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The Riemann Hypothesis as a Stability Principle

  The Riemann Hypothesis can be reformulated as an infinite sequence of explicit positivity tests. This is not a proof. It turns a statement about complex zeros into inequalities that can be calculated, falsified at finite order, and compared with operator or physical models.   The completed zeta function is $$\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s),$$ with $$\Xi(t)=\xi\left(\frac12+it\right).$$ The function $\Xi$ is even and real on the real axis. A nontrivial zero $\rho=\beta+i\gamma$ corresponds to $$t_\rho=\gamma-i\left(\beta-\frac12\right).$$ Therefore $$t_\rho\in\mathbb R \quad\Longleftrightarrow\quad \beta=\frac12,$$ and hence $$\boxed{\mathrm{RH}\quad\Longleftrightarrow\quad\text{every zero of }\Xi\text{ is real}.}$$ Folding the Zeros Because $\Xi$ is even, write $$\Xi(t)=\sum_{n\ge0}c_nt^{2n}.$$ Now define $$G(z)=\frac{\Xi(i\sqrt z)}{\Xi(0)}.$$ Although this contains $\sqrt z$, the even expansion makes $G$ entire: $$G(z)=...
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The Riemann Hypothesis as a Stability Principle

 

The Riemann Hypothesis can be reformulated as an infinite sequence of explicit positivity tests.

This is not a proof. It turns a statement about complex zeros into inequalities that can be calculated, falsified at finite order, and compared with operator or physical models.

 

The completed zeta function is

$$\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s),$$

with

$$\Xi(t)=\xi\left(\frac12+it\right).$$

The function $\Xi$ is even and real on the real axis. A nontrivial zero $\rho=\beta+i\gamma$ corresponds to

$$t_\rho=\gamma-i\left(\beta-\frac12\right).$$

Therefore

$$t_\rho\in\mathbb R \quad\Longleftrightarrow\quad \beta=\frac12,$$

and hence

$$\boxed{\mathrm{RH}\quad\Longleftrightarrow\quad\text{every zero of }\Xi\text{ is real}.}$$

Folding the Zeros

Because $\Xi$ is even, write

$$\Xi(t)=\sum_{n\ge0}c_nt^{2n}.$$

Now define

$$G(z)=\frac{\Xi(i\sqrt z)}{\Xi(0)}.$$

Although this contains $\sqrt z$, the even expansion makes $G$ entire:

$$G(z)=\frac1{\Xi(0)} \sum_{n\ge0}(-1)^nc_nz^n.$$

If $t_\rho$ is a zero of $\Xi$, then the corresponding zero of $G$ is $z_\rho=-t_\rho^2$. Thus

$$z_\rho\in(-\infty,0) \quad\Longleftrightarrow\quad t_\rho\in\mathbb R,$$

so

$$\boxed{\mathrm{RH}\quad\Longleftrightarrow\quad G\text{ has only negative-real zeros}.}$$

The folding replaces each pair $\pm\gamma_n$ by one positive number $\lambda_n=\gamma_n^2$. Since $\Xi$ has order one, $G$ has order one half and therefore genus zero. Under RH its product is

$$G(z)=\prod_{n\ge1}\left(1+\frac{z}{\lambda_n}\right).$$

From Zeros to a Positive Measure

Take the logarithmic derivative:

$$R(z)=\frac{G'(z)}{G(z)} =\sum_{n\ge1}\frac1{z+\lambda_n}.$$

If every $\lambda_n$ is positive, then $R$ is the Stieltjes transform of a positive discrete measure:

$$R(z)=\int_0^\infty\frac{d\mu(\lambda)}{z+\lambda}, \qquad d\mu(\lambda)=\sum_{n\ge1}\delta_{\lambda_n}(d\lambda).$$

For $x>0$,

$$(-1)^kR^{(k)}(x) = k!\sum_{n\ge1}\frac1{(x+\lambda_n)^{k+1}} \ge0.$$

The derivatives alternate in sign. Expanding at the origin gives

$$R(z)=\sum_{k\ge0}(-1)^km_kz^k,$$

where

$$m_k=\sum_{n\ge1}\lambda_n^{-k-1} =\sum_{n\ge1}\gamma_n^{-2k-2}.$$

Choose a scale $0<a<\lambda_1$, set $u_n=a/\lambda_n$, and define $M_k=a^{k+1}m_k$. Then $0<u_n<1$, and

$$M_k=\int_0^1u^k,d\nu(u), \qquad d\nu(u)=\sum_{n\ge1}u_n,\delta_{u_n}(du).$$

Applying the forward-difference operator $\Delta M_k=M_{k+1}-M_k$ gives

$$(-1)^r\Delta^rM_k = \sum_{n\ge1}u_n^{k+1}(1-u_n)^r \ge0.$$

Therefore

$$\boxed{\mathrm{RH} \Longrightarrow (-1)^r\Delta^rM_k\ge0 \quad\text{for every }k,r\ge0.}$$

The Riemann Hypothesis has become an infinite positivity problem.

The Converse Is the Difficult Part

The Hausdorff moment theorem says that a sequence ${M_k}$ comes from a finite positive measure on $[0,1]$ exactly when $(-1)^r\Delta^rM_k\ge0$ for every $k,r\ge0$. But this theorem alone does not prove that a specified entire function has only negative-real zeros. One must also establish the analytic hypotheses that connect the reconstructed measure to $G'/G$ and control its poles.

For the relevant genus-zero criterion, these include: order strictly below one; positivity of the Taylor coefficients; summability of the reciprocal zeros; and a uniform geometric condition on the transformed zeros.

The order and genus follow from the known growth of $\Xi$. The remaining conditions must be proved separately. They cannot be inferred merely from a finite or formal collection of moment inequalities.

Subject to those hypotheses,

$$\boxed{\begin{aligned} \mathrm{RH} &\Longleftrightarrow G(z)=\frac{\Xi(i\sqrt z)}{\Xi(0)} \text{ has only negative-real zeros} \ &\Longleftrightarrow (-1)^r\Delta^rM_k\ge0 \quad\text{for every }k,r\ge0. \end{aligned}}$$

The second equivalence is valid only after the required analytic conditions have been established.

Computing the Moments Without the Zeros

The moments can be extracted directly from derivatives of $\Xi$. Write

$$G(z)=\sum_{n\ge0}b_nz^n, \qquad b_n= \frac{(-1)^n\Xi^{(2n)}(0)} {(2n)!,\Xi(0)}.$$

If $G'(z)/G(z) = \sum_{k\ge0}r_kz^k$, then coefficient comparison gives

$$r_k=(k+1)b_{k+1} -\sum_{j=0}^{k-1}r_jb_{k-j}.$$

The moments are $m_k=(-1)^kr_k$ and $M_k=a^{k+1}m_k$. Every finite-difference test can therefore be calculated from Taylor derivatives of $\Xi$, without supplying a list of zeros.

Numerically, $\Xi(0)=\xi\left(\frac12\right) \approx0.49712077818831410991$, and $\gamma_1^2\approx199.79045483238685946$. Taking $a=100$, low-order calculations produce positive values for all tested pairs with $r\le4$ and $k+r\le7$. This confirms the coefficient recursion and low-order positivity. It does not establish the infinite family.

A single negative finite difference would disprove the proposed positivity criterion at that order. Positive results at finitely many orders remain evidence only.

The Operator Target

If there were an independently constructed positive self-adjoint operator $A$ with eigenvalues $\lambda_n$, then formally

$$G(z)=\det\left(I+zA^{-1}\right)$$

and

$$R(z)=\operatorname{Tr}(A+z)^{-1}.$$

The positive measure would then arise from the spectral resolution of the resolvent trace. Positivity of $A$ would place the zeros of $G$ on the negative real axis.

But defining $A$ to have eigenvalues $\gamma_n^2$ would be circular. A genuine construction must establish its domain, boundary conditions, self-adjointness, positivity, discrete spectrum, and arithmetic origin without assuming RH. This states the operator problem precisely:

$$\boxed{\text{Construct }A\ge0\text{ independently such that } \operatorname{Tr}(A+z)^{-1}=\frac{G'(z)}{G(z)}.}$$

What Is New Here?

The reformulation supplies four concrete advances.

A global zero problem becomes an explicit family of inequalities. Each $M_k$ is computable from derivatives of $\Xi$, and every finite-order condition can be tested directly.

The moment normalization is exact. The correct finite measure is $d\nu(u)=\sum_nu_n,\delta_{u_n}(du)$, so that $M_k=\int_0^1u^k,d\nu(u)$.

The operator target becomes explicit. A spectral construction must reproduce $G'/G$ independently of the zero locations.

The main logical gap is isolated. The unresolved step is to derive the required positive measure, together with the analytic hypotheses for the converse, directly from arithmetic information.

What This Does Not Prove

The Hausdorff theorem cannot be used by itself to determine the zeros of an arbitrary entire function. The scale must satisfy $0<a<\lambda_1$. An operator defined by inserting the Riemann zeros as its spectrum proves nothing about their location. Quantum systems that encode selected zeta values or zeros do not independently exclude off-line zeros. Finite numerical positivity does not establish positivity for every $k$ and $r$.

The remaining problem is therefore

$$\boxed{\frac{G'(z)}{G(z)} = \int_0^\infty\frac{d\mu(\lambda)}{z+\lambda}, \qquad d\mu\ge0,}$$

derived directly from primes, Weil's quadratic form, a trace formula, or another independent arithmetic construction.

If such a global positive-measure or positive-operator identity can be proved, the zeros are forced onto the required axis. Until then, the reformulation provides a precise programme rather than a proof:

$$\boxed{\text{Riemann zeros} \quad\longrightarrow\quad \text{folded entire function} \quad\longrightarrow\quad \text{moments} \quad\longrightarrow\quad \text{positivity}.}$$

 

References

R. Zhang, An Application of Hausdorff Moment Problem, arXiv:2303.09396v6, 2023.

A. Connes, The Riemann Hypothesis: Past, Present and a Letter Through Time, arXiv:2602.04022v1, 2026.

D. Schumayer and D. A. W. Hutchinson, Physics of the Riemann Hypothesis, Reviews of Modern Physics 83, 307, 2011.

P. Betzios, N. Gaddam and O. Papadoulaki, Black Holes, Quantum Chaos, and the Riemann Hypothesis, arXiv:2004.09523v4, 2021.

S. Wei et al., The Riemann Hypothesis Emerges in Dynamical Quantum Phase Transitions, arXiv:2511.11199v1, 2025.

M. Planat, A Theta-Kernel Reformulation of Riemann-$\Xi$ Growth and the Obstruction to Blockwise Positivity, Symmetry 18, 1283, 2026. 

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