{"id":"118bb196-c3fe-4e53-b89e-891c70e23040","arxiv_id":"1908.00668","paper_version":2,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"A counterexample proves that the multilinear fractional integral operator Iγ does not map H^1(R) × H^p(R) into H^q(R) for 0 < p ≤ γ^{-1} and 1/q = 1 + 1/p − γ.","lead":"This short paper constructs a counterexample showing that a multilinear fractional integral operator, previously known to send products of Hardy spaces into Lebesgue spaces, does not send them into Hardy spaces in a certain range. The proof works by integrating the operator on explicit test functions and showing the result has nonzero integral.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The counterexample's core is sound, but Lemma 1's second term is printed with the wrong sign; the proof's version is correct, so the paper needs a typo fix rather than a mathematical rethink.","rationale":"The reader's weakest-assumption analysis correctly locates Lemma 1's sign error as the main soft spot. I agree that the printed lemma is inconsistent with its proof and that this inconsistency affects the derivation of Corollary 2. However, I would phrase the consequence slightly differently: taking the printed sign literally changes the constant in the identity to -(α+1)/α, which is still nonzero, so the counterexample's broad strategy would survive after correcting the corollary's constant. The real problem is that the manuscript as written does not let a reader verify the stated identity from the stated lemma. The proof's own computation of II, plus an independent regularization of ∫(|x-s|+|x-t|)^{α-1}dx, confirms the intended coefficient (α−1)/α. The rest of the argument is sound: Theorem 1.1 of [1] gives the L^1 estimate needed for the density approximation, the dense subspace of functions with sufficiently many vanishing moments is available for the required range, and (S2) gives the final contradiction because I(a1,b) is in L^1 with nonvanishing integral. The theorem's range p≤γ^{-1} is consistent with q≤1 and with the proof; the abstract's p≤γ−1 is an additional typo. These are revision-level issues, so I do not move the reader's conditional verdict.","tokens_in":4532,"tokens_out":28832,"duration_ms":287844,"concrete_test":"Recompute the Fourier transform of the interval term: \\mathcal{F}(χ_{(s,t)})(ξ)=∫_s^t e^{-ixξ}dx, and substitute this into the proof's II. Then evaluate lim_{ε→0}[\\hat K_{s,t}(εξ)+2^αΓ(α)cos(απ/2)|εξ|^{-α}] at ξ=0; with the corrected sign it equals ((α−1)/α)|t-s|^α, and with the printed sign it equals -(α+1)/α|t-s|^α. Also verify the explicit double integral for a1=χ_{(-1,0)}-χ_{(0,1)}, a2=a1(·−2) is nonzero for 0<α<1; both constants are nonzero, so the counterexample survives after the sign correction.","verdict_should_be":"UNCHANGED","load_bearing_attack":"Lemma 1 is the engine of Corollary 2. For s<t, K_{s,t}(x)=d^{α-1} on (s,t), so its Fourier transform contains +d^{α-1}∫_s^t e^{-ixξ}dx; the proof's computation of II says exactly this. The printed formula instead has |d|^{α-1}sgn(t-s)∫_t^s e^{-ixξ}dx, which equals -d^{α-1}∫_s^t e^{-ixξ}dx. Evaluating the regularized limit in Corollary 2 at ξ=0, the correct interval term contributes +d^α and the third term -d^α/α, giving ((α−1)/α)d^α; the printed sign would give -d^α - d^α/α = -(α+1)/α d^α. Thus the stated proof of Corollary 2 is not consistent with the printed lemma. This is load-bearing because the nonzero integral in the counterexample comes from this identity. The proof's version is nevertheless correct, so this is a fixable typo. Separately, the abstract's range p≤γ−1 should be p≤γ^{-1} as in Theorem 3; the proof supports the theorem's range. These are typos, not structural failures.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper constructs an explicit counterexample to show that the bilinear fractional integral operator I_γ, for 1 < γ < 2, is not bounded from H^1(R) × H^p(R) into H^q(R) on the range 0 < p ≤ γ^{-1}, 1/q = 1 + 1/p − γ. The proof computes the distributional Fourier transform of the kernel (|x−s|+|x−t|)^{α−1}, uses it to derive the identity ∫_R I_{α+1}(a_1,a_2)(x) dx = ((α−1)/α) ∫∫_{R^2} a_1(s)a_2(t)|t−s|^α dsdt for compactly supported bounded a_1,a_2 with a vanishing moment, and then chooses explicit functions a_1 ∈ H^1 and a_2 ∈ H^{(α+1)^{-1}} for which the double integral is nonzero. The case of smaller p is obtained by a density argument that moves the example from H^{(α+1)^{-1}} to H^p.","tokens_in":4795,"tokens_out":8410,"duration_ms":83973,"significance":"If the corrected version of Lemma 1 and the identity in Corollary 2 are accepted, the paper resolves a natural question: while L^q estimates for multilinear fractional integrals were proved by Lin–Lu and Cruz-Uribe–Moen–van Nguyen, the corresponding Hardy-space target H^q fails, and the explicit counterexample demonstrates the optimality of the stated range. The argument is elementary and fully explicit, with no fitted parameters, and the nonzero double integral is computed directly; the negative result follows from an external boundedness theorem without assuming the target unboundedness. The main weaknesses are several typographical errors and a missing justification of a limit interchange, all of which are local and fixable.","major_comments":[{"comment":"The displayed statement of Lemma 1 contains a sign error in the second term: for s < t it reads |t−s|^{α−1} sgn(t−s) ∫_t^s e^{-ixξ} dx, but the proof computes II = (t−s)^{α−1} ∫_s^t e^{-ixξ} dx. Since ∫_t^s = −∫_s^t, the printed formula has the opposite sign. The proof's version is correct, but as printed the statement is inconsistent with its proof and would change the coefficient in Corollary 2 to −(α+1)/α instead of (α−1)/α. This must be fixed before the paper can be verified.","section":"Lemma 1 (Section 2)"},{"comment":"The third equality in the displayed chain in Corollary 2 passes the limit ε → 0+ inside the double integral over s and t, after a change of variables in the inner distributional pairing. This interchange is not justified. The authors should state the regularized identity explicitly and give a dominated-convergence argument, showing that the ε^{-α}|ξ|^{-α} singular term is controlled by the moment condition on a_1 (or a_2) and that the remaining integrands converge uniformly on the compact supports of a_1 and a_2.","section":"Corollary 2 proof (Section 2)"},{"comment":"The abstract states 0 < p ≤ γ−1, but the theorem and the proof require 0 < p ≤ γ^{-1} (since γ = α+1, the two ranges are not equivalent: for 1<γ<2, γ^{-1}<1 while γ−1 is in (0,1) and is larger than γ^{-1}). The abstract should read γ^{-1} to match Theorem 3 and the argument in Section 3.","section":"Abstract and Theorem 3"}],"minor_comments":[{"comment":"The notation \\widehat{φ_ε}(ξ) is not defined at first use; under the Fourier convention \\widehat{f}(ξ)=∫ f(x)e^{-ixξ}dx, one has \\widehat{φ_ε}(ξ)=ε^{-1}\\widehat{φ}(ξ/ε), and this change of variables should be displayed explicitly.","section":"Section 2, proof of Lemma 1 and Corollary 2"},{"comment":"The sentence beginning \"For 0 < p < (α + 1)−1\" should state explicitly that the approximating function b is chosen from the dense class of bounded, compactly supported functions with vanishing moments up to N while approximating a_2 in H^{(α+1)^{-1}}; the inclusion of that class in H^p then gives b ∈ H^p.","section":"Section 3, density argument"},{"comment":"The numerator 4·3^{α+2} − 4^{α+2} − 6·2^{α+2} + 4 is asserted to be nonzero without proof; since this is a nontrivial claim controlling the whole counterexample, a brief justification (for instance, evaluating at α=0 and α=1 or showing monotonicity) would be helpful.","section":"Section 3, computation of the double integral"}],"recommendation":"minor_revision","confidential_remarks":"The manuscript is short and its main mathematical idea is sound after correcting the printed sign error in Lemma 1. The referee sees no circularity or fitted-parameter issue, and the counterexample is genuinely explicit. The required changes are local: fix the lemma statement, correct the abstract's range, and add a dominated-convergence justification in Corollary 2. These should not require new mathematics."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Short version: the counterexample works, and the paper deserves a referee, but not in its current printed form. The two things everyone will trip on — the sign error in Lemma 1's statement and the abstract's range p ≤ γ−1 — are real, and both are typos rather than structural cracks. Lemma 1's proof computes the interval term with the opposite sign from the printed formula, and the proof's version is the correct one. A reader who follows the proof gets Corollary 2's identity with the right coefficient; a reader who trusts the printed lemma gets a coefficient that would kill the counterexample. The abstract should read p ≤ γ^{−1}, matching the proof.\n\nWhat the paper does genuinely well: the natural question after Lin-Lu and Cruz-Uribe-Moen-van Nguyen is whether the L^q (weighted Hardy) bounds upgrade to H^q in the range p ≤ 1/γ, where q ≤ 1. They do not, and this note shows it by computing the distributional Fourier transform of the kernel in x, extracting the zero-frequency contribution, and getting ∫ I_{α+1}(a1,a2) dx = ((α−1)/α) ∫∫ a1(s)a2(t)|t−s|^α dsdt under a moment condition. The explicit a1, a2 and the stated nonzero value of the double integral check out (I spot-checked numerically; it is positive inside (0,1), vanishing only at the endpoints). The density argument from the endpoint p = 1/γ down to all p < 1/γ is standard and goes through. No fitting, no circularity, and the citations — Gelfand-Shilov for the |x|^{α−1} transform, Stein for Hardy-space facts, the two positive-result papers — are appropriate. The kernel identity is a genuinely reusable tool for endpoint questions of this type.\n\nSoft spots in order of size. One: the printed Lemma 1 sign error is load-bearing as printed but repaired by the paper's own proof; a referee should insist the statement match the proof. Two: the abstract range. Three, which neither the reader nor the stress-test flagged: Corollary 2's proof has two canceling Fourier-convention slips. The displayed equality ∫ K φ_ε dx = ∫ K̂ φ̂_ε dξ is off by 1/2π (the pairing identity runs the other way), and the limit is evaluated as if ∫ φ̂ = φ(0) = 1, whereas ∫ φ̂ = 2π. The two errors cancel, so the stated identity is correct, but the proof as written is not self-consistent on constants.\n\nWho this is for: specialists in multilinear fractional integrals and Hardy spaces. It settles a boundary case in a short note; it is not a breakthrough and opens no new line. With the fixes above I would publish it in a mainstream harmonic analysis journal and would cite it for the endpoint failure. Recommendation: send it to peer review, with acceptance conditional on the typo and constants fixes.","headline":"The counterexample is right and the note is publishable after fixes; the printed sign error in Lemma 1 and the abstract's range typo are load-bearing but repairable from the paper's own proof.","tokens_in":5202,"tokens_out":36648,"would_cite":true,"duration_ms":303314,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["42B20","42B30"],"pacs":[],"model":"deepseek-v4-flash","headline":"A counterexample shows that the bilinear fractional integral operator $I_\\gamma$ is not bounded from $H^1(\\mathbb{R})\\times H^p(\\mathbb{R})$ into $H^q(\\mathbb{R})$ when $0<p\\le 1/\\gamma$ and $1/q=1+1/p-\\gamma$.","keywords":["multilinear fractional integral","Hardy spaces","counterexample","bilinear fractional integral operator","moment conditions","distributional Fourier transform","unboundedness"],"falsifier":"Numerically evaluate the identity of Corollary 2 for the paper's explicit functions $a_1(s)=\\chi_{(-1,0)}(s)-\\chi_{(0,1)}(s)$ and $a_2(t)=a_1(t-2)$ at $\\alpha=1/2$: the formula predicts a nonzero value, so a direct quadrature of $\\int\\int a_1(s)a_2(t)|t-s|^{1/2}\\,ds\\,dt$ should match the closed form; if it instead matches the sign-altered version of Lemma 1, the counterexample collapses.","tokens_in":4360,"feed_emoji":"","tokens_out":11978,"duration_ms":107930,"temperature":0.7,"pith_summary":"This paper constructs a counterexample to show that the bilinear fractional integral operator $I_\\gamma$, defined by $I_\\gamma(f_1,f_2)(x)=\\int\\int f_1(s)f_2(t)(|x-s|+|x-t|)^{\\gamma-2}\\,ds\\,dt$, is not bounded from $H^1(\\mathbb{R})\\times H^p(\\mathbb{R})$ into $H^q(\\mathbb{R})$ when $1<\\gamma<2$, $0<p\\le 1/\\gamma$, and $1/q=1+1/p-\\gamma$. Earlier theorems gave estimates from products of Hardy spaces into Lebesgue spaces, so the nature of the target space was the open question. The proof computes the integral of the output exactly and exhibits test functions for which that integral is nonzero. Since an integrable function in a Hardy space $H^q$, $q\\le1$, must have zero integral, the nonzero value forces the output outside $H^q$. The result settles that this operator has no Hardy-space-to-Hardy-space boundedness in the stated range.","feed_headline":"Counterexample: no Hardy-space bound for bilinear fractional integrals","feed_subtitle":"For 0 < p ≤ 1/γ, the operator's output integral is nonzero, so it cannot land in the target Hardy space.","key_machinery":"The carrying mechanism is the distributional Fourier transform, in the $x$ variable, of the kernel $K_{s,t}^{\\alpha}(x)=(|x-s|+|x-t|)^{\\alpha-1}$. Lemma 1 computes this transform; feeding it through a limiting argument with a Schwartz function and using the moment condition of one input yields Corollary 2's identity, which reduces the integral of $I_{\\alpha+1}(a_1,a_2)$ to a constant times $|t-s|^\\alpha$ weighted by $a_1(s)a_2(t)$. This identity is what turns the abstract question of Hardy-space membership into the concrete question of whether that weighted double integral can be nonzero. The explicit test functions are chosen precisely so that the double integral is nonzero.","core_discovery":"For $\\gamma=\\alpha+1$ with $0<\\alpha<1$, the paper's central claim is enforced by an exact identity: whenever $a_1,a_2$ are bounded and compactly supported and at least one has integral zero, $\\int_{\\mathbb{R}} I_{\\alpha+1}(a_1,a_2)(x)\\,dx = \\frac{\\alpha-1}{\\alpha}\\int_{\\mathbb{R}^2} a_1(s)a_2(t)|t-s|^\\alpha\\,ds\\,dt$. With the explicit pair $a_1(s)=\\chi_{(-1,0)}(s)-\\chi_{(0,1)}(s)$ and $a_2(t)=a_1(t-2)$, the double integral equals $[4\\cdot 3^{\\alpha+2}-4^{\\alpha+2}-6\\cdot 2^{\\alpha+2}+4]/((\\alpha+1)(\\alpha+2))$, which is nonzero for every $0<\\alpha<1$. The integral of the operator output is therefore nonzero, and by the Hardy-space moment conditions no element of $L^1\\cap H^q$ with $q\\le1$ can have a nonzero integral. The proof then approximates the special $a_2$ by functions in $H^p$, preserving the nonzero obstruction, and concludes that the operator is unbounded from $H^1(\\mathbb{R})\\times H^p(\\mathbb{R})$ into $H^q(\\mathbb{R})$ for $0<p\\le1/\\gamma$.","pith_inferences":["The abstract and theorem print the range as $p\\le\\gamma-1$, but the proof establishes $p\\le1/\\gamma$; reading the printed exponent as $-1$ is necessary for the stated counterexample, and the paper's own proof follows the $1/\\gamma$ version.","Lemma 1's displayed Fourier transform has a sign in its second term opposite to the one obtained in the proof; if the displayed sign were correct, the coefficient in Corollary 2's identity would change and the nonzero-integral conclusion would fail, although the proof's computation appears to be the correct one.","The identity suggests a necessary condition for any Hardy-space bound: the weighted double integral $\\int\\int a_1(s)a_2(t)|t-s|^\\alpha\\,ds\\,dt$ must vanish for all admissible compactly supported inputs; test functions with separated supports and opposite signs are natural obstructions.","A similar moment-obstruction argument should extend to $m$-linear versions of $I_\\gamma$, since the kernel's Fourier transform would produce an analogous weighted integral over all $m$ variables."],"forward_implications":["For every $1<\\gamma<2$, the operator $I_\\gamma$ is not bounded from $H^1(\\mathbb{R})\\times H^p(\\mathbb{R})$ into $H^q(\\mathbb{R})$ when $0<p\\le1/\\gamma$ and $1/q=1+1/p-\\gamma$.","The earlier positive estimates from products of Hardy spaces into $L^q$ cannot be upgraded to a Hardy-space target at these endpoint parameters, even in the one-dimensional bilinear case.","For $p=1/\\gamma$, where $q=1$, the failure is a genuine $H^1$-to-$H^1$ failure: the integral of the output is nonzero, so the output is not in $H^1$.","The obstruction is stable: the special second factor $a_2$ can be replaced by a dense class of $H^p$ functions without making the integral of the output vanish, so the unboundedness is not an isolated example."],"supporting_citations":[{"why":"Supplies the $H^1\\times H^{(1/\\gamma)}$ to $L^1$ bound used to control the error term $I_{\\alpha+1}(a_1,a_2-b)$ when replacing $a_2$ by an $H^p$ approximant.","marker":"[1]"},{"why":"Provides the Fourier transform of $|x|^{\\alpha-1}$ used in Lemma 1's derivation of the kernel transform.","marker":"[2]"},{"why":"Proves the Hardy-space-to-Lebesgue estimates for $I_\\gamma$ that serve as the positive background result the counterexample tests.","marker":"[3]"},{"why":"Supplies the Hardy-space moment conditions (S1), (S2), and the density statement 5.2(b) used to select the approximant $b$ and to conclude integrable $H^q$ functions have zero integral.","marker":"[4]"}],"fun_headline_variants":["Nonzero integral blocks Hardy-space bound for fractional operator","Bilinear fractional operator unbounded on Hardy spaces","Counterexample: no Hardy-space boundedness for fractional integrals","Hardy-space bound fails: fractional integral's output has nonzero mean","Fractional operator escapes Hardy space via nonzero integral"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The load-bearing premise is Corollary 2's exact identity, which depends on the distributional Fourier transform of the kernel; if Lemma 1 is read with its printed sign error, the identity's coefficient changes and the nonzero-integral conclusion fails.","fun_headline_variants_meta":{"raw":{"variants":["Nonzero integral blocks Hardy-space bound for fractional operator","Bilinear fractional operator unbounded on Hardy spaces","Counterexample: no Hardy-space boundedness for fractional integrals","Hardy-space bound fails: fractional integral's output has nonzero mean","Fractional operator escapes Hardy space via nonzero integral"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000217,"raw_usage":{"total_tokens":1393,"prompt_tokens":862,"completion_tokens":531,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":478,"completion_tokens_details":{"reasoning_tokens":453}},"tokens_in":478,"tokens_out":531,"duration_ms":5549,"temperature":1.0,"reasoning_tokens":453,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T15:40:46.098752+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Numerically evaluate the identity of Corollary 2 for the paper's explicit functions $a_1(s)=\\chi_{(-1,0)}(s)-\\chi_{(0,1)}(s)$ and $a_2(t)=a_1(t-2)$ at $\\alpha=1/2$: the formula predicts a nonzero value, so a direct quadrature of $\\int\\int a_1(s)a_2(t)|t-s|^{1/2}\\,ds\\,dt$ should match the closed form; if it instead matches the sign-altered version of Lemma 1, the counterexample collapses.","supporting_citations":[{"cited_title":"Multilinear fractional Calder\\'on-Zygmund operators on weighted Hardy spaces","cited_arxiv_id":"1903.01593","evidence_quote":"Supplies the $H^1\\times H^{(1/\\gamma)}$ to $L^1$ bound used to control the error term $I_{\\alpha+1}(a_1,a_2-b)$ when replacing $a_2$ by an $H^p$ approximant."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Provides the Fourier transform of $|x|^{\\alpha-1}$ used in Lemma 1's derivation of the kernel transform."},{"cited_title":"Lin and S","cited_arxiv_id":null,"evidence_quote":"Proves the Hardy-space-to-Lebesgue estimates for $I_\\gamma$ that serve as the positive background result the counterexample tests."},{"cited_title":"Stein: Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory In- tegrals","cited_arxiv_id":null,"evidence_quote":"Supplies the Hardy-space moment conditions (S1), (S2), and the density statement 5.2(b) used to select the approximant $b$ and to conclude integrable $H^q$ functions have zero integral."}],"review_version":1}