{"id":"31800c54-ed85-46d3-870f-a3d54143acdd","arxiv_id":"2508.21789","paper_version":2,"verdict":"REJECT","confidence":"MODERATE","novelty_score":3.0,"correctness_risk":"high","formal_verification":"none","parameter_count":2,"one_line_summary":"The claimed Salem-type equivalence for the Riemann hypothesis is not established; the proof assumes the hypothesis in the crucial step.","lead":"This paper proposes a new kernel and a Titchmarsh theorem reformulation that would yield a Salem-type equivalence for the Riemann hypothesis. The main theorem is not a valid equivalence as stated, because the key condition is conditional on the Riemann hypothesis and one step of the proof invokes the conclusion.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The proof never establishes the Salem-type converse: Theorem 1.2's (iii) is conditional on RH, and Section 3 admits the one-one property was not applied.","rationale":"Reader's REJECT is correct. I would not rest the rejection solely on the residue constant; the theorem's logical structure is the deeper problem. The reader's weakest_assumption identifies the coefficient error, which can be repaired; my concern identifies the absent converse, which is the advertised point. Even with a corrected kernel, the paper's Theorem 1.2 remains conditional on RH, and Section 3's equivalence is a bare assertion. A good-faith reader could reconstruct the converse via f=e^{-iγx}, but that reconstruction is not in the paper and requires showing h∈L1, so as submitted the central claim is unproved. No issue with the person; the argument itself is incomplete.","tokens_in":5947,"tokens_out":24669,"duration_ms":288886,"concrete_test":"To settle the converse, take an alleged off-critical zero ρ=β+iγ with 1/2<β<1, set σ=β, and test the candidate solution f(x)=e^{-iγ x}. Compute Iσ(z)=∫ f(x)e^{σ(z−x)}k(e^{z−x})dx = e^{−iγ z} times the Fourier transform of e^{σu}k(e^{u}) at −γ, which is a nonzero constant times Γ(β+iγ)ζ(β+iγ)ζ(β+iγ−1/2)=0. If this f is an allowed bounded measurable function and the integral converges absolutely once the (1.4) coefficient is corrected, the Section 3 claim is true but unproved; if the kernel is not in L1 as printed, the Plancherel/Titchmarsh argument fails. Either outcome locates the missing load-bearing step.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The central claimed equivalence is unsupported. Theorem 1.2 lists condition (iii) as 'Ff(t)=0 for t<0, if the Riemann hypothesis is true'; the proof in §2 then shows only that, assuming RH, the multiplier Γ(σ−it)ζ(σ−it)ζ(σ−1/2−it) is nonzero for t<0, so (ii) and (iii) coincide. It does not prove the converse needed for a Salem-type equivalence: that if no nonzero bounded f satisfies Iσ(f)=0, then RH holds. Section 3 concedes 'the one-one property was not applied in Theorem 1.2 to relate f directly to the Riemann hypothesis', and then simply asserts the converse. No construction from an off-critical zero, no tauberian argument, and no distribution argument is provided. This is not a minor omission: without it, Theorem 1.2 is a conditional reformulation of RH, not a new equivalence. A separate arithmetic issue in (1.4)–(2.2) (the residue coefficient at s=3/2 is (√π/2)ζ(3/2), not √(π/2)ζ(3/2)) would also break L2, but it is logically independent of the missing converse.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper claims to extend Salem's equivalence for the Riemann hypothesis by applying Titchmarsh's theorem on Fourier transforms and Hilbert transforms. It introduces a kernel k(x) in (1.4), defines a convolution integral I_sigma(f), and states Theorem 1.2 as a set of alternative necessary and sufficient conditions for a complex-valued function to be a boundary value of an analytic function in the upper half-plane with L2 growth. Theorem 1.2 lists three conditions: (i) Hilbert-transform pairing, (ii) vanishing of the Fourier transform of the modified convolution for t < -m, and (iii) vanishing of Ff(t) for t < 0 'if the Riemann hypothesis is true.' Section 3 then asserts a Salem-type converse and gives a related sufficient condition involving logarithmic integrals. The central claim is that these Fourier-analytic conditions are equivalent to RH.","tokens_in":6297,"tokens_out":2516,"duration_ms":30622,"significance":"If the claimed Salem-type equivalence were rigorously established, it would provide a new Fourier-analytic reformulation of the Riemann hypothesis, connecting RH to support properties of Fourier transforms and Hilbert transforms. This would be a noteworthy contribution. The paper also aims to use Titchmarsh's classical theorem in a new way. However, as written the central equivalence is not proved: the converse direction is merely asserted, and parts of the proof are circular or rely on an incorrect residue computation. The paper contains no machine-checked proofs, reproducible code, or fully parameter-free derivations that would independently support the claims. Its value is therefore contingent on substantial additional work rather than being established in this manuscript.","major_comments":[{"comment":"The claimed Salem-type equivalence is asserted, not proved. The sentence 'the Riemann hypothesis would be true should no bounded measurable function f satisfy ... = 0, other than the trivial case f ≡ 0' is stated without a proof of the converse direction. Theorem 1.2's condition (iii) is explicitly conditional on RH, and the proof in §2 only shows that, assuming RH, the multiplier is nonzero, so that (ii) and (iii) coincide. No construction from a hypothetical off-critical zero of ζ is given, and no Tauberian, distributional, or approximation argument establishes that such a zero would force a nonzero bounded f annihilating the convolution. This missing converse is the core of the paper's claimed equivalence and is not a minor omission.","section":"Section 3"},{"comment":"The derivation of condition (ii) from condition (i) is circular. The text reads: 'To see how condition (ii) follows from (i), observe that by (ii) if x < -m, F Ī_σ,m(x) = ... = 0'. This assumes the very condition that is being proved. The subsequent Parseval/L2-growth argument is a separate attempt, but it is not developed rigorously and contains variable confusion between x and t. A correct proof must derive the vanishing of the Fourier transform for t < -m from the analyticity and growth condition alone, without invoking (ii).","section":"§2, proof of Theorem 1.2, condition (ii)"},{"comment":"The residue subtraction in the definition of k(x) is numerically incorrect. The residue of Γ(s)ζ(s)ζ(s−1/2) at s = 3/2 is (√π/2) ζ(3/2), not √(π/2) ζ(3/2) as printed in (1.4). If this is not a typo, then the singular term is not cancelled and e^{σx}k(e^x) is not in L²(−∞,∞), invalidating the Plancherel argument and the applicability of Titchmarsh's theorem. If it is a typo, it must be corrected, but as written it is a load-bearing numerical error in the kernel construction.","section":"§1.4 and §2.2"},{"comment":"Condition (iii) is not an 'alternative necessary and sufficient condition' in the sense of the theorem. It reads 'Ff(t) = 0 for t < 0, if the Riemann hypothesis is true.' This makes the theorem conditional rather than an equivalence, and it does not establish a Salem-type reformulation. Moreover, no definition is given that relates the vanishing of Ff(t) for t<0 to the absence of bounded solutions of the convolution equation. The theorem as stated is therefore not a coherent equivalence statement.","section":"Theorem 1.2, condition (iii)"}],"minor_comments":[{"comment":"There are frequent typos and notational inconsistencies, e.g., 'Riemman' in the Introduction, and the repeated use of 'sinc(x) := e^{-imx} sin(x)/x = e^{-imx} sinc(x)' which is a tautology. These should be corrected in revision.","section":"Throughout"},{"comment":"In the Plancherel computation, the variable of integration is written inconsistently (sometimes x, sometimes t). For example, the equality involving |Γ(σ−ix)ζ(σ−ix)ζ(σ−ix−1/2)|² should use a consistent Fourier variable. This makes the argument harder to follow and should be fixed.","section":"§2, after Eq. (2.2)"},{"comment":"The integral formula is stated imprecisely. The standard evaluation is ∫_0^∞ x^{ε−1}/(1+x²) dx = π/(2 sin(πε/2)) for 0<ε<2, hence the full integral over R is π/sin(πε/2). The text 'is twice π/ sin(επ/2)' appears to contain a factor error or a typo.","section":"§3, Eq. (3.4)"}],"recommendation":"reject","confidential_remarks":null},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Colleague,\n\nThe quick take: this is a short note that tries to add a Titchmarsh-theorem version of Salem's RH equivalence. The kernel built from zeta(s)zeta(s-1/2) is new, and the idea of applying the shifted Hilbert-transform criterion to it is not something I've seen. But the central claim doesn't hold up. Theorem 1.2 lists three conditions as 'alternative necessary and sufficient', but condition (iii) is itself conditional on RH — so the theorem is not an equivalence. In the proof, condition (ii) is 'proved' from (i) by invoking (ii) on the same line. That's circular.\n\nSection 3 then asserts the Salem-type converse: if no bounded measurable f satisfies the convolution equation, then RH. But the author concedes the one-one property was not applied in Theorem 1.2, and no argument is given to pass from the support condition to a zero of zeta. So the claimed equivalence is exactly the missing piece.\n\nThere is also an arithmetic issue. The residue of Gamma(s)zeta(s)zeta(s-1/2) at s=3/2 is (sqrt(pi)/2) zeta(3/2), not sqrt(pi/2) zeta(3/2). If the constant in (1.4) is not a typo, the pole isn't cancelled and e^{sigma x}k(e^x) is not in L2, which breaks the Plancherel argument.\n\nWhat works: the Mellin-transform setup is standard, the L2 estimate via Stirling is fine, and the literature is cited and engaged. The kernel construction is original. But the load-bearing parts — the equivalence and the L2 claim — fail.\n\nThis is for a reader who wants a compact warning example of how easy it is to repackage RH in Fourier language without adding content. Not for citation. If I were the editor, I'd desk reject: the flaws are evident and the salvage would require rewriting the main theorem and fixing the kernel. I would not spend referee time on it.\n\nRecommendation: reject.","headline":"A short note that recycles Salem's equivalence through a new kernel; the central theorem is conditional, circular, and contains an arithmetic slip, so it should not be accepted as is.","tokens_in":6787,"tokens_out":6941,"would_cite":false,"duration_ms":78083,"reading_group":"maybe","serious_thinker":"no","would_accept_peer_review":false},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["26A33","30B40"],"pacs":[],"model":"deepseek-v4-flash","headline":"The Riemann hypothesis is tied to a Fourier-support condition on a modulated convolution built from a kernel whose Mellin transform is Gamma(s) zeta(s) zeta(s-1/2).","keywords":["Riemann hypothesis","Salem equivalence","Titchmarsh theorem","Fourier integrals","Hilbert transform","Mellin transform","zeta function"],"falsifier":"Check whether e^{sigma x} k(e^x) is square-integrable by evaluating the integral of its square numerically for sigma=0.75 using the series in (1.4); if it diverges with the printed constant, the Plancherel step collapses. Alternatively, compute the Mellin transform of the explicit kernel and see if it equals Gamma(s) zeta(s) zeta(s-1/2) in the strip.","tokens_in":5842,"feed_emoji":"🧮","tokens_out":9140,"duration_ms":86675,"temperature":0.7,"pith_summary":"This paper extends Salem's 1953 equivalence for the Riemann hypothesis by applying Titchmarsh's boundary-value theorem. The author constructs a new kernel from a Dirichlet series so that its Mellin transform is Gamma(s) zeta(s) zeta(s-1/2), and forms a convolution that lies in L2. The main theorem ties the Fourier-support condition of a modulated version of this convolution to the absence of zeros off the critical line: under RH the Fourier transform of the original function must vanish on the negative half-axis, and any off-line zero would give a nontrivial bounded solution. If correct, this gives a new Fourier-analytic reformulation of RH.","feed_headline":"New convolution condition is equivalent to the Riemann hypothesis","feed_subtitle":"A modulated convolution's Fourier support edge would settle the Riemann hypothesis.","key_machinery":"The kernel k(x) in (1.4) is the sum over d_{1/2}(n)e^{-nx} minus the singular terms x^{-3/2} sqrt(pi/2) zeta(3/2) and x^{-1} zeta(1/2), chosen so its Mellin transform is Gamma(s) zeta(s) zeta(s-1/2) in the critical strip. The function e^{sigma x}k(e^x) is square-integrable, which brings Titchmarsh's Theorem 95/96 about boundary values of L2 analytic functions into play; the e^{-imz} factor shifts the Fourier support so the symmetry pairs of off-critical zeros are covered.","core_discovery":"The paper constructs a kernel k(x) from the Dirichlet series sum_{n>=1} d_{1/2}(n)e^{-nx} minus two singular terms, so that its Mellin transform is Gamma(s) zeta(s) zeta(s-1/2). For f in L1, the convolution I_sigma(z) = (f * e^{sigma x} k(e^x))(z) lies in L2, and e^{-imz} I_sigma(z) satisfies Titchmarsh's boundary-value conditions. Theorem 1.2 states that the Fourier-support condition for this modulated convolution (null for t<-m) is equivalent, under RH, to Ff(t)=0 for t<0, and that RH is true iff no bounded measurable f solves the convolution equation identically except f=0.","pith_inferences":["The apparent typo in the residue constant, if corrected, likely yields a family of kernels indexed by r in (0,1) from the divisors d_r(n), each giving a distinct Salem-type equivalence; testing with a simple f (e.g., a Gaussian) could numerically probe the sharpness of the support condition.","The paper leaves open whether the condition F bar-I_{sigma,m}(t)=0 for t<-m alone, without assuming RH, forces Ff(t)=0 for t<0; proving that unconditional implication would give an independent RH criterion not explicitly demonstrated.","The Hilbert-transform formulation suggests RH might be rephrased as a statement about conjugate functions: the real and imaginary parts of the modulated convolution are Hilbert transforms of each other exactly when the zero symmetry of the critical line holds, potentially connecting to other Hilbert-space reformulations of RH."],"forward_implications":["If the paper's equivalence is correct, RH is equivalent to the statement that the convolution equation (f * e^{sigma x} k(e^x))(z)=0 has no bounded measurable solution other than zero.","The Fourier-support condition in Theorem 1.2 provides a concrete test: for RH, any f whose modulated convolution is analytic in the upper half-plane with Fourier support in t<-m must have Ff(t)=0 for t<0.","The asymptotic relation (3.1) shows the support edge m can be read off from the exponential growth rate of the L2 norm of the analytic extension, offering a quantitative handle.","Using Wiener's theorem, the paper concludes that under RH the translates of e^{sigma x}k(e^x) are dense in L1.","The Paley–Wiener integral identity (3.2) together with the logarithmic integrability condition gives a sufficient criterion for RH in terms of decay of the modulated convolution."],"supporting_citations":[{"why":"Supplies the original Salem equivalence that the paper extends.","marker":"[5]"},{"why":"Provides Titchmarsh's Theorems 95 and 96, the Hilbert transform definition, and Parseval's identity used to characterize analytic boundary values.","marker":"[6]"},{"why":"Gives the Mellin transform of the Dirichlet series, the functional equation, and Stirling's estimate used in the L2 argument.","marker":"[7]"},{"why":"Used for Plancherel's theorem, Minkowski's inequality, and the isometric property of the Fourier transform on L2.","marker":"[3]"},{"why":"Supplies the Paley–Wiener theorem used in the integral identity (3.2).","marker":"[4]"},{"why":"Supplies Wiener's Tauberian theorem used to state the closedness of translates under RH.","marker":"[8]"}],"fun_headline_variants":["Convolution condition is a new RH equivalence","Riemann hypothesis rests on a convolution's support","Modulated convolution gives RH a test","No bounded solution means RH is true","Fourier support edge decides Riemann hypothesis"],"cache_read_input_tokens":2688,"weakest_assumption_plain":"The load-bearing premise is that the kernel in (1.4) exactly equals the contour integral after residue subtraction, with a correct constant in the x^{-3/2} term, so that e^{sigma x} k(e^x) is square-integrable and Titchmarsh's Hilbert transform machinery applies; the printed sqrt(pi/2) may be a typo for (sqrt(pi)/2) zeta(3/2).","fun_headline_variants_meta":{"raw":{"variants":["Convolution condition is a new RH equivalence","Riemann hypothesis rests on a convolution's support","Modulated convolution gives RH a test","No bounded solution means RH is true","Fourier support edge decides Riemann hypothesis"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000152,"raw_usage":{"total_tokens":945,"prompt_tokens":554,"completion_tokens":391,"prompt_tokens_details":{"cached_tokens":256},"prompt_cache_hit_tokens":256,"prompt_cache_miss_tokens":298,"completion_tokens_details":{"reasoning_tokens":326}},"tokens_in":298,"tokens_out":391,"duration_ms":5237,"temperature":1.0,"reasoning_tokens":326,"cache_read_input_tokens":256,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-05T13:57:54.437676+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Check whether e^{sigma x} k(e^x) is square-integrable by evaluating the integral of its square numerically for sigma=0.75 using the series in (1.4); if it diverges with the printed constant, the Plancherel step collapses. Alternatively, compute the Mellin transform of the explicit kernel and see if it equals Gamma(s) zeta(s) zeta(s-1/2) in the strip.","supporting_citations":[],"review_version":1}