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

1. Introduction

The task is to prove that $N_{T_{\rho }}=\left\{0\right\}$ where $T_{\rho }$ is the integral operator in the Alcantara-Bode equivalence having the kernel $\rho \in L^2{(0,1)}^2$ defined by the fractional part of the ratio $(y/x)$:

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

(1)

The main result (see an earlier formulation in [8], Theorem 1) states: an linear, bounded operator on a separable Hilbert space strict positive on a dense set, is injective.

To make it useful for practical purposes, we consider the dense set as the union of finite dimension subspaces organized in a multi-level (multigrid) structure on which we could define approximations of the linear operator. If on such subspaces the operator approximations are (strict) positive definite, then the operator has not zeros in the dense set and so, we could define the rules for the injectivity based on the relationship between the residuum of the eligible zeros of the operator and the operator positivity parameters on subspaces. When the subspaces are spanned by indicator interval functions having disjoint supports, we could interpret the method obtained as a ’degenerate’ finite element method on a multigrid architecture. As result, the matrix representation of an integral operator on a such subspace is sparse diagonal, whose minimum valued entry is used in the definition of the operator positivity parameter. If such parameter sequence is bounded, the injectivity of the operator is obtained due to an impossible relation between the potential zeros of the operator and the positivity parameters on the approximation subspaces.

2. Two Generic Theorems of Injectivity

Let H be a separable Hilbert space and T be a linear bounded operator on H positive definite on a dense set $S\subset H$, $\langle Tv,v\rangle >0\forall v\in S$. Then T has no zeros in S and its ’eligible’ zeros are all in the difference set $E:=H\setminus S$, i.e. $N_T\subset E$. The next theorem (see also an earlier version in [8]) proved that if T strict positive definite on the dense set then it has no zeros in the difference set either, obtaining $N_T=\left\{0\right\}$.

Theorem 1.

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

Proof.

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

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

(2)

Consider w an eligible element from the unit sphere, $\parallel w\parallel =1$.

Given ${\epsilon }_n$, there exists at least one element in the dense set $u_{{\epsilon }_n,w}\in S$ such that $\parallel u_{{\epsilon }_n,w}-w\parallel <{\epsilon }_n$ holds. Follows from (2), taking ${\epsilon }_n=1/n$: ∣ 1 - $\parallel u_{{\epsilon }_n,w}\parallel \mid$$<1/n$ showing that, for any choices of the sequence approximating w, $u_{{\epsilon }_n,w}\in S,n\ge 1$, the sequence $\parallel u_{{\epsilon }_n,w}\parallel \to 1$ with $n\to \infty$.

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

Suppose that there exists $w\in E,\parallel w\parallel =1$ a zero of $T,\mathrm{i}.\mathrm{e}.w\in N_T$ and consider a sequence of approximations of w, $u_{{\epsilon }_n,w}\in S,n\ge 1$ that, as we showed, is 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 to 1. Or, this happen for any choice of the sequence of approximations of w, verifying $\parallel u_{{\epsilon }_n,w}-w\parallel <{\epsilon }_n,n\ge 1$, 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 a union of a family F of finite dimension subspaces: $S={\cup }_{n\ge 1}{S}_n,\overline{S}=H$. It is not mandatory but will ease our proofs considering that the subspaces are including: $S_n\subset S_{n+1}$. In fact, that happen with the family of indicator interval functions in $L^2(0,1)$ by halving the mesh of the partitions of the domain (0,1): $S_h\subset S_{h/2},nh=1,n\ge 2$.

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 above 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. □

For T a linear bounded operator strict positive on each subspace $S_n$, there exists a sequence of positivity parameters ${\alpha }_n>0,n\ge 1$ such that $\langle Tv,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2\forall v\in S_n$. Being strict positive definite on every finite dimension subspace of the family, the operator has no zeros in the family subspaces and, its zeros if there are, are in the difference set $E:=H\setminus S$. Denoting the orthogonal projection of an eligible u onto $S_n\mathrm{by}{u}_n:=P_{n}u$ there exists $n_0:=n_0\left(u\right)$ due to the density of S, such that $u_n\ne 0$ for any n $\ge n_0$ and, ${\beta }_n:={\beta }_n\left(u\right)\to 0$.

Let $T:=T_{\phi }$ be an generic Hilbert-Schmidt integral operator on a separable Hilbert space H. Because a such operator is compact, it can be approximated by finite rank integral operators. If $P_n^o,n\ge 1$ is an orthogonal projection operator, then the finite rank approximations of our operator denoted by $T_n:=P_n^o\left(T_{\phi }\right),n\ge 1$, verifies ${ϵ}_n:=\parallel T_n-T\parallel \to 0$ for $n\to \infty$.

A sufficient condition for an Hilbert-Schmidt integral operator to verify the generic injectivity theorem is given below.

Theorem 2.

Let $T_{\phi }$ be an Hilbert-Schmidt integral operator on a separable Hilbert space and the dense set be the union of the finite dimension subspaces $S_n\in F$. If the positivity parameters of finite rank approximations $T_n$ on a dense family of including subspaces is inferior bounded, i.e. there exists $\alpha >0$ such that ${\alpha }_n\ge \alpha >0$, then $T_{\phi }$ is injective, equivalently its null space is $N_{T_{\phi }}=\left\{0\right\}$.

Proof.

Let $T:=T_{\phi }$ be the integral operator on H and $\{T_n;n\ge 1\}$ its sequence of finite rank operator approximations on the family F of subspaces. Suppose $S_n\subset S_{n+1},n\ge 1$ and there exists $\alpha$ such that ${\alpha }_n\ge \alpha >0$ where ${\alpha }_n,n\ge 1$ are the positivity parameters of the finite rank approximations $T_n$, $n\ge 1$, satisfying ${\alpha }_n{\parallel u_n\parallel }^2\le \langle T_n{u}_n,u_n\rangle$. We will split the proof in two parts: one, showing that the operator has no zeros in the dense set and next, it is itself strict positive definite on the dense set $S\subset H$ verifying the requests of the Theorem 1.

A) If there exists a zero of T in $S:=U_{n\ge 1}{S}_n$ then it should be in one of the subspaces of F, i.e. there exists $u_n\in S_n$, $u_n\ne 0$ such that $T{u}_n=0$. Then:

${\alpha }_n{\parallel u_n\parallel }^2\le \langle T_n{u}_n,u_n\rangle =\langle (T_n-T)u_n,u_n\rangle$$\le \parallel T-T_n\parallel \parallel u_n{\parallel }^2$ i.e. for the zeros of T in the dense set, ${\alpha }_n\le \parallel T-T_n\parallel$. Because T is an Hilbert-Schmidt operator, it is compact, so ${ϵ}_n:=\parallel T-T_n\parallel \to 0$ with $n\to \infty$. Follows: $\alpha \le {ϵ}_n\to 0$ with $n\to 0$, that is a contradiction, showing that $N_T\cap S=\left\{0\right\}$.

B) Now, 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 compactness parameter ${ϵ}_0$ depends only on the Hilbert-Schmidt operator, so is independent from any $v\in S$.

Given $v\in S$ there exists a subspace $S_n\ni v$. If $n\le n_0$ then from the including property of the subspaces of the family, $v\in S_{n_0}$. Follows for $v\in S_n$:

$\langle Tv,v\rangle =\langle T_{n}v,v\rangle -\left(\langle (T_n-T)v,v\rangle \right)\ge |\langle T_{n}v,v\rangle -|\left(\langle (T_n-T)v,v\rangle \right)\left|\right|$

$\ge (\alpha -\parallel T-T_n{\parallel )\parallel v\parallel }^2\ge {\alpha }_n\left(T\right){\parallel v\parallel }^2$

where ${\alpha }_n\left(T\right)=\alpha -{ϵ}_n$. Then, with $\alpha \left(T\right)=\alpha -{ϵ}_0$ we obtain the strict positivity of the operator on the dense set S: for very $v\in S,\langle Tv,v\rangle \ge {\alpha \left(T\right)\parallel v\parallel }^2$.

Thus $N_T=\left\{0\right\}$, by applying Theorem 1. □

Observation. 

The integral operator in (1) is a Hilbert-Schmidt on $L^2(0,1)$ ([2]) and so, compact. Taking the dense family as the union of the finite dimension subspaces $\left\{S_h\right\},nh=1,n\ge 2$ spanned by the indicator interval functions of the semi-open disjoint intervals of length h in (0,1), we showed in section 5.A) (Theorem 3) that using a finite rank discretization like in [5], the positivity parameters of the finite rank operator approximations are all constant ${\alpha }_h=\alpha :=(3-2\gamma )/4$ (10), where $\gamma$ is the Euler-Mascheroni constant. So, the integral operator in (1) is injective. Thus, by Alcantara-Bode equivalence RH holds. For ones interested only in the proof of RH, the rest of the article could be skipped.

3. The Associated Methods

Our hypothesis in the previous theorem has been the existence of a bound for the finite rank operator positivity parameters. But, an linear,bounded operator strict positive definite on every finite dimension subspace of the family, has no zeros in the family subspaces and, with or without a bound of the positivity parameters, its zeros if there are, should be in the difference set $E:=H\setminus S$.

Then, we have to take in consideration the case in which the sequence $\left\{{\alpha }_n\right\},n\ge 1$ converges to zero and, another case when instead of the finite rank operator approximations we will take simply, the restrictions of the linear bounded operator on the finite dimension subspaces $S_n,n\ge 1$.

The relationship between the residuum on finite dimension subspaces of the eligible zeros and the operator positivity parameters on these subspaces is exploited in defining the methods for investigation of its injectivity: the finite rank operator approximations and the operator restrictions methods.

Lemma 1. (Corollary of the Theorem 2)

Let T have its sequence of the finite rank operator approximations $T_n,n\ge 1$ be strict positive on the subspaces $S_n\in F$: $\langle T_n{u}_n,u_n\rangle \ge {\alpha }_n\left(T_n\right){\parallel u_n\parallel }^2$$\forall u_n\in S_n$, where ${\alpha }_n\left(T_n\right)\ge \alpha >0$ for $n\ge 1$. Then T is injective.

Proof. 

For $u\in H\setminus S$, $\parallel u\parallel =1$ having the not null orthogonal projections $P_{n}u,n\ge n_0:=n_0\left(u\right)$, from Observation 1 we have its residuum sequence on the approximation subspaces ${\beta }_n:={\beta }_n\left(u\right)=\parallel u-P_{n}u\parallel \to 0$.

If $u\in N_T$, we obtain:

$\alpha \parallel P_n{u\parallel }^2\le \langle T_n{P}_{n}u,P_{n}u\rangle \le \parallel T_n{P}_{n}u\parallel \parallel P_{n}u\parallel$$\le (\parallel T_n{P}_{n}u-T{P}_{n}u+T{P}_{n}u-Tu\parallel )\parallel P_{n}u\parallel$$\le (\parallel T-T_n\parallel \parallel P_{n}u\parallel +\parallel T\parallel \parallel u-P_{n}u\parallel )\parallel P_{n}u\parallel$.

Then:

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

where ${ϵ}_n:=\parallel T-T_n\parallel \to 0$ and ${\beta }_n/\sqrt{1-{\beta }_n^2}\to 0$. The inequality is violated from an $n_0$, $n\ge n_0$ meaning, $u\notin N_T$, for any zero of T in E. Once T has no zeros in the dense set, $N_T=\left\{0\right\}$. □

Lemma 2. (Injectivity Criteria)

Let T be a linear bounded operator strict positive on each subspace $S_n$, $\langle Tv,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2\forall v\in S_n$, ${\alpha }_n>0,n\ge 1$. Suppose that ${\alpha }_n\to 0$ and $\parallel T^{*}v\parallel \le {\omega }_n\parallel v\parallel$ for every $v\in S_n,n\ge 1$.

If ${\mu }_n:={\alpha }_n/{\omega }_n\ge C>0$ then T is injective.

Proof. 

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

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

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: $N_T=\left\{0\right\}$. □

Note. 

With these two lemmas, we have an unitary view on the methods through the relationship between the residuum on finite dimension subspaces of the eligible zeros and the corresponding operator positivity parameters. Moreover, if there exists $\alpha$ such that ${\alpha }_n\ge \alpha >0$ for $n\ge 1$, then in Lemma 2 we could remove the involvement of the adjoint operator approximations observing that $\parallel T^{*}{u}_n\parallel \le \parallel T^{*}\parallel \parallel u_n\parallel$ or from:

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

and the injectivity of the operator is coming by replacing the ${\omega }_n$ parameter with the operator or its adjoint norm. So, we could see Lemma 1 as a particular case of Lemma 2. With these lemmas, we also fixed a flaw in the injectivity criteria introduced in [1], mentioning that at that time we did not have the generic theorem of injectivity (Theorem 1) that has been introduced later ([8]).

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

The intervals of equal lengths $h=2^{-m},m\ge 1$, nh = 1, ${\Delta }_{h,k}=((k-1)/2^m,k/2^m],k=1,n-1$ together with ${\Delta }_{h,n}$ define for $m\in N$ 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)

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

(4)

The family F of finite dimensional subspaces $S_h$ that are the linear spans of interval indicator functions of the h-partitions defined by (4) with disjoint supports, $S_h=span\{I_{h,k};k=1,n,nh=1\}$, built on a multi-level structure, are including $S_h\subset S_{h/2}$ by halving the mesh h. In fact, due to (4) any $I_{h,k}\in S_h$ is embedded in $S_{h/2}$ by $I_{h,k}\left(t\right)=I_{h/2,(2k-1)}\left(t\right)+I_{h/2,2k}\left(t\right)$ for any $t\in \left(0.1\right)$, k=1,n, nh=1 (or 2nh/2=1). So, $S_h\ni u_h={\sum }_{k=1,n}{c}_k^h{I}_{h,k}={\sum }_{j=1,2n}{c}_j^{h/2}{I}_{h/2,j}\in S_{h/2}$ where $c_k^h=c_{2k-1}^{h/2}=c_{2k}^{h/2},k=1,n$. The family F will be used for finite rank approximations as in [5].

Let ${\Delta }_h^k=((k-1)/2^m,k/2^m),k=1,n$ be the set of open intervals that differ from the semi-open intervals ${\Delta }_{h,k}$ of the defined partitions only by the end-points $kh,k=1,n,n\ge 2$. Then, denoting with $I_h^k,k=1,n,nh=1$ the interval indicator functions, $\parallel I_h,k-I_{h,k}\parallel =0$ in $L^2\left(0.1\right)$, for $nh=1,n\ge 2$. Let observe that for any partition of size h, the points in which $I_h^k\ne I_{h,k},k=1,n$ are the right endpoints in the partition of the interval (0,1) and, their collection is of the Lebesgue measure zero. Thus, for any kernel function $\phi \in L^2{(0,1)}^2$ because any pair $I_h^k,I_h^j;k\ne j$ is orthogonal because of their disjoint supports ${\Delta }_h^k\cap {\Delta }_h^j=\varnothing$, we have: $I_h^k\left(y\right)\phi (y,x)I_h^k\left(x\right)=0$ for $k\ne j$.

Then $\left(T_{\rho }{I}_h^k\right)\left(y\right)=\left({\int }_0^1\phi (y,x)I_h^k\left(x\right)dx\right)\left(y\right)=({\int }_{{\Delta }_h^k}\phi (y,x)dx$)(y). So, denoting $d_{kk}^h=\langle T_{\rho }{I}_h^k,I_h^k\rangle$, for k=1,n, nh=1, we obtain

$d_{kk}^h:=\langle \left(T_{\rho }{I}_h^k\left(x\right)\right)\left(y\right),I_h^k\left(y\right)\rangle =\langle \left({\int }_{{\Delta }_h^k}\phi (y,x)I_h^k\left(x\right)dx\right)\left(y\right),I_h^k\left(y\right)\rangle$

= ${\int }_{{\Delta }_h^k}{\int }_{{\Delta }_h^k}{I}_h^k\left(y\right)\phi (y,x)I_h^k\left(x\right)dxdy$

= ${\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}{I}_{h,k}\left(y\right)\phi (y,x)I_{h,k}\left(x\right)dxdy$

Denoting by $F^o$ the family of the finite dimension subspaces that are linear span of indicator open-interval functions we have: $S^o$ is dense if and only if S is dense.

Proof. Suppose that $S^o$ is dense, well known in literature. If there exists $f\in L^2(0,1)$ such that $\langle f,I_{h,k}\rangle =0;k=1,n,nh=1$ then

$|\langle f,I_h^k\rangle |=|\langle f,I_h^k-I_{h,k}\rangle |\le \parallel f\parallel \parallel I_h^k-I_{h,k}\parallel =0$, meaning f is orthogonal to any element from the dense set $S^o$ and should be zero. Thus, S is dense. In the same manner, supposing $S^p$ dense we obtain S dense. □.

Now, because any pair $(I_h^k,I_h^j;k\ne j)$ is orthogonal because of their disjoint supports ${\Delta }_h^k\cap {\Delta }_h^j=\varnothing$, given $T_{\phi }$ an integral operator on $L^2(0,1)$ its matrix representation on the subspaces $S_h$ and $S_h^o$ generated by the families of the corresponding indicator interval functions are both sparse diagonal matrices coincides having the diagonal entries $d_{kk}^h,k=1,n,nh=1$ the only difference consisting in the orthonormality factor introduced with the finite rank approximations, $\left(h^{-1}\right)$ as we will see later: $M_h\left(T_{{\rho }_h}\right)=h^{-1}{M}_h\left(T_{\rho }\right):=h^{-1}diag[d_{kk}^h;k=1,n,nh=1]$.

5. RH Holds

The integral operator $T_{\rho }$ defined in (1) having the kernel function $\rho (y,x)$ continuous everywhere on ${(0,1)}^2$ excepting a set of measure Lebesgue zero - it is a countable set of unidimensional lines in ${(0,1)}^2$ of the form y = kx, k$\in N$ - is a linear, bounded, Hilbert-Schmidt operator ([2]) and so, compact.

Then, it could be approximated by projections on the finite dimension subspaces (see also [3]) of the $L^2(0,1)$.

A. Finite rank operator approximations.

Below, we follow the steps from [5] for defining the orthogonal projection of the integral operator $T_{\rho }$ by finite rank integral operators considering the dense set S the union of the including finite dimension subspaces spanned by indicator semi-open interval functions (4) as follows. The orthogonal projections of the kernel function are performed by the finite rank integral operators ([5], pg. 986) whose kernel functions are:

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

(5)

As result, the integral operator projection on $S_h$, $T_{{\rho }_h}$ has 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)

that is a sum of kernel pieces with disjoint supports in which $I_{h,k}\left(y\right)\rho (y,x)I_{h,j}\left(x\right)$ are taken to be 0 for $k\ne j$ as a consequence of the property of the basis: any pair from the basis on the subspace $S_n$ is orthogonal due to their disjoint supports. Thus, the matrix representation of the finite rank operator approximation $T_{{\rho }_h}$ is a sparse diagonal matrix

$M_h\left(T_{{\rho }_h}\right)=h^{-1}{M}_h\left(T_{\rho }\right):=h^{-1}diag[d_{kk}^h;k=1,n,nh=1]$, where

$$d_{kk}^h:=\langle T_{\rho }{I}_{h,k},I_{h.k}\rangle ={\int }_{{\Delta }_{h,k}}{\int }_{{\Delta }_{h,k}}\rho (y,x)dxdy$$

(7)

From $\langle T_{{\rho }_h}{I}_{h,k},I_{h.k}\rangle =h^{-1}{d}_{kk}^h$, like in [1] for $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}\in S_h$,

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

(8)

The values of the entries $d_{kk}^h,k=1,n,nh=1$ are given in ([1]):

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

(9)

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

for $k\ge 2$, where $\gamma$ is the Euler-Mascheroni constant.

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

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

(10)

Theorem 3. 

The Riemann Hypothesis holds.

Proof.

From (10), the positivity parameters of the finite rank operator approximations $T_{{\rho }_h}$ on the dense family F are inferior bounded, obtaining from Theorem 2 or Lemma 1 that $N_{T_{\rho }}=\left\{0\right\}$. Invoking the Alcantara-Bode’s equivalence, the Riemann Hypothesis holds. □

B. Operator restrictions on subspaces. Using this method, we will consider the the dense set of indicator open interval functions $S^o$.

Having the convergence to zero of the residuum, see Observation 1, next step is to compute the positivity parameters of our operator on the approximation subspaces. For this, we have to evaluate the inner product between $\left(T_{\rho }{u}_h\right)\notin S_h$ and $u_h\in S_h$ where $u_h$ is the not null orthogonal projection of an eligible $u\in E:=H\setminus S$.

Denoting with $P_h:E\subset H\to S_h\subset H$ the orthogonal projection onto $S_h$, then:

$E\ni u\to u_h:=P_{h}u=$${\sum }_{k=1,n}{I}_h^k{\displaystyle \frac{1}{\mid {\Delta }_h^k\mid }}{\int }_{{\Delta }_h^k}u\left(s\right)ds\in S_h$ holds. Or,

$$u_h\left(t\right):=\left(P_{h}u\right)\left(t\right)=h^{-1}\sum _{k=1,n}\left({\int }_{{\Delta }_h^k}u\left(s\right)ds\right)I_h^k\left(t\right)$$

(11)

With the basis $\{I_h^k;k=1,n,nh=1\}$ on the linear span $S_h$ the matrix representation of the integral operator $T_{\rho }$ is a sparse diagonal because the orthogonality of the basis elements is a consequence of the disjoint supports between any pair $I_h^k,I_h^j,k\ne j$: $I_h^k\left(y\right)\rho (y,x)I_h^j\left(x\right)=0\mathrm{for}k\ne j$. Then, the matrix $M_h\left(T_{\rho }\right)$ has the diagonal entries $d_{kk}^h,k=1,n,nh=1$ given by (9).

In fact, $\langle T_{\rho }{I}_h^k,I_h^k\rangle :=\langle P_h{T}_{\rho }{I}_h^k,I_h^k\rangle$ applying (11), as per

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

$h^{-1}\left[{\int }_{{\Delta }_h^k}{\int }_{{\Delta }_h^k}\rho (y,x)dxdy\right]I_h^k\left(y\right)$ = $h^{-1}{d}_{kk}^h{I}_h^k\left(y\right)$. So,

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

(12)

that we already knew, valued in (9). Follows for $k=1,n,nh=1$:

$\langle T_{\rho }{I}_h^k,I_h^k\rangle =\langle h^{-1}{d}_{kk}^h{I}_h^k,I_h^k\rangle =h^{-1}{d}_{kk}^h{\parallel I_h^k\parallel }^2=d_{kk}^h$ again and, we will define

${\alpha }_h\left(T_{\rho }\right)=h^{-1}{d}_{11}^h=h(3-2\gamma )/4$ observing that ${\alpha }_h\left(T_{\rho }\right)\to 0$.

The value of the positivity parameter of the operator defined on $S_h$, remains unchanged for any $u_h\in S_h$ as per following estimations: let $u_h={\sum }_{k=1,n}{c}_k{I}_h^k$ then applying (11) to $\left(T_{\rho }{u}_h\right)\notin S_h$ in order to evaluate the inner product $\langle T_{\rho }{u}_h,u_h\rangle$ as we will do now:

$\left(P_h{T}_{\rho }{\sum }_{k=1,n}{c}_k{I}_h^k\right)\left(y\right)=h^{-1}{\sum }_{k=1,n}{c}_k{d}_{kk}^h{I}_h^k\left(y\right)$.

So, we are able to compute the inner product in $S_h$:

$\langle T_{\rho }{u}_h,u_h\rangle =\langle \left(P_h{T}_{\rho }{\sum }_{k=1,n}{c}_k{I}_h^k\right)\left(y\right),{\sum }_{k=1,n}{c}_k{I}_h^k\left(y\right)\rangle$

$=h^{-1}{\sum }_{k=1,n}{d}_{kk}^h{c}_k\overline{c_k}{\parallel I_h^k\parallel }^2\ge h^{-1}{d}_{11}^h\left({\sum }_{k=1,n}{c}_k\overline{c_k}{\parallel I_h^k\parallel }^2\right)$

$\ge h^{-1}{d}_{11}^h\parallel v_h{\parallel }^2\ge {\alpha }_h\left(T_{\rho }\right){\parallel v_h\parallel }^2$, $nh=1,n\ge 2$

Thus, for $u\in N_{T_{\rho }}$:

$\langle T_{\rho }{u}_h,u_h\rangle =\langle T_{\rho }(u_h-u),u_h\rangle =\langle (u_h-u),T_{\rho }^{*}{u}_h\rangle \le {\beta }_h\parallel T_{\rho }^{*}{u}_h\parallel$. Or,

$${\alpha }_h\parallel u_h{\parallel }^2=h^{-1}{d}_{11}^h\parallel u_h{\parallel }^2\le {\beta }_h\parallel T_{\rho }^{*}{u}_h\parallel \le {\beta }_h{\omega }_h\parallel u_h\parallel$$

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that becomes: $0\leftarrow h(3-2\gamma )/4\le {\omega }_h{\beta }_h/\sqrt{1-{\beta }_h^2}$. Then, we have to evaluate ${\omega }_h$.

We could observe that $T_{\rho }$ is strict positive definite on any $S_h$ but, on S it is only positive definite. For investigating the injectivity, we should involve the adjoint operator $T_{\rho }^{*}$ whose kernel function is defined by

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

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

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

$={\sum }_{k=1,n}{c}_k{\int }_{{\Delta }_h^k}{\rho }_h^k(x,y)dy$,

a formula obtained from the property $I_h^j\left(x\right)\rho (x,y)I_h^k\left(y\right)=0$ for $k\ne j$. Then

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

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

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

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

$\parallel T_{\rho }^{*}{v}_h\parallel \le h\parallel v_h\parallel$. Taking ${\omega }_h\left(T_{\rho }^{*}\right)=h$, follows:

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

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With these estimations of the positivity and injectivity parameters, we have the following result applying the Injectivity Criteria.

Theorem 4. (Injectivity Criteria.)

The Hilbert-Schmidt operator $T_{\rho }$ defined in (1) has its restrictions on the approximation subspaces spanned by indicator open-interval functions with the injectivity parameters sequence ${\mu }_h,n\ge 2,nh=1$ inferior bounded. Consequently, $T_{\rho }$ is injective and RH holds due to the Alcantara-Bode equivalence.

Proof.

Because ${\mu }_h$ is a constant (see (14)) for any $h,nh=1$, Lemma 2 could be applied in order to have $N_{T_{\rho }}=\left\{0\right\}$. That means, half from equivalent formulation of the Alcantara-Bode is true so, the other half should be true i.e.: the Riemann Hypothesis holds. □