{"id":"65870126-66c8-4281-ae36-b9d0ac5b1706","arxiv_id":"2608.00067","paper_version":1,"verdict":"ACCEPT","confidence":"HIGH","novelty_score":7.0,"correctness_risk":"low","formal_verification":"none","parameter_count":2,"one_line_summary":"Every non-symmetric 1/3-convex function on R is 1/3-convex (hence midconvex), answering Páles' question in the negative.","lead":"This paper proves that a function satisfying the asymmetric inequality f((x+2y)/3) ≤ (f(x)+2f(y))/3 for all x ≤ y must be midconvex, settling an open question by Páles. The proof introduces a general criterion for deducing two-sided convexity from one-sided convexity and shows that the set of such parameters is dense.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The central claim relies on a cited equivalence (two-sided t-convex => midconvex) that is neither proved nor obviously true for a single rational t such as 1/3.","rationale":"The internal proof of Theorem 1.3 is algebraically sound, and the reader's verification of the t=1/3 computation is accurate. However, the advertised resolution of Páles's question requires more than non-symmetric => symmetric t-convexity: it requires the further theorem that a symmetric 1/3-convex function is midconvex. The manuscript cites this as well-known, but for a fixed rational t the implication is not a formal consequence of finite t-convexity interactions, because the midpoint is not finitely reachable from the 1/3-convexity semigroup. The reader's weakest assumption about the domain R is real but not the main vulnerability; the unexamined cited equivalence is. I therefore recommend conditionally accepting the paper: the central claim should be accepted only after the cited equivalence is verified (or replaced by a direct proof that non-symmetric 1/3-convexity implies midpoint convexity). If the equivalence turns out to be false, the conclusion of the paper is no longer supported.","tokens_in":4749,"tokens_out":59306,"duration_ms":524641,"concrete_test":"Verify the cited equivalence directly. Check whether [5, Cor. 3] states the implication for a single rational t or rather for all t in (0,1). Independently, attempt to construct a two-sided 1/3-convex f with f(0)=0, f(1)=1, and f(1/2)>1/2 by assigning values on a Hamel complement of S={m/3^k} and solving the induced linear inequalities. If such an f exists (or if the cited corollary is misquoted), the paper's final conclusion fails.","verdict_should_be":"CONDITIONAL","load_bearing_attack":"Theorem 1.3 proves only the first half of the advertised conclusion: every non-symmetric 1/3-convex f:R->R is two-sided t-convex. The final step \"hence midconvex\" depends entirely on the cited equivalence from the Introduction: \"a function f:R->R is midconvex if and only if it is t-convex for some rational t in (0,1)\" [3,5, Cor. 3]. This is the load-bearing step. For fixed t=1/3, finite applications of the t-convexity operation (x+2y)/3 to the endpoints 0,1 generate only weights in the semigroup S = {m/3^k}∩[0,1]. The midpoint 1/2 is not in S. By Farkas-type reasoning, any finite nonnegative combination of one-point convexity inequalities that yields the midpoint inequality would give a finite representation of 1/2 as an element of S. No such representation exists. Thus the standard equivalence is not automatic for a single fixed rational t. If it is actually false, then a two-sided 1/3-convex function that fails midpoint convexity would also be non-symmetric and would answer Páles's question affirmatively, directly contradicting the paper's abstract. The paper does not address this dependence.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper studies non-symmetric t-convex functions f:R→R satisfying f(tx+(1−t)y)≤tf(x)+(1−t)f(y) for all x≤y, and proves that for t=1/3 every such function is necessarily t-convex (Theorem 1.3). The proof is based on a general criterion (Proposition 2.1) that provides a sufficient condition for the implication, and on an explicit finite identity with nonnegative coefficients and a constant C>1−t. The paper also shows that the set T of t∈(0,1) for which the implication holds is dense, using a construction with iterated contractions. The abstract interprets the result as answering Páles's question negatively, relying on the standard equivalence between t-convexity for rational t and midconvexity.","tokens_in":5084,"tokens_out":31017,"duration_ms":250486,"significance":"If correct, the paper settles an open question in generalized convexity: Páles's question about the existence of a non-symmetric 1/3-convex function that is not t-convex (equivalently, not midconvex) is answered negatively. The main proof is self-contained and gives a concrete certificate (C, η, and ordered pairs) for the application of Proposition 2.1, which is a useful criterion. The density result for T is a notable structural addition. The paper is concise and the algebra is verifiable; the proof of Proposition 2.1 is logically sound and the explicit identity in the proof of Theorem 1.3 checks out.","major_comments":[],"minor_comments":[{"comment":"The statement 'similar computations show that 1/k ∈ T also for k ∈ {4,5,6} (we omit the details)' is unsupported. Since no certificates are provided, this side remark is not verifiable. Please either supply the computations or remove/reword the claim.","section":"Section 3"},{"comment":"The abstract's 'answer is negative' for the midconvex formulation depends on the cited equivalence that rational two-sided t-convexity is equivalent to midconvexity ([3], [5, Cor. 3]). The proof of Theorem 1.3 gives the two-sided t-convex conclusion directly, but the link to midconvexity is external. Please state the exact theorem used and, if space permits, include a short derivation for t=1/3 so that the abstract's claim is self-contained.","section":"Introduction / Abstract"},{"comment":"Minor formatting: the operator 'J t,φ' should be typeset as J_{t,φ}. Also, the phrase 'We now show that the numberstw are dense' in Section 3 has a missing space between 'numberst' and 'w'.","section":"Proof of Proposition 2.1"}],"recommendation":"minor_revision","confidential_remarks":"The main theorem is correct and the proof is verifiable. The side remark about k=4,5,6 and the reliance on the cited equivalence are the only points needing attention; neither undermines the central result. I recommend minor revision to address these presentation issues."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Juan, quick take on Leonetti (arXiv:2608.00067). The paper answers Páles's open question: every non-symmetric 1/3-convex function f:R→R is actually 1/3-convex, hence midconvex. The main theorem is new and the proof is genuinely worth a look. The key is Proposition 2.1, a sufficient criterion that reduces the task to finding a finite nonnegative combination of one-sided t-convexity inequalities that yields a particular identity (2). The author then supplies an explicit certificate for t=1/3: C=8/3, eta=10/9, three pairs with coefficients. I checked the algebra; it works. The contradiction argument in Prop 2.1 is also sound: the growth of H_n and the lower bound on f(-alpha) force a linear term to blow up, which can't happen since the coefficient is positive. The proof genuinely uses the whole real line: scaling by h>0 and iterations going to -infinity are essential. That is worth flagging but it's not a flaw; the theorem states R.\n\nThe density result Prop 3.1 is a pleasant addition. The construction with words over A,B and the polynomials p_w is standard but clearly written. The observation that T is symmetric around 1/2 is correct. The open question about algebraic t is natural.\n\nSoft spots: the side remark that 1/k ∈ T for k=4,5,6 is asserted without details. The author says 'we omit the details' so it's not a claim that carries the paper, but a referee should ask whether that can be made into an appendix or verified with the same method. There are also a few typos/uncomfortable notations (the map J_{t,φ} is defined for all reals but the inequality only for u≤v; no issue in practice).\n\nOne thing I checked because it worried me: the final step depends on the standard equivalence: full t-convexity for a rational t implies midconvexity. This is cited to Kuhn and Nikodem–Páles. The stress-test note raised a semigroup objection about generating 1/2 from 1/3 operations. That objection does not land, because Theorem 1.3 establishes full t-convexity on all of R, and the classical theorem covers that case. The proof never tries to derive the midpoint inequality from the one-sided inequalities; it first upgrades to the two-sided inequality and then invokes the known equivalence. So the dependence on [3,5] is real but legitimate.\n\nVerdict: this deserves a serious referee. It is a short, self-contained proof that settles a named open question. I would send it to review with confidence; the only request would be to either prove or delete the k=4,5,6 remark.","headline":"Páles's 1/3-convexity question is settled negatively with a clean certificate argument; the proof is sound and the density add-on is nice.","tokens_in":5539,"tokens_out":2346,"would_cite":true,"duration_ms":24410,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["26A51","39B62","39B22"],"pacs":[],"model":"deepseek-v4-flash","headline":"Every non-symmetric 1/3-convex function on the real line is midconvex.","keywords":["non-symmetric t-convexity","t-convex functions","midconvexity","functional inequalities","Jensen convexity","one-sided convexity","real functions"],"falsifier":"Verify the finite identity that powers the proof: for t=1/3, C=8/3, η=10/9, and the three pairs (1/9,4/9), (0,2/3), (0,1), the combination 3J(·)+2J(·)+(4/3)J(·) evaluated on any φ must equal φ(1/9)−3φ(1/3)+8/9φ(1)+10/9φ(0); testing with a nonlinear φ such as φ(x)=x^2 should confirm equality, and any mismatch would invalidate the criterion step.","tokens_in":4650,"feed_emoji":"📐","tokens_out":8542,"duration_ms":89514,"temperature":0.7,"pith_summary":"The paper closes an open question in generalized convexity by proving that a one-sided weighted Jensen inequality at t=1/3 is actually two-sided. Specifically, if a real function f satisfies f((x+2y)/3) ≤ (f(x)+2f(y))/3 for all x≤y, then it satisfies the same inequality for every pair x,y, which is the standard midpoint convexity condition. The proof introduces an algebraic criterion: when a finite nonnegative combination of the one-sided error terms can be rewritten as a specific expression in four values of the function, every function obeying the one-sided inequality is forced to obey it in all directions. The paper also shows that the set of t-values for which this implication holds is symmetric about 1/2 and dense in (0,1), with explicit examples beyond 1/3.","feed_headline":"One-sided 1/3-convexity forces midpoint convexity","feed_subtitle":"Functions satisfying the one-way Jensen inequality at 1/3 on R turn out to satisfy it both ways, closing the question.","key_machinery":"The key object is the error functional J_{t,φ}(u,v)=tφ(u)+(1-t)φ(v)-φ(tu+(1-t)v), which is nonnegative exactly when φ is non-symmetric t-convex. Proposition 2.1 shows that if some finite nonnegative combination of J at fixed pairs equals φ(t^2)-(t+C)φ(t)+Ctφ(1)+ηφ(0) with C>1-t, then every non-symmetric t-convex φ is t-convex. The proof then substitutes a scaled version φ(hx), uses the inequality to get a geometric growth bound on the differences f(t^{n+1})-t f(t^n), and compares that to a telescoping bound that forces the opposite positivity condition to fail.","core_discovery":"The central claim is that for t=1/3, non-symmetric t-convexity—the inequality f(tx+(1-t)y) ≤ t f(x)+(1-t)f(y) restricted to x≤y—implies full t-convexity, i.e., the inequality holds for all real x,y. Since t-convexity at 1/3 is equivalent to midconvexity, this gives a negative answer to the question whether a one-sided 1/3-convex but non-midconvex function exists. The paper introduces a criterion (Proposition 2.1) under which a finite nonnegative linear combination of the error terms J_{t,φ}(u_i,v_i) reduces to a specific expression in values of φ at 0, 1, t, and t^2; when that expression holds with a constant C>1-t, scaling and a telescoping argument force a contradiction unless f is t-conve","pith_inferences":["The proof's reliance on points tending to −∞ suggests that on bounded intervals the one-sided inequality may admit non-midconvex solutions, which would be a natural sharpening.","The explicit identity for t=1/3 is a certificate that could be generated systematically for other rational t, potentially creating an algebraic classification of t-values with the same implication.","Since T is dense and symmetric, the non-T set is nowhere dense; a complete description of T, for instance whether it consists exactly of algebraic numbers, remains an open problem this machinery could attack."],"forward_implications":["The open question about one-sided 1/3-convexity is closed: no counterexample exists on the real line.","For t=1/3, the non-symmetric convexity class coincides with the midconvex class, so all standard regularity results for midconvex functions apply automatically.","The set of t for which the implication holds is dense in (0,1) and contains 1/2, 1/3, and the two golden-ratio numbers (√5−1)/2 and (3−√5)/2.","The criterion provides a concrete recipe to certify additional t-values: any finite identity of the form (2) with C>1-t proves the implication for that t."],"fun_headline_variants":["One-sided t-convexity at 1/3 implies midconvexity","Restricted Jensen inequality at 1/3 forces full convexity","Páles question: one-sided 1/3-convexity is actually full convexity","One-way inequality at 1/3 becomes two-way for real functions","No non-midconvex function satisfies one-sided 1/3-Jensen"],"cache_read_input_tokens":2304,"weakest_assumption_plain":"The proof relies on the domain being the entire real line so that the scaling points −α t^n, which tend to −∞, are valid inputs for f; on a bounded interval the contradiction step would fail.","fun_headline_variants_meta":{"raw":{"variants":["One-sided t-convexity at 1/3 implies midconvexity","Restricted Jensen inequality at 1/3 forces full convexity","Páles question: one-sided 1/3-convexity is actually full convexity","One-way inequality at 1/3 becomes two-way for real functions","No non-midconvex function satisfies one-sided 1/3-Jensen"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000289,"raw_usage":{"total_tokens":1474,"prompt_tokens":635,"completion_tokens":839,"prompt_tokens_details":{"cached_tokens":256},"prompt_cache_hit_tokens":256,"prompt_cache_miss_tokens":379,"completion_tokens_details":{"reasoning_tokens":735}},"tokens_in":379,"tokens_out":839,"duration_ms":8766,"temperature":1.0,"reasoning_tokens":735,"cache_read_input_tokens":256,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-04T01:23:13.133824+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Verify the finite identity that powers the proof: for t=1/3, C=8/3, η=10/9, and the three pairs (1/9,4/9), (0,2/3), (0,1), the combination 3J(·)+2J(·)+(4/3)J(·) evaluated on any φ must equal φ(1/9)−3φ(1/3)+8/9φ(1)+10/9φ(0); testing with a nonlinear φ such as φ(x)=x^2 should confirm equality, and any mismatch would invalidate the criterion step.","supporting_citations":[],"review_version":1}