On Injectivity of an Integral Operator Connected to Riemann Hypothesis

Using the equivalent formulation of RH given by Beurling ([4], 1955), Alcantara-Bode showed ([2], 1993) that Riemann Hypothesis holds if and only if the integral operator on the Hilbert space L 2 (0 , 1) having the kernel deﬁned by fractional part function of the expression between brackets { y/x } , is injective. Since then, the injectivity of the integral operator used in equivalent formulation of RH has not been addressed nor has been dissociated from RH and, a pure mathematics solution for RH is not ready yet. Here is a numerical analysis approach of the injectivity of the linear bounded operators on separable Hilbert spaces addressing the problems like the one presented in [2]. Apart of proving the injectivity of the Beurling - Alcantara-Bode integral operator, we obtained the following result: every linear bounded operator (or its associated Hermitian), strict positive deﬁnite on a dense family of including approximation subspaces in L 2 (0,1) built on simple functions, is injective if the rate of convergence to zero of its unbounded sequence of inverse condition numbers on approximation subspaces is o ( n − s ) for some s ≥ 0. When s = 0, the sequence is inferior bounded by a not null constant, that is the case in the Beurling - Alcantara-Bode integral operator. In the Theorem 4.1 we addressed with numerical analysis tools the injectivity of the integral operator in [2] claiming that - even if a solution in pure mathematics is desired, together with the Theorem 1, pg. 153 in [2], the RH holds.

In the paper of Alcantara-Bode [2], the author proved (Theorem 1, pg. 152) that RH holds if and only if the following Hilbert-Schmidt integral operator T ρ defined on L 2 (0, 1) by: has its null space N T ρ = {0} where the kernel function is the fractional part function of the expression between brackets.
The linear bounded integral operator has been introduced in [4] (Beurling, 1955) for providing first equivalent formulation of RH in terms of functional analysis. This equivalent formulation of RH has been used in [2] (Alcantara-Bode, 1993) for providing second equivalent formulation of Riemann Hypothesis in terms of injectivity of the operator in (1) on L 2 (0,1).
We address in this paper the injectivity of linear bounded operators on separable Hilbert spaces. The Theorem 2.1 from §2 on generic separable Hilbert spaces as well its versions adapted for the integral operators on L 2 (0,1) ( §3, Lemmas 3.1 and 3.2), are the criteria using the operator approximations on a dense family of approximation subspaces.
Let H be a separable Hilbert space. The norm considered on H is the norm induced by its inner product. Because a linear bounded operator T and its associated Hermitian and positive definite operator (T*T) have the same null space N T , we will consider its associated Hermitian operator when we do not have enough information about positivity of T on H. Then, we take T or (T*T) in order to have its positivity T u, u ≥ 0 on H as we need. In the following T is positive definite on H and our aim is to provide methods for investigating its strict positivity on H, in other words, in finding if its null space N T = {0}.
We will see later that a weaker property than the positivity on H is enough and is mandatory for investigating the injectivity of the linear operators on H, that is the strict positivity of the operators on the dense family of approximation subspaces (a-positivity property).
Let B := {u ∈ H; u = 1} be the unit sphere in H. Because a not null element u is in N T if and only if u u ∈ N T , we will consider the normalized elements of N T for convenience.
A family of finite dimension subspaces ℑ := {S n , n ∈ N } ⊂ H is a proper family of approximation subspaces if the following properties hold on the subspaces of the family: (a) S n ⊂ S n+1 , n ∈ N (b) β n (u) := u -u n → 0, for n → ∞ for every u ∈ B and with u n its orthogonal projection on S n , (∀) n ∈ N .
A proper family subspaces is dense in H: ∪ ∞ n=1 S n = H, property easy to show. The main reason in defining a dense family of subspaces in H being a proper family, is to evidentiate the two properties used extensively in this paper: the inclusion of the approximation subspaces and, the increasing monotone convergence in norm to 1 of the sequence of the projections on approximation subspaces defined for every element from unit sphere B ⊂ H.
We say that the bounded linear operator T is almost strict positive definite on H or a-positive, if there exists a proper family ℑ on which, for every subspace S n ∈ ℑ, there exists α n > 0 such that: We name the set {α n , n ≥ 1}, the set of a-positive parameters associated to T on ℑ. In facts, a-positive parameters are the smallest eigenvalues of the operator restrictions to the corresponding subspaces. We will use the term a-positive also associated to the family ℑ for convenience.
Let L ℑ be the set of linear bounded operators on H that are a-positive with respect to the family ℑ.
Let T ∈ L ℑ . If there exists u ∈ B ∩ N T then u / ∈ S n for every n ≥ 1, i.e. u / ∈ ℑ 0 . For this reason we will consider in our analysis only u ∈ H that are eligible, in the sense that these elements from H could be the normalised zeros of T, u ∈ B 0 .
We will refer in this paper to B 0 as the set of eligible zeros of the apositive operators given a proper family ℑ.
Observations: we point out two extreme situations that we will exclude from our analysis: 1.) -If T is positive definite on H but is not a-positive on a proper family, then T should have a zero inside the family: there exists S n ∈ ℑ and v ∈ S n not null, such that T v, v = 0. So T is not injective and the investigations on its injectivity ends here. If T is not positive definite on H then the observation is valid for its associate Hermitian. That means, a-positivity property of an linear bounded operator is mandatory for its injectivity.
2.) -If T is a-positive and the approximation subspaces of ℑ are invariant subspaces of T, then T is injective and so, no investigations for its injectivity needed. As a consequence, from now on the linear bounded operators in considera-tion would be a-positive operators satisfying T (S n ) S n , (∀)n ≥ 1 §2. Injectivity Criteria. Let ℑ be a proper family of approximation subspaces on H and T ∈ L ℑ .
Suppose that there exists w ∈ S ⊥ n for which Tw / ∈ S ⊥ n . For a such element w and for every v ∈ S n both not null, the inner product T w, v is well defined in H. Now, because an strict positive operator on a finite dimension subspace has the same eigenvalues as its adjoint, in our case T ⋆ |Sn and T |Sn have both the same largest eigenvalue λ max n on S n , we obtain the following estimation for the considered inner product: If there exists u ∈ N T ∩ B 0 , we take w := (u n -u) ∈ S ⊥ n and v = u n , the projection of u on S n . Obviously, Tw = Tu n / ∈ S ⊥ n (because of the a-positivity of T T u n , u n = 0 and, because there exists an index n 0 (u) from which u is not orthogonal to S n from the density of the proper family). Then: α n u n 2 ≤ T u n , u n =| T (u n − u), u n | and, together with the previous relations we obtain: where µ n := λ min n λ max n is the ratio between the smallest and largest eigenvalues of the restriction of T to the subspace S n , inequality obtained using the a-positivity property and replacing α n with λ min n .
Pointing out: if there exists u ∈ N T ∩ B 0 , then (2) holds on every S n ∈ ℑ n ≥ n 0 (u) where n 0 (u) is the finite index from where the orthogonal projections of u are not null -in virtue of the properties of the proper family given u ∈ B 0 . So, the property (2) for an eligible u ∈ N T , is valid starting from a finite index n 0 (u); however, it is validated also for indexes n < n 0 (u) because for such indexes u n = 0 and β n (u) = 1.
For T ∈ ℑ, given u ∈ B 0 we define the expression θ n (u) := θ T n (u) on every subspace of the proper family S n , n ≥ n 0 (u) by: with u n the orthogonal projection of u on S n . The last member in the equality is obtained using the relationship for every u ∈ B and its orthogonal projections on S n and S ⊥ n : u n 2 + (u − u n ) 2 = u 2 = 1.
We will split the set of eligible elements B 0 in two disjoint subsets: Note. Because for every normalised u ∈ N T the inequality (2) holds globally on the proper family ℑ, this behaviour will be propagated from now on for each statement involving (2); it is the reason we would omit carrying out every time the index specifications referring to all subspaces of the family, when it is evident from context. We would do it until the end of the paragraph.
A.) If the sequence {µ n }, n ≥ 1 is inferior bounded by a strict positive constant independent of n, µ n ≥ C > 0, for every n ≥ 1 then T is injective.
B.) If Q is a linear, bounded and injective operator on H such that Q (or its associate Hermitian) verifies µ n (Q) ≤ µ n (T ), n ≥ 1 then T is injective.

Proof of A.)
Suppose that the sequence {µ n }, n ≥ 1 is inferior bounded by a constant C independent of n, strict positive. Let u ∈ B 0 . 1) If there exists an infinite subsequence of subspaces from the family for which θ nm (u) ≥ 0, n m ≥ n 0 (u) we obtain: and lim m→∞ u nm 2 < 1 (1 + C 2 ) < 1. Or, {u nm }, n m ≥ n 0 (u) should converge in norm to 1 on the approximation subspaces for every u ∈ B. Thus, the inequality θ nm (u) ≥ 0 could not take place on an infinity of subspaces of the family. We should have instead at most only a finite number of subspaces verifying it.
If there exists a finite number of subspaces on which θ nm (u) ≥ 0, n m ≥ n 0 (u) then we will shrink the family including these subspaces inside of the subspaces already verifying the opposite relationship, obtaining u / ∈ B 1 0 when the sequence {µ n }, n ≥ 1 is inferior bounded by a not null constant. Then, we are in the situation θ n (u) < 0 for every n ≥ n 0 (u) on the ℑ, i.e. u ∈ B 2 0 .
2) If θ n (u) < 0 holds for n ≥ n 0 (u), then β n (u) < µ n u n , n ≥ n 0 (u). But, u satisfying such relationship could not be in N T because a reverse inequality given in (2) holds for eligible u in N T . The only restriction we put on u ∈ H, has been its eligibility, u ∈ B 0 . Thus, if {µ n }, n ≥ 1 is inferior bounded, does not exists u ∈ B 0 that is in N T . Or, B 0 contains all normalized zeros of T, so N T = {0}.

Proof of B.)
Because Q is injective, then Q or (Q ⋆ Q) is strict positive on H and so, is a-positive on ℑ. Let µ n (Q), n ≥ 1 be the injectivity parameters of Q or of its associate Hermitian.
For every u ∈ B 0 , u ∈ B 2 0 (Q) having θ Q n (u) < 0 otherwise, u should be in N Q that is a null set. Let u ∈ B 0 . From µ n (Q) ≤ µ n (T ) we obtain θ T n (u) ≤ θ Q n (u) < 0 , valid for n ≥ 1 showing that u ∈ B 2 0 (T ). Because u is chosen an arbitrary eligible element, that is an orthogonal projection in L 2 (0,1) on S h having the orthogonal eigenfunctions {χ n,k }, k = 1, n. In fact, S h ∈ ℑ χ are generated by the families of the orthogonal eigenfunctions of the projection operators {P h := P | S h }, nh =1, n ≥ 2. Moreover, because the orthogonality of eigenfunctions is dictated by the disjoint interval support, χ n,k (x)χ n,j (y) = δ k,j the entries in the matrix representation (5) bellow, of the operator restrictions on approximation subspaces will be zero outside the diagonal and, it explain the form of the discrete approximations of the kernel functions on approximation subspaces defined bellow.
Thus, the linear integral operator (like in (1)) has the discrete approximations of the kernel function ρ like in [5], [6]: Obviously, (∀) u ∈ H, its projection on S h take the form of a simple function: P | S h u = n k=1 u, χ n,k χ n,k .
Accordingly, taking any v h ∈ S h not null, of the form v h (t) = n k=1 c k χ n,k (t) the discrete approximation of the integral operator, verifies the following equality: Taking a look at (5) and at the matrix representation M h of the T h ρ on S h , if the matrix is strict positive definite, we have: λ min (M h ) = min k=1,n d h kk , from where: So, the a-positivity parameters associated to T ρ should have the form like α n = h −2 λ min (M h ), nh = 1, n ≥ 1.

Remark 3.1) The linear bounded integral operator T ρ is a-positive on ℑ χ if and only if on approximation subspaces its kernel function is verifying
for k = 1,n, nh = 1, i.e. the diagonal entries in the diagonal matrices M h are strict positive.
is inferior bounded, µ h (T ρ ) > C by a strict positive constant C independent of n, nh = 1, then T ρ is injective. Proof.
In the view of the previous paragraphs, if T ∈ L ℑχ is a linear bounded integral operator its a-positivity parameters are given by α n = h −2 min k=1,n d h kk , nh =1, n≥ 2. Thus, the injectivity criteria (Theorem 2.1.A) could be used in determining the injectivity of T ρ observing that in this case {µ n , n ≥ 1} are given as in (7).
We will introduce a method for analysing the the injectivity of apositive operators having the injectivity parameters sequence inferior unbounded.
Let define the I-multiplier operator, the operator obtained as a product of identity by a power of x, {I s }, s ∈ N on L 2 (0,1) as follows: for u ∈ L 2 (0,1) It is easy to show that for every s, I s is linear, bounded, positive definite and injective on L 2 (0, 1) and so, a-positive on ℑ χ . The injectivity and positivity properties on H are coming from I s u, u = 1 0 w(x)w(x)dx > 0 for u = 0 where w(x) = x s 2 u(x). Its injectivity parameters applying the formula (7), are for n ≥ 1 µ n(Is) = 1/( i=0,s Lemma 3.2 (I-multipliers Rule.) Let T ∈ L ℑχ . If exists an I-multiplier operator I s for some s ≥ 1 such that µ n (I s ) ≤ µ n (T ) for n ≥ 1, then T is injective. Proof Suppose that the relationship exists. Then, taking Q = I s in the Theorem 3.1.B, we obtain T injective.
As a consequence of Lemmas 3.1 and 3.2, we could give the following suficient condition for injectivity of a linear bounded operator on L 2 (0, 1) extending the range of s from s ∈ N to s ≥ 0: Remark 3.2) Let T ∈ L ℑχ . If its injectivity parameters verifies µ n (T ) = o(n −s ) for some s ≥ 0, then T is injective. The case s = 0 is specific for the operators having the sequence µ n (T ), n ≥ 1 inferior bounded (Lemma 3.1) and the cases for s > 0 are entering under the I-multipliers Rule (Lemma 3.2). We remember that the a-positivity on a proper family of approximation subspaces is necessary for injectivity of the linear bounded operators on H. We do not have a response yet for the question if it is also a sufficient condition at least for operators ∈ L ℑχ . §4. On the injectivity of the integral operators with discontinue kernel functions. We will investigate the injectivity of two integral operators having the kernel functions discontinue on H := L 2 (0,1). We will use the proper family of approximation subspaces introduced in previous paragraph, ℑ χ on which, for a-positivity of integral operators we will apply the Remark 3.1. The I-multiplier Rule is the right method to apply for the injectivity of the integral operator in Example 1 and, the Lemma 3.1 for the injectivity of the integral operator in the Example 2.   (7): µ n (T σ ) = (1/2 − n + n 2 ln((n + 1)/n))/(1 − π 2 /12) n ≥ 2, nh = 1.
Hence T σ is a-positive but, µ n (T σ ) is not inferior bounded. Then we will apply the I-multipliers Rule (Lemma 3.2).
Observing that (f(t) -1/t) and (f(t) -1/t 2 ), both → 0 for t → ∞ and, 1/t 2 < f (t) < 1/t for t 4 we could consider the following modulo the factor (1 − π 2 /12), as I-multipliers I 1 and I 2 (s =1 and s = 2 being the powers of x in the definition of the I-multiplier I x s u(x) = x s u(x) on L 2 (0, 1). Then by Lemma 3.2, taking the I-multiplier I s , s = 2, we proved that T σ is injective.

Example 2:
The proof of the injectivity of the integral operator used in RH Equivalence ( [2]). Let the linear bounded integral operator T ρ defined by on L 2 (0,1) having the kernel function ρ(y, x) := { y x } . This kernel function is the fractional part function of the quantity between brackets having jumps on the lines like y = k·x, k ≥ 1 resulting a discontinue function.
Because (T ρ χ n,k )(y) / ∈ span( {χ n,k }, 1 ≤ k ≤ n, nh = 1), the approximation subspaces are not invariant subspaces of T ρ . Computing the entries of the diagonal matrices M h , (see (6)) of the kernel approximations on the subspaces {S h }, nh = 1, n ≥ 2, we obtain: Hence the sequence µ n (T ρ ), n ≥ 1, is bounded by a constant strict positive on every approximation subspace of the family. By the Lemma 3.1 follows that the integral operator T ρ is injective, equivalently, N Tρ = {0}. We just proved the following theorem: Claim: Based on the Theorem 4.1 and the Theorem 1, pg. 152 in [2], we could claim that the Riemann Hypothesis holds.