The barycenter conjecture for one-dimensional free Gibbs measures

MathDB #365102: the logarithmic-growth barycenter conjecture is false

Result

The literal Conjecture 1 of arXiv:2410.02715v2 is false. A smooth counterexample is

$$f(x)=2\log\bigl(1+(x-1)^2\bigr). \tag{1}$$

This potential satisfies the source's logarithmic confinement assumption. Its free Gibbs measure exists and has barycenter $1$, not zero. Moreover, for every nonzero linear tilt $f+\lambda\operatorname{id}$, the free-energy variational problem is unbounded above. Consequently there is no $\lambda$ whose associated free Gibbs equilibrium has barycenter zero.

1. The source hypothesis holds

The one-dimensional existence assumption numbered (14) in the source is

$$\lim_{|x|\to\infty}\bigl(u(x)-2\log|x|\bigr)=+\infty. \tag{2}$$

The function in (1) is real-valued, smooth, and

$$f(x)-2\log|x| =2\log\left(\frac{1+(x-1)^2}{|x|}\right) \longrightarrow +\infty, \tag{3}$$

because the fraction in (3) is asymptotic to $|x|$. For an elementary bound, if $|x|\geq4$, then

$$1+(x-1)^2\geq \frac{|x|^2}{4},$$

so the expression in (3) is at least $2\log(|x|/4)$.

The existence-and-uniqueness theorem quoted immediately before (2) in the source therefore supplies a unique compactly supported maximizer $\nu_f$ of

$$\chi_f(\mu)=\chi(\mu)-\int f\,d\mu. \tag{4}$$

Here the additive normalization in $\chi$ will play no role.

2. The un-tilted equilibrium has barycenter one

Let $r(x)=2-x$, reflection about $1$. Equation (1) gives

$$f(r(x))=f(x). \tag{5}$$

For every probability measure $\mu$, reflection preserves the logarithmic interaction because

$$|r(s)-r(t)|=|s-t|.$$

Thus $\chi(r_\#\mu)=\chi(\mu)$, and (5) also gives

$$\int f\,d(r_\#\mu)=\int f\,d\mu.$$

The functional (4) is invariant under $r_\#$. By uniqueness of its maximizer,

$$r_\#\nu_f=\nu_f. \tag{6}$$

The measure is compactly supported, so its barycenter $m$ is finite. Using (6),

$$m=\int x\,d\nu_f(x) =\int r(x)\,d\nu_f(x) =2-m.$$

Hence

$$\int x\,d\nu_f(x)=1. \tag{7}$$

In particular, $\lambda=0$ does not center the equilibrium measure.

3. Every nonzero tilt destroys finite-barycenter equilibrium

Fix $\lambda\ne0$, put $s=\operatorname{sgn}(\lambda)$, and let $\mu_0$ be normalized Lebesgue measure on $[-1,1]$. Its logarithmic energy is finite; in fact

$$\iint\log|x-y|\,d\mu_0(x)d\mu_0(y)=\log2-\frac32.$$

For $t>0$, translate it by $-st$:

$$\mu_t=(x\mapsto x-st)_\#\mu_0.$$

Free entropy is translation invariant, so $\chi(\mu_t)=\chi(\mu_0)$. Also, for $y\in[-1,1]$,

$$|y-st-1|\leq t+2$$

and therefore

$$f(y-st) \leq2\log\bigl(1+(t+2)^2\bigr) \leq4\log(t+3). \tag{8}$$

Since $\mu_0$ is centered, the tilted functional satisfies

$$\begin{aligned} \chi_{f+\lambda\operatorname{id}}(\mu_t) &=\chi(\mu_0)-\int_{-1}^{1}f(y-st)\,d\mu_0(y) -\lambda(-st)\\ &\geq \chi(\mu_0)+|\lambda|t-4\log(t+3). \end{aligned} \tag{9}$$

The right side tends to $+\infty$. Thus

$$\sup_\mu\chi_{f+\lambda\operatorname{id}}(\mu)=+\infty \qquad(\lambda\ne0). \tag{10}$$

Every probability measure with a barycenter has finite first absolute moment. For such a measure the positive part of the logarithmic interaction is finite, since

$$\log^+|x-y|\leq \log(1+|x|+|y|)\leq |x|+|y|,$$

and the potential term is finite. Its value in (10) is therefore finite or $-\infty$, never $+\infty$, so it cannot maximize a functional whose supremum is $+\infty$. In the standard equilibrium convention, this unbounded variational problem has no free Gibbs measure at all. Under any extended-value convention, it has no free Gibbs maximizer with a finite barycenter, and in particular none with barycenter zero.

Combining (7) and (10), no $\lambda\in\mathbb R$ has the property asserted in Conjecture 1.

Scope of the refutation

This counterexample targets the source's literal quantifier over every continuous $f$ satisfying only (2). It grows like $4\log|x|$, so a linear tilt is nonconfining in one direction.

The surrounding proposed application considers functions $f,g$ satisfying $f(x)+g(y)\geq xy$ for every $x,y$. That stronger premise forces $f$ and $g$ to grow faster than every linear function. The example above therefore does not refute a repaired conjecture restricted to superlinear potentials, or more generally to potentials for which every linear tilt remains confining.

Exact verification

The file verify_counterexample.py checks the reflection and confinement algebra on exact rational grids, the uniform escape bound for both tails, and an exact diverging family behind (9). The all-real-variable implications and the variational argument are proved above; the program is a regression check for their algebraic inequalities.