ being the inverse of so called condition number of the matrices associated to restriction operators. □