REVIEW
On variants of P\'olya's conjecture
Number Theory · Submitted 27 Aug 2026 · 22 references
Pith's verdict: A new sufficient condition for RH via a weighted Liouville sum, cleanly proved. The converse is conditional and heavy, but the forward direction is the real result.
T0 · AI review
Why Pith says this
Highlight any text to comment on it.
Referee report
The paper
The paper claims
If a certain weighted sum of the Liouville function never turns negative, the Riemann hypothesis follows.
The reading
What carries the argument
The carrying mechanism is a Mellin-transform identity paired with Landau's sign-singularity theorem. The identity $\int_1^\infty f(x)x^{-s+\frac12}\frac{dx}{x}=\frac{F'(s)}{(s-\frac12)^2}$ with $F(s)=\zeta(2s)/\zeta(s)$ ties the sign of $f$ to the analytic location of a zeta quotient: at $s=1$ the factor $1/\zeta(s)$ has a simple zero while $\zeta(2s)$ is regular, so $F$ vanishes there, and the only pole of $\zeta(2s)$ sits at $s=\frac12$, leaving $\frac{F'(s)}{(s-\frac12)^2}$ free of real singularities on $(\frac12,\infty)$. Landau's theorem (Lemma 2.4) supplies the converse direction of the logic: a non-negative function $A(x)$ forces its Mellin abscissa of convergence to be a singularity, so the abscissa for $x\mapsto f(x)x^{1/2}$ must lie at or below $\frac12$, which is exactly the analyticity that excludes off-line zeros. For the converse theorem, the pole of $F$ at $s=\frac12$ with leading coefficient $1/(2\zeta(\frac12))$ generates the cubic main term through a residue computation using the first three derivatives of $\zeta$ at $\frac12$ and the constants from $\zeta$'s Laurent expansion at $1$, while each simple zero $\rho=\frac12+i\gamma$ contributes the residue $\frac{\zeta(1+2i\gamma)e^{i\gamma L}}{\gamma^2\zeta'(\frac12+i\gamma)}(L+\frac{2i}{\gamma})$, whose pairings over conjugate zeros sum into the bounded functions $\Phi$ and $\Psi$; the assumption H3 is precisely the absolute convergence that makes those sums uniform in $L$.
Extended reading notes
Core claim
The core claim is the pair of theorems linking the sign of $f$ to the zeros of $\zeta$. Theorem 1.1 proves that eventual non-negativity of $f(x)=-\sum_{n\le x}\lambda(n)\frac{\log n}{\sqrt{n}}\log\frac{x}{n}$ implies RH: the Mellin transform of $f$ is $F'(s)/(s-\frac12)^2$ with $F(s)=\zeta(2s)/\zeta(s)$, Lemma 2.4 (Landau's theorem) forces a non-negative function's abscissa of convergence to be a real singularity, and the quotient has no real singularity in $(\frac12,\infty)$ — hence the transform is analytic in $\Re s>\frac12$, which rules out any zero $\rho$ of $\zeta$ with $\Re\rho>\frac12$, because such a zero would be a pole of $F'$. Theorem 1.2 is the conditional converse: assuming RH, the Simple Zero Conjecture, and the condition H3 (that $\sum_{\gamma>0}\frac{|\zeta(1+2i\gamma)|}{\gamma^2|\zeta'(\frac12+i\gamma)|}<\infty$), the paper derives the explicit formula $f(x)=P(\log x)+(\log x)\Phi(\log x)-\Psi(\log x)+O_\varepsilon(x^{-1+\varepsilon})$, where $P(L)=-\frac{L^3}{12\zeta(\frac12)}+\alpha L+\beta$ is an explicit cubic polynomial and $\Phi,\Psi$ are bounded functions built from the zeta zeros. Since $\zeta(\frac12)<0$, the leading coefficient is positive, and the corollary states that $f(x)\sim\frac{(\log x)^3}{12|\zeta(\frac12)|}$ as $x\to\infty$, so $f$ tends to $+\infty$ and is eventually positive.
Load-bearing premise
The unconditional direction needs only the sign hypothesis it assumes, but the converse theorem, and with it the eventual positivity of $f$, rests on the unproved condition H3: that the sum $\sum_{\gamma>0}\frac{|\zeta(1+2i\gamma)|}{\gamma^2|\zeta'(\frac12+i\gamma)|}$ over the positive imaginary parts of the zeta zeros converges absolutely, and the paper never verifies this sum converges or shows it follows from RH.
Editorial extensions
If this is right
- Eventual non-negativity of this one weighted Liouville sum implies RH; contrapositively, if any zero of $\zeta$ lies off the critical line, $f(x)$ must be negative at arbitrarily large $x$.
- Under RH, the Simple Zero Conjecture, and H3, $f(x)\sim\frac{(\log x)^3}{12|\zeta(\frac12)|}$, so the sum grows to $+\infty$ with only bounded oscillations and is eventually positive.
- The leading coefficient is fixed by the single number $\zeta(\frac12)$; the zeta zeros affect only the bounded fluctuation terms $\Phi$ and $\Psi$, not the main trend.
- The sufficient condition joins the Pólya–Turán family of sign criteria: unlike the original Pólya conjecture on $\sum_{n\le m}\lambda(n)$, which is known to be false, this weighted variant's non-negativity is strong enough to force RH.
Reading between the lines
- The paper leaves open whether H3 follows from RH or from the Simple Zero Conjecture; because the zeros are known to very large height, the plausibility of H3 is directly testable numerically, and verifying it would be the natural first step toward removing the auxiliary condition.
- Read through the Chebyshev-bias lens, the corollary predicts a one-sided bias of fixed sign growing like $(\log x)^3$, in contrast to biases that hold only on sets of logarithmic density, suggesting weighted Liouville sums can carry biases of a qualitatively different strength.
- Replacing the weight $\sqrt{n}$ by a general $n^\alpha$ would produce a one-parameter family of sums interpolating between the $L_\alpha$ criteria of the predecessor literature and this Riesz-type $f$; locating the range of $\alpha$ for which non-negativity still implies RH is a natural extension the paper does not pursue.
- Taken together, Theorem 1.1 and Corollary 1.3 show that if RH, the Simple Zero Conjecture, and H3 all hold, the hypothesis of Theorem 1.1 is in fact satisfied — the two directions are mutually consistent, and the sufficient condition can only be false if RH itself fails.
Original abstract
In this paper, we study a Riesz-type weighted sum of the Liouville function, $f(x):=-\sum_{n\le x}\frac{\lambda(n)\log n}{\sqrt n}\log\frac{x}{n}$ for sufficiently large $x$. Motivated by a recent research on the sign criteria for the Riemann Hypothesis arising from weighted prime-counting functions, we investigate the sign behavior of $f(x)$ and its relation to the zeros of the Riemann zeta function. We first prove that if $f(x)$ is non-negative for all sufficiently large $x$, then the Riemann Hypothesis holds. The proof is based on the Mellin transform of $f$ and the analytic properties of $\frac{\zeta(2s)}{\zeta(s)}$. Conversely, assuming the Riemann Hypothesis, the Simple Zero Conjecture, and an absolute convergence condition involving the nontrivial zeta zeros, we derive an explicit formula for $f(x)$ in terms of these zeros. In particular, we show that $f(x)\sim \frac{(\log x)^3}{12|\zeta(\frac{1}{2})|}$ as $x\to \infty$. Consequently, we show that under these hypotheses, $f(x)$ is eventually positive.
The compiled record
The paper's own sentences, arranged, kept beside the review and separate from it. Each label is drawn from a closed vocabulary; the number beside it is the probability the classifier gave that label. Numbers in the tables are the authors', checked by code. No sentence in this record was written by a model.
The abstract, sentence by sentence
In this paper, we study a Riesz-type weighted sum of the Liouville function, $f(x):=-\sum_{n\le x}\frac{\lambda(n)\log n}{\sqrt n}\log\frac{x}{n}$ for sufficiently large $x$. Motivated by a recent research on the sign criteria for the Riemann Hypothesis arising from weighted prime-counting functions, we investigate the sign behavior of $f(x)$ and its relation to the zeros of the Riemann zeta function. We first prove that if $f(x)$ is non-negative for all sufficiently large $x$, then the Riemann Hypothesis holds. The proof is based on the Mellin transform of $f$ and the analytic properties of $\frac{\zeta(2s)}{\zeta(s)}$. Conversely, assuming the Riemann Hypothesis, the Simple Zero Conjecture, and an absolute convergence condition involving the nontrivial zeta zeros, we derive an explicit formula for $f(x)$ in terms of these zeros. In particular, we show that $f(x)\sim \frac{(\log x)^3}{12|\zeta(\frac{1}{2})|}$ as $x\to \infty$. Consequently, we show that under these hypotheses, $f(x)$ is eventually positive.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | In this paper, we study a Riesz-type weighted sum of the Liouville function, $f(x):=-\sum_{n\le x}\frac{\lambda(n)\log n}{\sqrt n}\log\frac{x}{n}$ for sufficiently large $x$. | contribution 0.98 | 0.78 | 0.03 | asserted 0.77 |
| s1 | Motivated by a recent research on the sign criteria for the Riemann Hypothesis arising from weighted prime-counting functions, we investigate the sign behavior of $f(x)$ and its relation to the zeros of the Riemann zeta function. | background 0.67 | 0.58 | 0.04 | asserted 0.60 |
| s2 | We first prove that if $f(x)$ is non-negative for all sufficiently large $x$, then the Riemann Hypothesis holds. | result 0.99 | 0.92 | 0.37 | asserted 1.00 |
| s3 | The proof is based on the Mellin transform of $f$ and the analytic properties of $\frac{\zeta(2s)}{\zeta(s)}$. | method 1.00 | 0.82 | 0.04 | asserted 0.99 |
| s4 | Conversely, assuming the Riemann Hypothesis, the Simple Zero Conjecture, and an absolute convergence condition involving the nontrivial zeta zeros, we derive an explicit formula for $f(x)$ in terms of these zeros. | result 0.69 | 0.95 | 0.11 | asserted 1.00 |
| s5 | In particular, we show that $f(x)\sim \frac{(\log x)^3}{12|\zeta(\frac{1}{2})|}$ as $x\to \infty$. | result 1.00 | 0.93 | 0.21 | asserted 1.00 |
| s6 | Consequently, we show that under these hypotheses, $f(x)$ is eventually positive. | result 0.96 | 0.94 | 0.36 | asserted 1.00 |
What was examined, and what was not
- arXiv listing abstract v1
- full text
- figures and tables
- references and citing sentences
- relations to other papers
- P abstract
- S sections
- T body sentences
- C citations
- R cross-paper relations
References
How the literature uses it
What this paper does with the arXiv papers it cites
Discussion
Disagree with the review, or know something it missed? Leave a comment, highlight any passage above to comment on it, or challenge this review formally.