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

1. Introduction

The Riemann Hypothesis (RH) claims (1859) that the Riemann Zeta function defined by the infinite sum $\zeta \left(s\right)={\sum }_{n=1}^{\infty }1/n^s$ has its non trivial zeros on the vertical line $\sigma =1/2$. The Hilbert-Schmidt integral operator $T_{\rho }$ defined on $L^2(0,1)$ having the kernel function $\rho (y,x)=\left\{{\displaystyle \frac{y}{x}}\right\}$, the fractional part of the quantity between brackets,

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

(1)

has been used by Alcantara-Bode ([2], pg. 151) in his theorem of the equivalent formulation of the RH. Denoting by $N_{T_{\rho }}$ the null space of this operator, the equivalent formulation consists in (Alcantara-Bode’s Equivalence of RH):

The Riemann Hypothesis holds if and only if N_Tρ = {0}.

Thus, the task is to prove that $N_{T_{\rho }}=\left\{0\right\}$ and for doing it, we provided a solution containing the theory and the associated methods for the investigation of the linear bounded operators on separable Hilbert spaces.

2. An Injectivity Theorem on Separable Hilbert Spaces

Let H be a separable Hilbert space and T be a linear bounded operator on H strict positive on a dense set $S\subset H$, i.e. there exists $\alpha >0$ such that $\langle Tv,v\rangle \ge {\alpha \parallel v\parallel }^2\forall v\in S$. 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 [9]) proved that T 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 an element $w\in H,w\ne 0$ and $w/\parallel w\parallel$ are both or are not both 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 (2.1), taking ${\epsilon }_n=1/n$: $\mid 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. □

For using the theorem for practical purposes, we have to ease the terms in which it works considering the positivity only of the operator on a dense set. Then additional requests should be imposed in order to remain in terms of the theorem.

Suppose that the dense set S is the result of a union of finite dimension subspaces: $S={\cup }_{n\ge 2}{S}_n,\overline{S}=H$.

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. □

Let T be a linear bounded operator strict positive on each subspace $S_n$, $\langle Tv,v\rangle \ge {\alpha }_n{\parallel v\parallel }^2\forall v\in S_n$, ${\alpha }_n>0,n\ge 1$. 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 $\ge n_0$.

The relationship between the residuum on finite dimension subspaces of an eligible zero of an linear bounded operator and the operator positivity parameters on these subspaces is exploited in defining the methods for investigation of its injectivity.

Supposing $u\in N_T$ is an eligible zero of T with $\parallel u\parallel =1$, the relation in (3) on $S_n$ involving its not null projection $u_n,n\ge n_0:=n_0\left(u\right)$, defines the inequalities corresponding to the methods for investigation of the injectivity:

a) If ${\alpha }_n\ge \alpha >0$ then

Corollary. The following inequality holds for $n\ge n_0:=n_0\left(u\right)$:

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

(4)

$$\alpha /\parallel T\parallel \le {\alpha }_n/\parallel T\parallel \le {\beta }_n/\parallel u_n\parallel ={\beta }_n/\sqrt{1-{\beta }_n^2}$$

This is a relationship in which the right side converges to zero and so, there exists a range $n_0:=n_0\left(u\right)$ such that for $n\ge n_0$ the inequality is violated for any potential zero of unitary norm of the operator, making the operator injective.

b) If ${\alpha }_n\to 0$ then we can use:

Injectivity Criteria. From (4) this time we evaluate the parameters from:

$\langle T(u_n-u),u_n\rangle =\langle (u_n-u),T^{*}{u}_n\rangle$. Then, ${\alpha }_n\parallel u_n{\parallel }^2\le {\beta }_n{\omega }_n\parallel u_n\parallel$ where ${\omega }_n$ is closest to the norm of the adjoint operator: $\parallel T^{*}v\parallel \le {\omega }_n\parallel v\parallel$ for every $v\in S_n$.

Defining the injectivity parameter ${\mu }_n={\alpha }_n/{\omega }_n$ the following relation holds:

$${\mu }_n\le {\beta }_n\parallel u_n\parallel :={\beta }_n/\sqrt{1-{\beta }_n^2}$$

(5)

Lemma 1.

If one from the sequences of the positivity parameters $\{{\alpha }_n,n\ge 1\}$ and the injectivity parameters $\{{\mu }_n,n\ge 1\}$ is inferior bounded by a not null positive constant, then T is injective on H.

Proof. 

a) (Corollary) Suppose that ${\alpha }_n\ge \alpha >0$ for $n\ge 1$. From (4), $\alpha /\parallel T\parallel \le {\beta }_n\parallel u_n\parallel \to 0$ with $n\to \infty$ that is a contradiction. Hence, $u\in E$ is not a zero of T. Because T has no zeros in S either, its null space contains only 0: $N_T=\left\{0\right\}$.

b) (Injectivity Criteria) If the sequence ${\alpha }_n$ is unbounded inferior, from (5), ${\mu }_n\le {\beta }_n\parallel u_n\parallel$. If there exists a constant $C>0$ for which ${\mu }_n\ge C$ for every $n\ge 1$ then from $0<C\le {\mu }_n\le {\beta }_n\parallel u_n\parallel$ we obtain $u\notin N_T$, valid for any zero of T eligible, so $N_T=\left\{0\right\}$ because T has no zeros from the dense set either. □

For now on, $H=L^2(0,1)$. 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 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 $L^2(0,1)$.

3. Algebraic - Functional Approach

Let $h:=2^{-m},m\ge 1$, nh =1 and define the set of disjoint and open intervals of length h, ${\Delta }_h^k=((k-1)/2^m,k/2^m)$ that are the supports of the interval indicator functions:

$$I_h^k\left(t\right)=1\mathrm{for}t\in {\Delta }_h^k\mathrm{and}0\mathrm{otherwise}.$$

(6)

The union S of the family of spans $S_h:=span\{I_h^k,k=1,n,nh=1,n\ge 2\}$ is dense in H, well known in literature. Then by Observation 1, for any $u\in E:=H\setminus S$,

$${\beta }_h\left(u\right):=\parallel u-u_h\parallel \to 0$$

with $u_h$ the orthogonal projection of u on $S_h$.

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,

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

(7)

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 sparse diagonal because the orthogonality of the basis elements is a consequence of the disjoint supports between any pair $I_h^k,I_h^j,k\ne j$: $I_h^k\left(y\right)\rho (y,x)I_h^j\left(x\right)=0\mathrm{for}k\ne j$.

So, the matrix $M_h\left(T_{\rho }\right)$ has the diagonal entries $d_{kk}^h,k=1,n,nh=1$ given by:

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

(8)

$$\begin{array}{cc}\hfill d_{kk}^h& =\underset{(k-1)h}{\overset{kh}{\int }}\left[\underset{(k-1)h}{\overset{kh}{\int }}(y/x)dx-\underset{(k-1)h}{\overset{y}{\int }}dx\right]dy\hfill \\ \hfill & ={\displaystyle \frac{h^2}{2}}(-1+{\displaystyle \frac{2k-1}{k-1}}ln{\left({\displaystyle \frac{k}{k-1}}\right)}^{k-1})\hfill \end{array}$$

(9)

for $k\ge 2$, where $\gamma$ is Euler-Mascheroni constant, values taken from [1].

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 (7), 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,

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

(10)

and their values for our operator of interest are in (9). Then:

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

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

The positivity parameter value is the same on $S_h$ as per estimating $P_h\left(T_{\rho }{u}_h\right)$ with $u_h={\sum }_{k=1,n}{c}_k{I}_h^k$ then applying (7) 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 now, to compute the inner product $\langle T_{\rho }{u}_h,u_h\rangle$ in (4):

$$\langle \left(P_h{T}_{\rho }\sum _{k=1,n}{c}_k{I}_h^k\right)\left(y\right),\sum _{k=1,n}{c}_k{I}_h^k\left(y\right)\rangle =h^{-1}\sum _{k=1,n}{d}_{kk}^h{c}_k\overline{c_k}{\parallel I_h^k\parallel }^2$$

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

in order to obtain the sequence of the positivity parameter ${\alpha }_h$ valued like before.

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

$${\rho }^{*}(y,x)=\overline{\rho (x,y)}=\rho (x,y).$$

So, for $v_h={\sum }_{k=1,n}{c}_k{I}_h^k\in S_h$,

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

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

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

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

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

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

Because $\rho (x,y)$ is valued in [0,1),

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

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

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

(11)

In computing the values of the diagonal matrix entries, we used the observation made by Beurling ([4]): for $0<a<b<min\{2a,1\}$ the fractional part $\{b/a\}=(b/a-1)$. 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 $\to \infty$. Follows $\forall h,d_{kk}^h>h^2/2>d_{11}^h,k\ge 2$. With these estimations of the positivity and injectivity parameters, we have the following result applying the Injectivity Criteria.

Theorem 2.1

(Injectivity Criteria.) The Hilbert-Schmidt operator $T_{\rho }$ defined in (1) is injective and, consequently RH holds.

Proof. 

Because ${\mu }_h$ is a constant for any $h,nh=1$, there exists an $n1:=n1\left(u\right)$ such that for $n\ge n1,nh=1$, ${\beta }_h\left(u\right)\sqrt{1-{\beta }_h\left(u\right)}<{\mu }_h$ that contradicts the relationship in (5) meaning, Lemma 1 could be applied for obtaining $u\notin N_{T_{\rho }}$, valid for any eligible zero of $T_{\rho }$. Because $T_{\rho }$ has no zeros in the dense set S either, we obtain

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

(12)

Now, from (12) half from equivalent formulation of the Alcantara-Bode is true so, the other half is true obtaining: the Riemann Hypothesis holds. □

The involvement of the adjoint operator could be avoided by using the finite rank operator approximations together with the compactness property of the integral operator.

4. Finite Rank Operator Approximations

It is well known that on a Hilbert space, every compact operator is a limit of finite-rank operators. The converse is always true. The integral operator in (1) is Hilbert-Schmidt ([2]) and then, compact. Thus, it can be approximated by finite rank integral operators. If $P_n^r$ is a such projection operator, then the finite rank approximations of $T_{\rho }$ verifies $\parallel P_n^r\left(T_{\rho }\right)-T_{\rho }\parallel \to 0$ for $n\to \infty$. Let be $T_n:=P_n^r\left(T_{\rho }\right)$.

Theorem 2.

An Hilbert-Schmidt integral operator on a separable Hilbert space with the finite rank approximations on a family of subspaces whose union is dense having the positivity parameters inferior bounded, is injective.

Proof. 

Suppose $S_n\subset S_{n+1}$. Let ${\alpha }_n\ge \alpha >0$ be the positivity parameters of the finite rank approximations $T_n$, $n\ge 1$, satisfying ${\alpha }_n{\parallel u_h\parallel }^2\le \langle T_n{u}_n,u_n\rangle$

Because T is Hilbert-Schmidt operator, it is compact, so, $\parallel T_n-T\parallel \to 0$ with $n\to \infty$.

Now, for $u\in H\setminus U_{n\ge 1}{S}_n$, $\parallel u\parallel =1$, for the not null orthogonal projections $P_{n}u$ of u verifies ${\beta }_n=\parallel u-P_{n}u\parallel \to 0$. If $u\in N_T$ then we have:

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

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

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 ${\beta }_n\to 0$ being the residuum of u on $S_n$. Then, the inequality is violated from an $n_o$, $n\ge n_0$ meaning, $u\notin N_T$, for any zero of T in E.

Now, if there exists a zero of T in $S^{\prime }$ then it should be in one of the subspaces of $F^{\prime }$ i.er. there exists $u_n{\S }_n^{\prime }$ 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$ or, $\alpha \le \parallel T_n-T\parallel \to 0$ with $n\to 0$ a contradiction, shoving that the operator has no zeros in the dense set. Thus, $N_T=\left\{0\right\}$ and so injective. □

We will use the previous theorem as support for proving the injectivity of our integral operator on $L^2(0,1)$. The intervals of equal lengths $h=2^{-m},m\ge 1$, nh = 1, ${\Delta }_{h,k}^{\prime }=((k-1)/2^m,k/2^m],k=1,n-1$ 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:

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

(13)

The family $F^{\prime }$ of finite dimensional $S_h^{\prime }$ linear spans of interval indicator functions of the h-partitions defined by (13) with disjoint supports, $S_h^{\prime }=span\{I_{h,k}^{\prime };k=1,n,nh=1\}$, built on a multi-level structure, are including $S_h^{\prime }\subset S_{h/2}^{\prime }$ by halving the mesh h.

Let observe that the pairs of indicator open interval functions defined in (6) and the indicator semi-open interval functions defined in (13) $I_h^j$ and $I_{h,j}^{\prime }$ having the supports of size h, as functions in $L^2(0,1)$ differ as values only in the right endpoint $(k+1)h\in (0,1)$ of the support of the semi-open interval and so, being different on a set of measure Lebesgue zero containing only the right endpoint, verifies on $L^2(0,1)$: $\parallel I_h^j-I_{h,j}^{\prime }\parallel =0$, for j = 1,n, nh = 1, $n\ge 2$.

Now, suppose that there exists $f\in L^2\left(0.1\right)$ orthogonal to any $I_{h,j}^{\prime }$, j = 1,n, nh = 1, $n\ge 2$. Then from

$$|\langle f,I_h^j\rangle |=|\langle f,I_h^j-I_{h,j}^{\prime }\rangle |\le \parallel f\parallel \parallel I_h^j-I_{h,j}^{\prime }\parallel =0$$

we obtain that f is orthogonal to $I_h^j,j=1,n,nh=1$, $n\ge 2$ and so, because S is dense, then f = 0. Because f has been considered orthogonal in $L^2(0,1)$ to any $I_{h,j}^{\prime }\in S^{\prime }$ follows the density of $S^{\prime }$.

Moreover, because the pairs of indicator interval functions $I_h^j$ and $I_{h,j}^{\prime }$ j = 1, n, nh = 1, differ only on the endpoints of the h-partitions of the domain (0, 1), the entries in the matrix representations of the integral operator restrictions to $S_h^{\prime }$ coincide with the corresponding entries in the matrix representations of the integral operator restrictions to $S_h$, $n\ge 2,nh=1$ and so, both having the same diagonal entries $d_{kk}^h,k=1,n,nh=1$, i.e.

$$d_{kk}^h=\underset{{\Delta }_k^h}{\int }\underset{{\Delta }_k^h}{\int }\rho (y,x)dxdy=\underset{{\Delta }_{k,h}^{\prime }}{\int }\underset{{\Delta }_{k,h}^{\prime }}{\int }\rho (y,x)dxdy$$

valued in (8), (9).

Below, we follow the steps from [5] for defining the orthogonal projection of the integral operator $T_{\rho }$ by finite rank integral operators as follows.

The orthogonal projections of the kernel function is performed by the orthogonal projections of the integral operator through the finite rank integral operators ([5], pg. 986) whose kernel functions are:

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

(14)

As result, the integral operator projection on $S_h^{\prime }$, $T_{{\rho }_h}$ has the kernel function ([5])

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

(15)

that is a sum of kernel pieces with disjoint supports in which $I_{h,k}^{\prime }\left(y\right)\rho (y,x)I_{h,j}^{\prime }\left(x\right)$ are taken to be 0 for $k\ne j$. Due to the observations made between the indicator functions having the supports open intervals and semi-open intervals, on $S_h^{\prime }$ we have:

$$\langle T_{{\rho }_h}{I}_{h,h}^{\prime },I_{h,h}^{\prime }\rangle =h^{-1}\underset{{\Delta }_{h,k}^{\prime }}{\int }\underset{{\Delta }_{h,k}^{\prime }}{\int }{\rho }_h^k(y,x)dxdy$$

$$=h^{-1}{d}_{kk}^h=h^{-2}{d}_{kk}^h{\parallel I_{h,h}^{\prime }\parallel }^2,k=1,n,nh=1,n\ge 2$$

meaning, the matrix representation of the finite rank integral operator $T_{{\rho }_h}$ differs by the normalisation factor $h^{-1}$ from the matrix representation of the restriction of the integral operator to $S_h^{\prime }$ that, in turn coincides with the matrix representation of the restriction of the integral operator to $S_h$.

Then, the positivity parameter of the finite rank operator approximations verifies

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

If we proceed like in [1], we could verify that the positivity parameter ${\alpha }_h\left(T_{{\rho }_h}\right)$ is preserved on the whole family $F^{\prime }$. So, for $v_h={\sum }_{k=1,n}{c}_k{I}_{h,k}^{\prime }\in S_h^{\prime }$, let compute

$$\begin{array}{cc}\hfill \langle T_{{\rho }_h}{v}_h,v_h\rangle & =h^{-1}\underset{0}{\overset{1}{\int }}\left[\underset{0}{\overset{1}{\int }}\sum _{k=1,n}{I}_{h,k}^{\prime }\left(y\right)\rho (y,x)I_{h,k}^{\prime }\left(x\right)v_h\left(x\right)dx\right]\overline{v_h}\left(y\right)dy\hfill \\ \hfill & =h^{-1}\sum _{k=1,n}\underset{{\Delta }_{h,k}^{\prime }}{\int }\left[\underset{{\Delta }_{h,k}^{\prime }}{\int }{I}_{h,k}^{\prime }\left(y\right)\rho (y,x)I_{h,k}^{\prime }\left(x\right)v_h\left(x\right)dx\right]\overline{v_h}\left(y\right)dy\hfill \\ \hfill & =h^{-1}\sum _{k=1,n}{c}_k\overline{c_k}\underset{{\Delta }_{h,k}^{\prime }}{\int }\underset{{\Delta }_{h,k}^{\prime }}{\int }{I}_{h,k}^{\prime }\rho (y,x)I_{h,k}^{\prime }dxdy=h^{-1}\sum _{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h\hfill \\ \hfill & \ge h^{-2}\left(mi{n}_{k=1,n}{d}_{kk}^h\right)\parallel v_h{\parallel }^2\ge h^{-2}{d}_{11}^h{\parallel v_h\parallel }^2\hfill \end{array}$$

(16)

The computed values of the entries $d_{kk}^h,k=1,n,nh=1$ of the diagonal matrix representation $M_h\left(T_{{\rho }_h}\right)=h^{-1}{M}_h\left(T_{\rho }\right)$ are given by ([1]) and are identical with their values from the previous paragraph.

Follows $\forall h,d_{kk}^h>h^2/2>d_{11}^h,k\ge 2$ with the positivity parameter

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

(17)

a constant mesh independent, for any $n\ge 2,nh=1$.

From the convergence in norm of the finite rank approximations to the compact integral operator, $\parallel T_{{\rho }_h}-T_{\rho }\parallel \to 0\mathrm{for}n\to \infty ,nh=1$, we have:

$|\langle (T_{{\rho }_h}-T_{\rho })v,v\rangle |\le {ϵ}_h{\parallel v\parallel }^2$ with ${ϵ}_h\to 0$, where ${ϵ}_h:={ϵ}_h\left(v\right)$.

Let observe that the kernel $\rho$ is positive valued on (0,1)². Then $T_{\rho }$ is positive definite on S and has no zeros in the dense set as follows from:

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

= ${\int }_{{\Delta }_{h,k}^{\prime }}{\int }_{{\Delta }_{h,k}^{\prime }}\rho (y,x)dxdy:=d_{kk}^h>0$. Then, $u_h\notin N_{T_{\rho }}$ because from $0=\langle T_{\rho }{u}_h.u_h\rangle ={\sum }_{k=1,n}{c}_k\overline{c_k}{d}_{kk}^h$ results: $c_k\overline{c_k}=0,k=1,n$, i.e. $u_h=0$ for every $u_h={\sum }_{k=1,n}{c}_k{I}_{h.k}$. Then $\langle T_{\rho }{u}_h,u_h\rangle >0$, $n\ge 2,nh=1$.

Lemma 2.

The compact operator $T_{\rho }$ is positive definite on the dense family $S^{\prime }$ and without zeros in $S^{\prime }$.

Proof. 

Because ${ϵ}_h$ converges to 0 with $h\to 0,nh=1$, there exists $n_1:=n_1(u,ϵ)$ such that for $n\ge n_2=max(n_0,n_1)$ for which ${ϵ}_h<\alpha$, nh = 1, holds:

$\langle T_{\rho }{u}_h,u_h\rangle$$\ge |\langle T_{{\rho }_h}{u}_h,u_h\rangle -|\left(\langle (T_{{\rho }_h}-T_{\rho })u_h,u_h\rangle \right)\left|\right|\ge {\alpha }_h\left(T_{\rho }\right){\parallel u_h\parallel }^2>0$

where ${\alpha }_h\left(T_{\rho }\right)=\alpha -{ϵ}_h>0$ and so, $T_{\rho }$ is positive definite on the dense set $S\subset L^2(0,1)$, has no zeros in the dense set. Thus, its zeros if any, are in the difference set. □

Observation 2.

The finite rank approximations of $T_{\rho }$ are strict positive on the dense set and, their sequence of the positive parameters is inferior bounded.

The result obtained in Lemma 2 together with the compactness of the Hilbert-Schmidt operator, i.e. $\parallel T_{\rho }-T_{{\rho }_h}\parallel \to 0$ will be used for:

a) - obtaining the strict positivity of $T_{\rho }$ on the dense set $S^{\prime }$ in order to fulfil the demands of the Corollary (Lemma 1) of the Theorem 1;

b) - a shorter proof of the injectivity bypassing the request for the strict positivity of the operator on the dense set in the Theorem 1.

Theorem 3.1 (Corollary).

The Hilbert-Schmidt operator $T_{\rho }$ defined in (1) is injective and, consequently RH holds.

Proof. 

Supposing $u\in E\cap N_{T_{\rho }},\parallel u\parallel =1$, then (4) holds with ${\alpha }_h=\alpha -{ϵ}_0\left(u\right)$ a constant on $S^{\prime }$ for that zero of $T_{\rho }$ from previous estimation taking ${ϵ}_h<{\alpha }_h/2$ for $n\ge n_2:=n_2\left(u\right),nh=1$ in order to obtain ${\alpha }_h\left(T_{\rho }\right)=\alpha /2=(3-2\gamma )/8$. Then, from (Lemma 2)

$(3-2\gamma )/(8\parallel T_{\rho }\parallel )\le {\beta }_h/\sqrt{1-{\beta }_h^2}$ where right hand converges to zero, we obtain a contradiction. So $u\notin N_{T_{\rho }}$. Because the operator has no zeros in $S^{\prime }$ either (see Lemma 2), it is injective, equivalently $N_{T_{\rho }}=\left\{0\right\}$. □

In the hypotheses of the inferior bounded of the positivity parameters of the finite rank operators approximations (17) of the Hilbert-Schmidt integral operator $T_{\rho }$ we obtain again the injectivity of the operator in (1) through its compactness. The following theorem is a particular case of the generic Theorem 2.

Theorem 3.2.

Let the finite rank approximations of a separable Hilbert space $T_{\phi }$ on the family of finite dimension subspaces built on the indicator interval functions where the domain is partitioned of size h, $S_h^{\prime }\subset S_{h/2}^{\prime },nh=1,n\ge 2$ whose union is dense. If the sequence of the positivity parameters of the finite rank operator approximations is bounded inferior, then the operator is injective.

Proof. 

$T_{\rho }$ has its finite rank approximations strict positive definite with positivity parameters mesh independent. On $S_h^{\prime }$:

$\alpha \parallel u_h{\parallel }^2\le \langle T_{{\rho }_h}{u}_h,u_h\rangle \le \parallel T_{{\rho }_h}{u}_h\parallel \parallel u_h\parallel$.

Follows for $u\in E\cap N_{T_{\rho }},\parallel u\parallel =1$, reminding that ${\beta }_h\left(u\right)$ is the normed residuum of $u\in E$ on $S_h^{\prime }$ converging to zero (Observation 1):

$\parallel T_{{\rho }_h}{u}_h\parallel =\parallel (T_{{\rho }_h}-T_{\rho })u_h+T_{\rho }(u_h-u)\parallel$$\le \parallel T_{{\rho }_h}-T_{\rho }\parallel \parallel u_h\parallel +\parallel T_{\rho }\parallel {\beta }_h$

$\le {ϵ}_h\parallel u_h\parallel +\parallel T_{\rho }\parallel {\beta }_h$ where ${ϵ}_h\to 0,{\beta }_h\to 0,\parallel u_h\parallel \to 1.$ Thus, because $\parallel T_{\rho }\parallel \le 1$:

$$(3-2\gamma )/4\le \left({ϵ}_h+{\beta }_h/\sqrt{1-{\beta }_h^2}\right)\to 0$$

(18)

for any $n\ge n_0:=n_0\left(u\right),nh=1$, from an index $n_0<\infty$. But, this is a contradiction because both terms in the right side sum converges to 0 from a range $n_1\ge n_0$. The only supposition we made, has been $T_{\rho }u=0$, so, $u\notin N_{T_{\rho }}$ valid for any zero of $T_{\rho }$ if there are in E. Then $N_{T_{\rho }}=\left\{0\right\}$ once $T_{\rho }$ has no zeros in $S^{\prime }$ either. □