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Every non-symmetric 1/3-convex function on the real line is midconvex.

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2026-08-04 01:23 UTC pith:DQCMGXKV

load-bearing objection 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.

arxiv 2608.00067 v1 pith:DQCMGXKV submitted 2026-07-29 math.GM

On non-symmetric t-convexity

Paolo Leonetti This is my paper
classification math.GM MSC 26A5139B6239B22
keywords non-symmetric t-convexityt-convex functionsmidconvexityfunctional inequalitiesJensen convexityone-sided convexityreal functions
verification ladder T0 review T1 audit T2 compute T3 formal T4 reserved

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

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.

Core claim

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

What carries the argument

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.

Load-bearing premise

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.

What would settle it

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.

Watch this falsifier. Get emailed when new claim-graph text bears on it.

If this is right

  • 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.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • 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.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit.

Referee Report

0 major / 3 minor

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.

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.

minor comments (3)
  1. [Section 3] 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.
  2. [Introduction / Abstract] 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.
  3. [Proof of Proposition 2.1] 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'.

Circularity Check

0 steps flagged

No significant circularity: Theorem 1.3 is proved via an explicit algebraic certificate and a standard external equivalence.

full rationale

The derivation chain is synthetic and self-contained modulo an external theorem. Proposition 2.1 states a sufficient condition: if a certificate (C, pairs (u_i,v_i), coefficients μ_i, η) satisfies identity (2) for every function φ, then non-symmetric t-convexity implies t-convexity. The proof of Proposition 2.1 assumes failure of t-convexity, normalizes to f(0)=f(1)=0 and f(t)>0, applies (2) to scaled functions, and derives a contradiction from the growth of H_n; none of these steps assumes the conclusion. For t=1/3, the certificate is exhibited explicitly (C=8/3, η=10/9, three pairs) and identity (2) is verified by direct expansion, so the main technical claim is not an input renamed as an output. The final step from two-sided 1/3-convexity to midconvexity uses the cited external result [3,5, Cor. 3]; this is not a self-citation, is parameter-free, and is not equivalent to the paper's assumptions. No fitted parameter is called a prediction, no uniqueness theorem is imported from the authors, and no ansatz is smuggled in via citation. The paper itself flags an omitted proof ('we omit the details' for 1/k, k=4,5,6), but an omitted detail is a completeness concern, not circularity. Any doubt about the cited equivalence would be a correctness risk, not a circularity risk.

Axiom & Free-Parameter Ledger

2 free parameters · 3 axioms · 0 invented entities

The proof of the main theorem is self-contained; the listed axioms are standard background. The constants C=8/3 and η=10/9 are explicit algebraic certificates chosen so that identity (2) holds; they are not fitted to data. There are no invented entities.

free parameters (2)
  • C = 8/3
    Chosen to satisfy C>1−t and make the algebraic identity (2) hold for t=1/3. Exact value verified by direct substitution; not an empirical fit.
  • η = 10/9
    Chosen so that the right side of (2) matches the linear combination for t=1/3. Verified exactly; not a fitted parameter.
axioms (3)
  • domain assumption A function f:R→R is midconvex iff it is t-convex for some rational t∈(0,1)
    Cited as [3]/[5, Cor 3] and used in the introduction to equate 1/3-convexity with midconvexity, making the negative answer to Páles's question meaningful.
  • standard math Subtracting an affine function and scaling variables preserve non-symmetric t-convexity
    Used in Prop 2.1 to reduce to the case f(0)=f(1)=0 and f(t)>0. This is an algebraic consequence of the definition and is stated in the proof.
  • standard math Basic properties of real numbers and monotonicity of polynomials
    Used throughout the proof, e.g., the contraction argument and the monotonicity of Φ_w in Prop 3.1.

pith-pipeline@v1.3.0-alltime-deepseek · 4514 in / 17294 out tokens · 157983 ms · 2026-08-04T01:23:13.133824+00:00 · methodology

0 comments
read the original abstract

Zsolt P\'ales asked whether there exists a function $f: \mathbb{R}\to \mathbb{R}$ such that $$ f\left(\frac{x+2y}{3}\right) \le \frac{f(x)+2f(y)}{3} \quad \text{ for all }x\le y $$ which is not midconvex. We show that the answer is negative.

discussion (0)

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