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The Grommer inequalities are equivalent to RH and formulated on page 20 of Conrey - Riemann's hypothesis:

Let $$\Xi(t) := \xi(1/2+it).$$

Then RH is equivalent to : All zeros of $\Xi(t)$ are real. The functional equation for $\zeta$ is equivalent to $\Xi(t) = \Xi(-t)$. Let $Y(t):= \Xi(\sqrt{t})$ and let

$$-\frac{Y'(t)}{Y(t)} = s_1+s_2t+s_3t^2+\dotsb$$

and set $k(a,b) := s_{a+b}$. Let $K_N := ((k(m,n))_{1\le m,n\le N}$ and $D_N:=\det(K_N)$.

Then the Grommer inequalities are equivalent to RH:

$$\text{RH} \iff D_N > 0 \forall N \in \mathbb{N}.$$

But this is by the Sylvester criterion equivalent to thematrices $K_N$ being positive definite ($K_N>0$) for all $N \in \mathbb{N}$.

What is unclear to me: Is $K_N > 0$ $\forall N \in \mathbb{N}$ equivalent to the function $k$ being positive definite over the natural numbers?

If this is the case, one could have a candidate for the Hilbert–Pólya conjecture: By the Moore–Aronszajn theorem, there exists a mapping:

$$\phi: \mathbb{N} \rightarrow H$$

in some Hilbert space $H=l_2$, such that:

$$k(a,b) = \left\langle \phi(a), \phi(b) \right\rangle \forall a,b \in \mathbb{N}.$$

Define on a dense subspace $D :=\{ x | x \text{ has only finitely many nonzero entries } x_i\} \subset l_2$ the operator $R$ via:

$$R(e_n) := \phi(n)$$

and for $x = \sum_{i=1}^N x_i e_i \in D$:

$$R(x) := \sum_{i=1}^N x_i R(e_i), R: D \rightarrow D$$

where $(e_n)_{n \in \mathbb{N}}$ is an orthonormal basis in the Hilbert space $l_2$.

Set $O:= R^*R$. Then $O:D \rightarrow D$ is a positive, densely defined, self-adjoint operator, with the property:

$$\left\langle O(e_m),e_n \right\rangle = \left\langle R^*R (e_m) , e_n \right\rangle = \left\langle R(e_m),R(e_n)\right\rangle = k(m,n).$$

Then by Grommer inequalities and Sylvester's criterion (/modulo my question above can be answered positively) :

$O$ is a positive operator with the property $O=R^*R$ for some $R$ and $\left\langle O(e_m),e_n \right\rangle = k(m,n).$ $\iff$ $k$ is a positive definite function on the natural numbers $\iff$ RH.

Edit: There seems to be confusion about this question, so please let me clarify: I am not saying that this question is proving RH. What I am saying is, that $O>0 \iff k > 0 \iff \text{Grommer's ineq.} \iff RH$ and so, this makes $O$ an candidate for the Hilbert-Polya conjecture in some wide sense.

Second question: I tried to implement this idea in Sagemath to have some empirical backup, but without success since Sagemath cannot cope with the $\sqrt{t}$ in the series expansion, but it seems that Mathematica can, since I tried a little bit on Wolfram Alpha. Is it possible for someone with Mathematica to try and implement this idea for empirical backup? (I would define the matrix $K_N$ and see if it is positive definite for say $N=1,\dotsc,200$, and look at the eigenvalues of the matrix and compare them with OEIS sequence.)

Related question: Spectrum of "infinite-Gram matrix"?

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  • $\begingroup$ I think that your operator $O$ is positive semi-definite, i.e., its quadratic form is non-negative. I see no reason why it should be (strictly) positive definite. It follows that its compressions are also positive semi-definite but there is no argument for their strict positivity. $\endgroup$
    – Twi
    Commented Jul 21, 2023 at 10:50
  • $\begingroup$ @Twi: By the criterion of Sylvester on positive definite matrices? $\endgroup$ Commented Jul 21, 2023 at 10:57
  • $\begingroup$ No. I'm only saying that you know $O\geq0$. It follows that $O_{N}\geq0$ for all $N$. But you need $O_{N}>0$ for the Grommer ineq. $\endgroup$
    – Twi
    Commented Jul 21, 2023 at 11:17
  • $\begingroup$ @Twi I do not understand your comment. If $O>0$ then $O_n>0$ and so by Sylvester's criterion $\det(O_N)>0$. On the other hand: If by Grommer ineq. $\det(O_n)>0$ then by Sylvester criterion $O_N>0$ and so $O>0$. It would be nice if you could elaborate on your comment, as I really do not see what you mean. $\endgroup$ Commented Jul 21, 2023 at 12:12
  • $\begingroup$ If you are able to show that $O>0$ (it is not proven in your post) then the first chain of implications in your argument would be OK. My comment is that you only have $O\geq0$, and not $O>0$. This is why your post does not prove RH. $\endgroup$
    – Twi
    Commented Jul 21, 2023 at 12:45

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I will try to answer the first question on my own:

Since the matrices $K_n$ are assumed positive definite and since each matrix $K_n$ occurs as diagonal minor-submatrix of $K_{n+1}$, we can apply inductively inductive Cholesky decomposition to find an embedding $\phi(n)$ for each number $n$:

Equation 1): $$\phi(n+1) = \left(\begin{array}{r} {C_{n}^{-1} v_n} \\ \sqrt{K_{n+1,n+1}-{\left| {C_{n}^{-1} v_n} \right|}^{2} } \end{array}\right)$$

Equation 2):

$$v_n = \left(\begin{array}{r} K_{1,n+1} \\ K_{2,n+1} \\ \cdots\\ K_{n,n+1}\\ \end{array}\right)$$

Equation 3): $$C_n = \left(\begin{array}{r} \phi(1)^T \\ \phi(2)^T \\ \cdots\\ \phi(n)^T\\ \end{array}\right)$$

Equation 4): $$\phi(1) := \sqrt{K_{1,1}} e_1$$

Now we have found an embedding for each number $a$ in the Hilbert space of sequences $l_2$: $$\phi(a) = \sum_{i=1}^a f_{a,i} e_i$$

for some real numbers $f_{a,i}$, such that we have:

$$K_{a,b} = \left < \phi(a), \phi(b) \right > $$

But the value of $K_{a,b}$ is just $k(a,b)$ , so this proves that the function $k$ is positive definite over the natural numbers.

On the other hand, if the function $k$ is assumed to be positive definite over the natural numbers, then we have the matrices:

$$(k(a,b))_{1 \le a,b \le n}$$

will be positive definite for each $n$.

But these are just the $K_n$ matrices. So this proves the equivalence stated in the first question.

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