MathAudit

Paper metadata

Original submission
October 19, 2026
Claimed MSC
37A50, 37D35
arXiv categories
math.DS, math.PR
Paper license
CC BY 4.0 ↗
Comments
Comments are welcome!

Abstract

Effective intrinsic ergodicity (EIE) quantifies how the pressure defect of an invariant measure controls its distance from an equilibrium state. In this paper, we make explicit the connection between the Einsiedler-Kadyrov-Polo (EKP) inequality, which is central to EIE, and the transport-entropy inequality (TEI). We also extend the finite-state transport-entropy inequality of \cite{CMU19} to topologically mixing countable Markov shifts with the Big Images and Preimages (BIP) property, yielding an EIE estimate in Ornstein's dˉ\bar d-metric that strengthens \cite{Ruh21}. We also establish a local 12\frac{1}{2}-Hölder bound for the map sending a Hölder potential of finite pressure to its Bowen-Gibbs measure, with respect to the uniform norm on potentials and the dˉ\bar d-metric on measures. This result extends some aspects of Bhullar's continuity theorem \cite{B25}: it applies to proper subshifts, provides an explicit modulus of continuity with respect to the uniform norm, and does not require strong positive recurrence.

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Audit summary

Audited against arXiv v1

Not a correctness certificate. A green “no issues found” result may include yellow suspected typos or minor formal issues that do not affect substantive soundness. It means this audit found no unresolved substantive error under the stated criteria; it does not replace expert scrutiny or formal verification.

Current report

Detailed mathematical audit

Generated October 23, 2026
01Statements2 reported findingsNo statement issues found

The pressure-defect identity, sharp comparison of metrics, transport bound, and square-root continuity conclusion are supported. A local conditional-expectation equality in Proposition 3.1 needs weakening to an inequality, which is sufficient for the stated conclusion.

Proposition 3.2No issue found

The metric comparison has the claimed sharp universal constant

PDF pp. 10–11 · equations (34)–(36) · arXiv:2610.26588v1

The weighted coordinate metric lies between θdθ\theta d_\theta and θdθ/(1−θ)\theta d_\theta/(1-\theta). A stationary joining converts its expected cost into θ/(1−θ)\theta/(1-\theta) times the one-coordinate mismatch. Taking the infimum gives (34). For a constant-zero process versus Bernoulli(pp) zeros, the ratio is 1/(1−θp)1/(1-\theta p) and tends to the asserted optimal constant as pp tends to one.

Theorem 4.1; Corollaries 4.2–4.3No issue found

The normalized Hamming cost gives the required inverse-length entropy factor

PDF pp. 11–15 · arXiv:2610.26588v1

McShane extension from admissible words to the full finite-word space preserves the normalized Hamming Lipschitz constant. Each coordinate oscillation is at most ∣f∣d(n)/n|f|_{d^{(n)}}/n, so the Gaussian concentration exponent is Dt2∣f∣2/nDt^2|f|^2/n. Bobkov–Götze gives Ctei=2DC_{tei}=2\sqrt D. Taking the stationary finite-volume limit proves (39). The Gibbs ratios and pressure Lipschitz bound give relative entropy density at most 2∥ϕ−ψ∥∞2\|\phi-\psi\|_\infty; applying the estimate in both directions gives the minimum of the two constants in (46). The bounded-difference argument remains valid even when the individual potentials are unbounded.

02Proofs2 reported findingsSuspected incomplete or incorrect proofs

The main concentration and continuity arguments close. Proposition 3.1's conditional-probability calculation drops zero-denominator sets; its claimed equality should be an upper bound. That bound still supplies every estimate needed for the pressure-defect identity.

Proposition 3.1Suspected incorrect as written

The conditional-ratio integral can be strictly less than one

PDF p. 10 · equations (32)–(33) · arXiv:2610.26588v1

Writing Mka=ν(X0=a∣X1k)M_k^a=\nu(X_0=a\mid X_1^k), conditioning 1X0=aMka/M∞a1_{X_0=a}M_k^a/M_\infty^a on the whole future yields Mka1M∞a>0M_k^a1_{M_\infty^a>0}, not MkaM_k^a on every future. For example, on the binary full shift take ν=(δ0∞+δ1∞)/2\nu=(\delta_{0^\infty}+\delta_{1^\infty})/2 and k=0k=0. The integral on the left of (33) is 1/21/2, not 11. The required repair is to replace the equality by ≤1\leq1. Then the negative part of ℓk\ell_k is still integrable and ∫(1−Mk/M∞) dν≥0\int(1-M_k/M_\infty)\,d\nu\geq0, which still gives rk≥r∞r_k\geq r_\infty. Together with reverse Fatou and the positive-part tail bound, this proves the proposition without changing its statement.

Proposition 3.1; Theorem 4.1No issue found in the reviewed proof

The remaining limiting and extension steps retain their hypotheses

PDF pp. 8–13 · arXiv:2610.26588v1

The one-cylinder Gibbs bound makes the sum of the weights qaq_a finite. The stopped conditional-probability estimate therefore bounds the supremum of the positive parts by an integrable function, allowing the one-sided Fatou argument even when the limiting integral is minus infinity. The finite-word extension in Theorem 4.1 addresses inadmissible intermediate words before invoking separate-coordinate concentration. Full cylinder support makes the finite-volume absolute-continuity condition automatic; measures of infinite relative entropy satisfy the inequality trivially.

03Novelty0 reported findingsNo novelty concerns found

No non-novelty findings.

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Paper
arXiv:2610.26588v1
Authors listed
Mirmuhsin Maxmud
Audit date
October 23, 2026
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