Generalized Riemann Hypothesis for Dirichlet L-functions

Generalized Riemann Hypothesis for Dirichlet L-functions

OpenAI's claim

OpenAI presents the following claims relevant to this question, with manuscript numbers following paper order in Sources. In manuscript 1 (family 003; 2026-10-05), the asserted result is that zeta, Dirichlet functions, and finite-order Eisenstein-field Hecke families admit a zero-free region with boundary $11/12$; In manuscript 2 (family 003; 2026-09-30), the asserted result is that zeta, Dirichlet $L$-functions, and finite-order Hecke $L$-functions for $\mathbb Q(\sqrt{-3})$ have no zeros with real part greater than $7/8$, with poles excluded; In manuscript 3 (family 003; 2026-10-01), the asserted result is that an absolute positive constant bounds $(1-\beta)\log q$ away from zero for real zeros of primitive nonprincipal real Dirichlet characters of conductor $q\ge3$.

Mathematical statement of the claimed result

Manuscript 1.

For $K=\mathbb Q(\sqrt{-3})$ and every finite-order Hecke character $\chi$ over $K$, the claimed zero-free assertion, with meromorphic poles excluded from evaluation, is \[ \operatorname{Re}s>11/12\quad\Longrightarrow\quad L_K(s,\chi)\ne0. \] It includes every Dirichlet $L$-function and the Riemann zeta function. The accompanying prime-progression claim uses $\pi(x;q,a)$ and asserts, with an absolute effective constant, \[ \sup_{\substack{1\le q\le x\$a,q)=1}} \left|\pi(x;q,a)-\frac1{\varphi(q)}\int_2^x\frac{dt}{\log t}\right| \ll x^{11/12}\log x\quad(x\ge2). \] The critical-line requirement $\operatorname{Re}\rho=1/2$ is not a consequence of this half-plane claim.

Manuscript 2.

For every finite-order Hecke character of $F=\mathbb Q(\sqrt{-3})$, every Dirichlet character and the zeta function, the principal claim is \[ \operatorname{Re}s>7/8\quad\Longrightarrow\quad L(s)\ne0, \] where a principal character's pole at $s=1$ is allowed. The weaker displayed $11/12$ theorem follows within this scope. The cited arithmetic consequence gives absolute $A,C>0$ such that every odd prime $p$ has least positive quadratic nonresidue \[ n(p)\le C(\log p)^A. \] Thus $n(p)\ll_{\delta} p^{\delta}$ for each fixed $\delta>0$. Given binary $p$ and $a\in\mathbb F_p$, a deterministic algorithm of time polynomial in $\log p$ is further claimed to return a square root of $a$ or correctly declare it nonexistent. The half-plane assertion leaves zeros between the critical line and $7/8$ untreated.

Manuscript 3.

The manuscript's Theorem 1 asserts an absolute positive constant, uniform over all real primitive characters: \[ \exists c>0\ \forall q\ge3\ \forall\chi\text{ primitive, real, nonprincipal of conductor }q \ \forall\beta\in(0,1):\quad L(\beta,\chi)=0 \ \Longrightarrow\ (1-\beta)\log q\ge c. \] Thus these real zeros cannot approach one faster than a constant multiple of $1/\log q$. The assertion concerns real Dirichlet zeros; it does not place all nonreal zeros on the critical line.

Outline of the argument

For manuscript 1 (family 003; 2026-10-05), twisted Möbius sums are enlarged to a family before Poisson summation. A recursive estimate for the dual mean square, followed by inversion of cube factors, supplies the saving used to exclude a reciprocal-function singularity.

For manuscript 2 (family 003; 2026-09-30), smoothed character sums are estimated in two ways: directly by large-sieve and recursive moment bounds, and through a Mellin transform involving the reciprocal of an $L$-function. Poisson summation and sextic residue characters align the two expressions so that their savings force continuation of the reciprocal.

For manuscript 3 (family 003; 2026-10-01), a hypothetical exceptional zero forces a pronounced prime bias. Interpolation on a projected integer box produces a weighted conjugate determinant whose Archimedean size conflicts with its divisibility at primes selected by that bias.

Formal verification

For manuscript 2 (family 003; 2026-09-30), OpenAI lists Lean statements for this assertion: zeta, Dirichlet $L$-functions, and finite-order Hecke $L$-functions for $\mathbb Q(\sqrt{-3})$ have no zeros with real part greater than $7/8$, with poles excluded.

This coverage does not establish a formal proof of the complete problem as stated here.

For manuscript 3 (family 003; 2026-10-01), OpenAI lists Lean statements for this assertion: an absolute positive constant bounds $(1-\beta)\log q$ away from zero for real zeros of primitive nonprincipal real Dirichlet characters of conductor $q\ge3$.

This coverage does not establish a formal proof of the complete problem as stated here.

No Lean formalization is listed at this snapshot for manuscript 1 (family 003; 2026-10-05).

The linked family Comparator checks are DirichletSevenEighths.lean, HeckeSevenEighths.lean, QuasiRiemannHypothesis.lean, SiegelZeros.lean.

Their named declarations include OAI.DirichletCharacter.LFunction_ne_zero_of_seven_eighths_lt_re, OAI.SevenEighths.HeckeFamily.LFunction_ne_zero_of_seven_eighths_lt_re, OAI.SiegelZeros.WeightedTorusJets.dirichletRealZeroBound_proof, OAI.SiegelZeros.WeightedTorusJets.exists_absolute_real_zero_gap, OAI.riemannZeta_ne_zero_of_seven_eighths_lt_re.

This site has not executed the Lean code. Described statement coverage is not an independent proof check.

Status caveat

The manuscripts were produced by an internal OpenAI model. OpenAI cautions that results without formalization can contain mistakes. This site has not independently checked the proofs; the label records the scope of OpenAI's claim.

Sources

• OpenAI, The Quasi-Riemann Hypothesis, 2026-10-05; citation key OAI:The-Quasi-Riemann-Hypothesis-October-5-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main.
• OpenAI, The Quasi-Riemann Hypothesis: A Zero-Free Half-Plane Re (s)>7/8, 2026-09-30; citation key OAI:The-Quasi-Riemann-Hypothesis-September-30-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main.
• OpenAI, Uniform exclusion of Landau–Siegel zeros, 2026-10-01; citation key OAI:Uniform-exclusion-of-Landau-Siegel-zeros-October-1-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main, eq:main.
• Family 003: Lean scope.
• Comparator DirichletSevenEighths.json.
• Comparator DirichletSevenEighths.lean.
• Comparator HeckeSevenEighths.json.
• Comparator HeckeSevenEighths.lean.
• Comparator QuasiRiemannHypothesis.json.
• Comparator QuasiRiemannHypothesis.lean.
• Comparator SiegelZeros.json.
• Comparator SiegelZeros.lean.
• OpenAI catalogue entry.

Generalized Riemann Hypothesis for Dirichlet L-functions

Definitions and mathematical statement

A Dirichlet character modulo $q$ is a homomorphism $({\mathbb{Z}}/q{\mathbb{Z}})^\times\to{\mathbb{C}}^\times$, extended periodically to integers and by zero on nonunits. Initially, for $\Re s>1$, \[ L(s,\chi)=\sum_{n\ge1}\frac{\chi(n)}{n^s}=\prod_p(1-\chi(p)p^{-s})^{-1}. \] Take the meromorphic continuation. For every $q\ge1$ and every character, ask whether \[ L(\rho,\chi)=0,\quad0<\Re\rho<1\quad\Longrightarrow\quad\Re\rho=\tfrac12. \] Principal and imprimitive characters remain in scope. The restriction to the open strip excludes poles and zeros on its boundary caused by removed Euler factors.

Formulation and concept references: [S1].

Short English statement

Do all Dirichlet $L$-functions place their zeros inside the critical strip on its central vertical line?

Research attempt

Scope, literature check, and chosen attack.

This attempt retains every modulus, including the principal character modulo one. It therefore contains the ordinary Riemann Hypothesis as a special case. The analytic continuation, primitive-character functional equation, and imprimitive factorization used below are precisely the standard formulas in [S1], equations 25.15.2, 25.15.4, and 25.15.5. An imprimitive character induced by $\chi^*$ satisfies \[ L(s,\chi)=L(s,\chi^*)\prod_{p\mid q}(1-\chi^*(p)p^{-s}). \] Every nonconstant additional factor has its zeros on $\Re s=0$, since $|\chi^*(p)|=1$. Thus reduction to primitive characters loses no zero in the open strip. Principal characters reduce to $\zeta$ in the same way.

Reciprocal Dirichlet series provide a concrete arithmetic route to excluding zeros. Garg--Maji [E1], Theorem 2.2, develop established smoothed Möbius-sum criteria for GRH. The present calculation uses a different, unsmoothed weighted energy and proves only the sufficient implication it needs. It does not claim a new equivalence theorem. The catalogue's further references remain recorded, but no claimed resolution in [S3] is assumed. The immediate objective is to turn the assertion into an explicit cancellation estimate whose failure or proof can be investigated without starting from a presumed list of critical-line zeros.

Reciprocal coefficients and a rigorous energy criterion.

Write $\mu(n)=0$ when $n$ has a squared prime factor, and $\mu(n)=(-1)^r$ when $n$ is a product of $r$ distinct primes; $\mu(1)=1$. For a fixed character set \[ M_\chi(x)=\sum_{n\le x}\mu(n)\chi(n),\qquad \mathcal E_\chi(\sigma)=\int_1^\infty |M_\chi(x)|^2x^{-2\sigma-1}\,dx. \] The proposed missing estimate is \begin{equation}\label{wsref-0-72-61-3a-39-3a-65-6e-65-72-67-79} \mathcal E_\chi(\sigma)<\infty \quad\hbox{for every character $\chi$ and every $\sigma>1/2$.} \end{equation} Constants and the values of these integrals may depend on $q,\chi,\sigma$. No bound uniform in the conductor is required by this particular criterion.

Lemma 9.1. If $\mathcal E_\chi(\sigma)<\infty$ for one real $\sigma>0$, then $L(s,\chi)$ has no zero in $\Re s>\sigma$.

Proof. For $\Re s>1$, absolute convergence of the Euler product gives \[ \frac1{L(s,\chi)}=\sum_{n\ge1}\frac{\mu(n)\chi(n)}{n^s} =s\int_1^\infty M_\chi(x)x^{-s-1}\,dx. \] The last identity follows by summing $s\int_n^\infty x^{-s-1}\,dx=n^{-s}$, with absolute convergence. For $u=\Re s>\sigma$, Cauchy--Schwarz gives \begin{equation}\label{wsref-0-72-61-3a-39-3a-74-61-69-6c} \left|s\int_X^\infty M_\chi(x)x^{-s-1}\,dx\right| \le \frac{|s|X^{\sigma-u}}{\sqrt{2(u-\sigma)}} \left(\int_X^\infty |M_\chi(x)|^2x^{-2\sigma-1}\,dx\right)^{1/2}. \end{equation} Consequently the integral defines a holomorphic function $F_\chi$ on that half-plane: on each compact subset the right side tends uniformly to zero, so the truncated holomorphic integrals converge locally uniformly. The meromorphic identity $F_\chi(s)L(s,\chi)=1$ holds first to the right of both $1$ and $\sigma$, and then throughout the connected half-plane by the identity theorem. A zero of $L$ there would make the product zero, contradicting this identity. For a principal character, the pole at $1$ is harmless: the identity instead forces $F_\chi(1)=0$ when $1$ lies in the domain. No division by $L$ near an unknown zero was used. $\square$

If (ra:9:energy) holds, Lemma 9.1 excludes every strip zero with real part greater than $1/2$. For a primitive character the functional equation reflects a strip zero to one of the conjugate character across $1/2$; its gamma factors are finite and nonzero inside the strip. The family in (ra:9:energy) includes that conjugate character. Thus a zero to the left would give a forbidden zero to the right. Together with the imprimitive reduction, this proves that (ra:9:energy) suffices for the full catalogue assertion. Conversely, an off-line zero with real part $\beta>1/2$ forces $\mathcal E_\chi(\sigma)=\infty$ for every $1/2<\sigma<\beta$. This contrapositive is a useful diagnostic, but no converse from GRH to this energy condition is claimed here.

An arithmetic form of the missing cancellation.

For a reduced residue $a$ modulo $q$ define the integer \[ M_{q,a}(x)=\sum_{\substack{n\le x\\n\equiv a\ (q)}}\mu(n), \qquad V_q(x)=\sum_{a\in(\mathbb Z/q\mathbb Z)^\times}M_{q,a}(x)^2. \] Character orthogonality, applied to the finite sums, gives exactly \begin{equation}\label{wsref-0-72-61-3a-39-3a-70-61-72-73-65-76-61-6c} \sum_{\chi\bmod q}|M_\chi(x)|^2=\varphi(q)V_q(x). \end{equation} Indeed, after expanding the absolute square, the inner sum $\sum_\chi\chi(m)\overline{\chi(n)}$ is $\varphi(q)$ if $m,n$ are units with $m\equiv n\pmod q$, and zero otherwise. This also proves the identity for complex characters; replacing an absolute square by an ordinary square would not work.

A specific sufficient target is therefore \begin{equation}\label{wsref-0-72-61-3a-39-3a-76-61-72-69-61-6e-63-65} \forall q\ge1\ \forall\eta>0\ \exists C_{q,\eta}<\infty: \qquad V_q(x)\le C_{q,\eta}x^{1+\eta}\quad(x\ge1). \end{equation} For $\sigma=1/2+\delta$, take $\eta=\delta$. Each term on the left of (ra:9:parseval) is nonnegative, so its integral is bounded by a constant times $\int_1^\infty x^{-1-\delta}\,dx<\infty$. This proves (ra:9:energy). The estimate is required separately for every fixed $q$; a statement for most moduli, or for moduli growing with $x$, would not automatically supply it.

Expanding $V_q$ exposes the problem rather than resolving it: \begin{align*} V_q(x)&=D_q(x)+C_q(x),\\ D_q(x)&=\sum_{\substack{n\le x\$n,q)=1}}\mu(n)^2\le x,\\ C_q(x)&=2\sum_{\substack{1\le m<n\le x\\m\equiv n\ (q)\$mn,q)=1}} \mu(m)\mu(n). \end{align*} The diagonal already has the correct scale. The unresolved part is a signed off-diagonal estimate $C_q(x)\ll_{q,\eta}x^{1+\eta}$. Bounding every product by its absolute value gives only an $O_q(x^2)$ estimate. Positivity of $V_q$ supplies a lower bound $C_q\ge-D_q$, not the needed upper bound. Thus the sum-of-squares reformulation does not create the missing cancellation for free.

There is also no easy escape through averaging characters. If $\chi_0$ is principal, then \[ V_q(x)\ge\frac1{\varphi(q)} \left|\sum_{\substack{n\le x\$n,q)=1}}\mu(n)\right|^2. \] This is Cauchy--Schwarz on the residue sums. At $q=1$ it is an equality and the problem becomes the ordinary Mertens cancellation problem. A large family average cannot justify deleting the principal character or an exceptional member while retaining the stated target.

Stress tests of two tempting shortcuts.

For a nonprincipal $\chi$, the untwisted sums $\sum_{n\le x}\chi(n)$ are bounded by $q$, because complete periods sum to zero. This proves analytic continuation of the series for $L(s,\chi)$ into $\Re s>0$ by partial summation. It says nothing comparable about its reciprocal: $\mu(n)\chi(n)$ is not periodic. For instance, for the real character modulo four, $\mu(1)\chi(1)=1$ whereas $\mu(5)\chi(5)=-1$, despite $1\equiv5\pmod4$. Confusing these two coefficient sequences replaces zero-freeness by holomorphy, a strictly different assertion.

A second shortcut is to extrapolate an energy estimate from finite partial sums. For any observed prefix of a bounded sequence, extend its unseen tail first by alternating $+1,-1$, and alternatively by all $+1$. The first extension has bounded partial sums and finite energy for every $\sigma>0$; the second has partial sums asymptotic to $x$ and infinite energy for $\sigma\le1$. The two sequences agree on the entire prefix. These are auxiliary sequences, not alternative Möbius functions, but they show exactly why the omitted arithmetic tail needs a theorem. Even testing only dyadic endpoint values is inadequate for general bounded coefficients: on $[2^k,2^{k+1})$ put $+1$ on its first half and $-1$ on its second. Endpoint sums return to zero while interior sums grow linearly with the block length. The integral records those excursions.

Boundary and uniformity audit.

There is a useful concrete imprimitive check. Induce the character modulo four to modulus twelve. Its $L$-function is $L(s,\chi_4)(1+3^{-s})$, since the newly deleted Euler factor corresponds to $\chi_4(3)=-1$. The extra zeros satisfy $\Re s=0$ and $\Im s=(2k+1)\pi/\log3$, for integers $k$. They do not contradict either Lemma 9.1 with $\sigma>1/2$ or the catalogue's explicitly open-strip formulation. Trying to prove that the entire imprimitive function has only critical-line zeros would instead target a false statement. Similarly, allowing a pole of a principal $L$-function at one does not invalidate a criterion for its reciprocal; the reciprocal vanishes there, which is entirely compatible with holomorphy.

The order of quantifiers deserves a separate check. Suppose for each modulus an energy bound were available only for $\sigma>1/2+1/q$. For that fixed modulus the lemma would leave an uncontrolled strip of positive width. Sending $q$ to infinity does not repair the missing statement for the original character. Conversely, constants growing arbitrarily rapidly with $q$ would still suffice if the exponent condition held for every positive $\delta$ and every fixed $q$. This attempt needs neither a conductor average nor a single constant controlling all characters. Finally, (ra:9:tail) is an effective truncation bound only when its unknown tail energy is bounded independently. Computing the finite integral and reusing its value as a bound for the infinite tail would be circular.

Finite verification and reproducibility.

The accompanying standard-library Python script computes $\mu$ by exact prime factorization and checks (ra:9:parseval) at every integer $1\le n\le4096$, for $q=1,3,4,5$. The four characters modulo five include both genuinely complex characters, with values represented by Gaussian integers. The step-function identity \[ \int_1^X V_q(x)x^{-2}\,dx =\sum_{n=1}^{X-1}\frac{V_q(n)}{n(n+1)} \qquad(X\in\mathbb Z,\ X\ge2) \] allows exact rational summation. Rounded display values are \[ \begin{array}{c|rrrr} q\backslash X&64&256&1024&4096\\\hline 1&1.317265&1.407570&1.455461&1.505754\\ 3&2.400019&2.701077&2.870196&2.969291\\ 4&2.223627&2.805885&3.110989&3.272757\\ 5&3.023042&3.432138&3.681850&3.813208 \end{array} \] These are finite integrals at the endpoint exponent $\sigma=1/2$. The argument requires infinite integrals at every strictly larger exponent and asserts no endpoint convergence. The table is a check of the bookkeeping, not evidence establishing convergence or a zero count. The local record is _Local/top_problems/continuation_2026_09_27/analytic/checks.py and its results.json; it contains no unreported zero search.

First gap and outcome.

The analytic implication and the finite Fourier identity are proved here; no simplicity, nonvanishing central value, or conductor-uniform bound is assumed. The first missing assertion is (ra:9:variance), or a weaker estimate sufficient to make its weighted integral finite. Neither the diagonal calculation nor periodic character cancellation bounds the off-diagonal Möbius correlations at the required scale. A useful next attack would have to retain their signs, quantify the tail in (ra:9:tail), and cover each fixed modulus, including one. A disproof along this route would still require a certified off-line zero; divergence at $\sigma=1/2$ alone would not suffice.

Outcome: UNRESOLVED. This notebook gives a fully justified sufficient energy criterion, its exact residue-class reduction, and explicit failures of two proposed shortcuts. The needed arithmetic cancellation has not been established, and no counterexample to GRH has been produced. The process used direct derivation, primary-source checks, and finite exact diagnostics; no external mathematical review is claimed.

Sources

• S1 NIST DLMF25.15, Dirichlet L-functions, equations25.15.1–4 and zeros; GRH convention in Kandhil–Languasco–Moree, Mathematische Annalen394:43$2026$. https://dlmf.nist.gov/25.15. Formulation source. Consulted 24 September 2026: Dirichlet characters, L-functions and zeros.
• S2 Large values of quadratic character sums — Dong, Wang, Wang, Zhang and Zhao. https://arxiv.org/abs/2608.15773. Further reference; not a proof endorsement.
• S3 On the Extended, Generalized, and Grand Riemann Hypotheses: A Unified Approach Based on the General Properties of L-Functions — Weicun Zhang, v19. https://www.preprints.org/manuscript/202506.0481. Further reference; not a proof endorsement.

• E1 Meghali Garg and Bibekananda Maji, Hardy--Littlewood--Riesz type equivalent criteria for the Generalized Riemann hypothesis, arXiv:2208.07596v1 (16 August 2022), Theorem 2.2. https://arxiv.org/pdf/2208.07596v1. Primary research context for smoothed reciprocal-coefficient criteria. Consulted 27 September 2026; not used as a premise of Lemma 9.1.

Generalized Riemann Hypothesis for Dirichlet L-functions

Definitions and mathematical statement

A Dirichlet character modulo $q$ is a homomorphism $({\mathbb{Z}}/q{\mathbb{Z}})^\times\to{\mathbb{C}}^\times$, extended periodically to integers and by zero on nonunits. Initially, for $\Re s>1$, \[ L(s,\chi)=\sum_{n\ge1}\frac{\chi(n)}{n^s}=\prod_p(1-\chi(p)p^{-s})^{-1}. \] Take the meromorphic continuation. For every $q\ge1$ and every character, ask whether \[ L(\rho,\chi)=0,\quad0<\Re\rho<1\quad\Longrightarrow\quad\Re\rho=\tfrac12. \] Principal and imprimitive characters remain in scope. The restriction to the open strip excludes poles and zeros on its boundary caused by removed Euler factors.

Formulation and concept references: [S1].

Short English statement

Do all Dirichlet $L$-functions place their zeros inside the critical strip on its central vertical line?

Research attempt

Current frontier.

Research check: 27 September 2026. The primitive functional equation, induced-character Euler factors and open-strip zero conventions are recorded in DLMF Section 25.15. [S1]. The August 2026 paper on large quadratic character sums assumes GRH for the Omega results described in its abstract; it cannot be used as an unconditional proof of the assumption. [S2]. We pursue cancellation in the coefficients of $1/L$, rather than positivity of a product of $L$-functions. The integral criterion below is a sufficient condition whose implication is proved here; no new unconditional cancellation estimate is claimed.

Attack route.

A spectral model would need a self-adjoint operator with the actual completed $L$-function as its determinant. A direct character-sum argument controls $L$ but not necessarily its reciprocal. The chosen route is to seek a mean-square estimate for Mobius-twisted sums; it would give a holomorphic reciprocal across the right half of the critical strip. The missing lemma is quantitative cancellation in those sums, with careful attention to the principal and imprimitive cases.

Attempt: reduce to primitive characters without losing cases.

Let $\chi\bmod q$ be induced by a primitive character $\chi^*$ of conductor $q^*\mid q$. Euler products in $\Re s>1$ give \[ L(s,\chi)=L(s,\chi^*) \prod_{\substack{p\mid q\\p\nmid q^*}} (1-\chi^*(p)p^{-s}). \tag{9.1} \] Meromorphic continuation preserves this finite-product identity. For a prime in the product, $|\chi^*(p)|=1$, so a zero of the Euler factor satisfies $p^{-\Re s}=1$, hence $\Re s=0$. Therefore these factors create and cancel no open-strip zeros. The principal character reduces to the primitive character of conductor one, whose $L$-function is $\zeta(s)$; its pole at $s=1$ is not a zero. This is exactly the induced-character convention in [S1], equations 25.15.1--25.15.5. Proving the assertion for primitive characters, including conductor one, suffices for every character in the entry.

Attempt: an $L^2$ criterion for a holomorphic reciprocal.

For a fixed primitive $\chi$ set \[ M_\chi(x)=\sum_{1\leq n\leq x}\mu(n)\chi(n), \] where $\mu(n)=0$ when a square prime factor divides $n$, and $\mu(n)=(-1)^r$ when $n$ is a product of $r$ distinct primes. Consider the following proposed estimate, separately for every $\chi$: \[ \text{for every }\varepsilon>0,\qquad I_{\chi,\varepsilon}:= \int_1^\infty |M_\chi(x)|^2x^{-2-\varepsilon}\,dx<\infty. \tag{9.2} \] Constants may depend on $\chi$ and $\varepsilon$. A uniform bound over all conductors is not added to this sufficient condition.

Assuming $9.2$, define \[ H_\chi(s)=s\int_1^\infty M_\chi(x)x^{-s-1}\,dx \quad(\Re s>1/2). \tag{9.3} \] To justify this definition, write $\sigma=\Re s$ and choose $0<\varepsilon<2\sigma-1$. Cauchy--Schwarz gives \[ \int_1^\infty |M_\chi(x)|x^{-\sigma-1}\,dx \leq I_{\chi,\varepsilon}^{1/2} \left(\int_1^\infty x^{-2\sigma+\varepsilon}\,dx\right)^{1/2} =\frac{I_{\chi,\varepsilon}^{1/2}}{ \sqrt{2\sigma-\varepsilon-1}}. \tag{9.4} \] For a compact subset of the half-plane, choose a single $\varepsilon$ smaller than twice its distance from the boundary. The same argument, with $(\log x)^{2j}$ in the second integral, bounds every derivative of the integrand. Consequently $9.3$ is holomorphic, by locally uniform convergence and differentiation under the integral. There is no unjustified interchange at $\Re s=1/2$; that boundary is not included.

In $\Re s>1$, the absolutely convergent Euler products give \[ H_\chi(s)=\sum_{n\geq1}\frac{\mu(n)\chi(n)}{n^s} =\frac1{L(s,\chi)}. \tag{9.5} \] The first equality is summation by parts; $M_\chi(x)=O(x)$ is enough for its endpoint term to vanish there. Since $L$ has the continuation specified in the original statement, meromorphic identity extends $L(s,\chi)H_\chi(s)=1$ to $\Re s>1/2$. A zero of $L$ at a point where it is holomorphic contradicts this identity. At $s=1$ for conductor one, $H_\chi$ instead has a zero cancelling the pole, which is harmless. The primitive functional equation relates a zero $\rho$ to a zero $1-\rho$ of the conjugate character's completed function. Gamma factors are finite and nonzero in the open strip. Applying the right-half-plane argument to all primitive characters, including their conjugates, therefore excludes left-half-strip zeros as well. [S1]. Together with $9.1$, this proves the original GRH assertion under $9.2$.

Attempt: a dyadic estimate would close the analytic step.

It would suffice to show, for every $\varepsilon>0$ and $X\geq1$, \[ \int_X^{2X}|M_\chi(x)|^2\,dx \leq C_{\chi,\varepsilon}X^{2+\varepsilon/2}. \tag{9.6} \] Indeed the contribution of $[2^j,2^{j+1}]$ to $9.2$ is at most $C_{\chi,\varepsilon}2^{-j\varepsilon/2}$, a convergent geometric series. This mean-square condition allows spikes and is not silently replaced by a pointwise bound. The trivial estimate $|M_\chi(x)|\leq x$ gives a block integral of order $X^3$, which misses the needed exponent by nearly one. The integer-squarefree weight $\mu$ is thus the substantive cancellation issue; the Mellin analysis by itself does not improve it.

Stress test: cancellation of $\chi$ is not cancellation of $\mu\chi$.

For a nonprincipal character modulo $q$, summing over a complete period gives zero. Thus $\sum_{n\leq x}\chi(n)=O(q)$, yielding a direct continuation formula for $L(s,\chi)$ on $\Re s>0$. There is no corresponding periodicity for $\mu(n)\chi(n)$: divisibility by squares and parity of the number of prime factors are not periodic modulo $q$. Replacing $M_\chi$ by the ordinary character sum would prove continuation of $L$, not continuation of its reciprocal.

A small analytic countermodel pinpoints that logical difference. The Dirichlet polynomial \[ D(s)=1-\tfrac32\,2^{-s} \] has bounded coefficient partial sums and is entire, but vanishes at $\beta=\log(3/2)/\log2$, with $1/2<\beta<1$. Its reciprocal has a pole there. This polynomial is not asserted to be a Dirichlet $L$-function; it disproves only the inference from bounded coefficient sums and continuation to zero-freeness. Likewise, agreement of numerical partial sums with square-root growth up to a finite cutoff would not establish $9.2$, whose integral runs to infinity.

Outcome.

Outcome: Conditional reduction. The attempt gives a fully justified sufficient mean-square criterion, with explicit compact convergence, principal-character treatment, and finite-Euler-factor bookkeeping. A proof of $9.2$, or the stronger dyadic bound $9.6$, for every primitive character would settle the exact entry. Neither estimate is established here, and the argument does not claim a converse or historical novelty for the reciprocal-Mellin method. The obstruction is arithmetic cancellation, not an omitted analytic continuation step.

Sources

• S1 NIST DLMF25.15, Dirichlet L-functions, equations25.15.1–4 and zeros; GRH convention in Kandhil–Languasco–Moree, Mathematische Annalen394:43$2026$. https://dlmf.nist.gov/25.15. Catalogue formulation source. Consulted 24 September 2026: Dirichlet characters, L-functions and zeros.
• S2 Large values of quadratic character sums — Dong, Wang, Wang, Zhang and Zhao. https://arxiv.org/abs/2608.15773. Catalogue research/status reference; not a proof endorsement.
• S3 On the Extended, Generalized, and Grand Riemann Hypotheses: A Unified Approach Based on the General Properties of L-Functions — Weicun Zhang, v19. https://www.preprints.org/manuscript/202506.0481. Catalogue research/status reference; not a proof endorsement.