On non-symmetric $t$-convexity

Paolo Leonetti

On non-symmetric $t$-convexity

T0 review · 0 major / 3 minor · reviewed 2026-08-04 · deepseek-v4-flash

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Pith's one-line read Every non-symmetric 1/3-convex function on the real line is midconvex.

desk verdict 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. read the letter →

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

Paolo Leonetti This is my paper · ORCID

classification math.GM MSC 26A5139B6239B22

keywords non-symmetrict-convexityt-convexfunctionsmidconvexityfunctionalinequalitiesJensenconvexityone-sidedreal

verification ladder T0 review T1 audit T2 compute T3 formal

The pith

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

The reading

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.

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.

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

Extended reading notes

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

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.

Editorial extensions

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.

Reading between the lines

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.

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Editorial analysis

A structured set of objections, weighed in public.

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

Desk Editor's Note

read the 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.

The 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.

Soft 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).

One 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.

Verdict: 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.

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)

[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.
[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.
[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

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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.

Assumptions & free parameters 2 free parameters · 3 assumptions · 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.

assumptions (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.

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Pith. "Pith review of On non-symmetric $t$-convexity." pith.science (2026). https://pith.science/paper/DQCMGXKV

@misc{pith2026260800067,
  author       = {Pith},
  title        = {Pith review of: On non-symmetric $t$-convexity},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/DQCMGXKV}},
  note         = {Machine review of arXiv:2608.00067}
}

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.

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