No, not all quasi-Bernoulli measures are Gibbs measures. A counterexample can be constructed by considering a convex combination of two distinct Bernoulli measures.

Counterexample:

Let ( X = {0,1}^{\mathbb{N}} ). Define two Bernoulli measures ( \mu_p ) and ( \mu_q ) with symbol probabilities ( p ) and ( q ) respectively, where ( p \neq q ) and ( p, q \in (0,1) ). For ( \mu_p...

No.

A convenient way to see this is via the subadditive (almost additive) thermodynamic formalism. There are natural quasi-Bernoulli equilibrium measures that come from almost additive potentials, and in general these cannot be realised as Gibbs measures for any single continuous (additive) potential.

Outline.

1. Almost additive potentials and their equilibrium states.
   On the full shift X = {...

No – quasi‐Bernoulli is strictly weaker than Gibbs. In fact one can take any continuous, strictly positive “g‐function’’ whose cylinder‐variation tends to zero but is not summable, and its unique g–measure will be ψ–mixing (hence quasi–Bernoulli) but not Bowen‐Gibbs for any continuous potential. Here is the classical construction of Walters (1975):

1. Define (g:X\to(0,1)) on (X={0,1}^{...

No.
There are quasi–Bernoulli measures on the full 2–shift which are not Gibbs for any continuous potential.
A concrete example can be obtained from a very simple finite–state hidden Markov process (“the even–zero process’’).

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1 The process

Hidden states : E (“even’’) and O (“odd’’).

...

This is a very nice question about the relationship between quasi-Bernoulli and Gibbs measures on symbolic spaces.

The answer is no - not all quasi-Bernoulli measures are Gibbs measures, even for continuous potentials.

Counterexample

A concrete counterexample can be constructed using g-measures (also called chains with complete connections). Here's a sketch:

Consider a g-function $g...

<think> We are given a question from MathOverflow asking for an example of a quasi-Bernoulli measure that is not Gibbs. The user has provided definitions and context.

First, recall the definitions:

1. Quasi-Bernoulli measure: A measure μ on X = {0,1}^ℕ is quasi-Bernoulli if there exists a constant C ≥ 1 such that for any two finite sequences (blocks) i and j, we have: [ C^{-1} \mu[ij]...

Excellent question. The answer is no, not all quasi-Bernoulli measures are Gibbs measures for a continuous potential.

The distinction is subtle and gets to the heart of what "long-range order" means in one dimension. A Gibbs measure for a continuous potential has strong regularity properties, which turn out to be more restrictive than the quasi-Bernoulli condition.

The Key Idea for a Cou...

Example of a Quasi-Bernoulli Measure Which is Not Gibbs

This is an interesting question in dynamical systems and ergodic theory. Let me work through it systematically.

Understanding the Definitions

A measure μ on X = {0,1}^ℕ is:

• Quasi-Bernoulli if there exists C ≥ 1 such that for any finite sequences i and j: C^(-1) μ[ij] ≤ μ[i]μ[j] ≤ Cμ[ij]

• Gibbs measure with potential f i...

No.
A quasi–Bernoulli measure need not be a Gibbs measure.
Below we give an explicit invariant and ergodic measure on
X={0,1}^{ℕ} which satisfies the quasi–Bernoulli inequalities but for which no continuous potential can satisfy the Gibbs bounds.

1. Construction of the measure μ

Think of the configuration as a string of blocks
“several 0’s followed by a 1”.
Choose the succe...