Overlapping entropy obstruction

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The moment obstructions rule out fixed higher-moment bounds for an overlapping potential. The following strengthening also rules out uniform integrability, including every superlinear Orlicz bound. It does not refute the disjoint first-kill entropy target and is not a counterexample to Problem 25.

For a finite singleton system, let $F_q$ be its full first-kill cells, $e_q=\mu(F_q)$, and $c_q(X)=H_{E_q}(X)/H_{\lfloor X\rfloor}$. Consider on Haar space

$$Z_X(z)=\sum_q q c_q(X)\mathbf1_{[a_q]_q}(z).$$

There are zero-residue systems and cutoffs $X=X_k$ such that, for a fixed $\gamma>0$,

$$\liminf_{k\to\infty} \int_{\{Z_{X_k}\ge e^{\gamma k}\}} Z_{X_k}\,d\mu \ge\frac12. \tag{1}$$

In particular $\int Z_{X_k}\log_+ Z_{X_k}\,d\mu\gg k$.

Construction and the first-kill formula

Choose a finite set $P$ of primes exceeding $k^3$, stopping when

$$\lambda:=\sum_{p\in P}p^{-1}=2k+O(k^{-3}).$$

Then $\sum_{p\in P}p^{-2}=O(k^{-2})$. Take every product of $k$ distinct primes from $P$ as a modulus, with residue zero, and order these products numerically. Put $L=\prod_{p\in P}p$ and $X=L^2$. These moduli are distinct and primitive. Every positive multiple of a modulus is active, so $E_q=F_q\cap\mathbb N$ exactly.

Each cell is $L$-periodic with exactly $e_qL$ points in every block of $L$ consecutive positive integers. In the $j$th such block, each reciprocal is at least $1/(jL)$. Therefore

$$H_{E_q}(L^2)\ge e_qH_L, \qquad c_q(L^2)\ge e_q/2, \tag{2}$$

using $H_{L^2}\le2H_L$. The latter inequality follows, for example, by bounding the $j$th block after the first in $H_{mn}$ by $1/(j-1)$, which gives $H_{mn}\le H_m+H_n$.

Define

$$Y=\frac12\sum_q qe_q\mathbf1_{q\mid z}\le Z_X, \qquad \beta=\sum_qe_q.$$

If $N_P$ counts the dividing primes of $P$ in a Haar integer, then $\beta=\Pr(N_P\ge k)=1-O(k^{-1})$, by Chebyshev, since $\mathbb E N_P=2k+O(k^{-3})$ and $\operatorname{Var}N_P\le\mathbb E N_P$. Thus

$$\mathbb EY=\beta/2\longrightarrow1/2. \tag{3}$$

For a modulus $q$, write $u(q)=\max_{p\mid q}p$ and $T(q)=\sum_{p\in P,\ p\le u(q)}1/p$. The least numerical $k$-prime product dividing a Haar integer consists of its $k$ smallest dividing pool primes. Consequently

$$qe_q=\prod_{\substack{p\in P,\ p<u(q)\\p\nmid q}}(1-1/p). \tag{4}$$

If $q$ has law $e_q/\beta$, it is that first product conditional on $N_P\ge k$. Therefore

$$\Pr(0.9k\le T(q)\le1.1k)=1-O(k^{-1}). \tag{5}$$

Indeed, a first product before reciprocal time $0.9k$ requires at least $k$ successes in a prime prefix of mean at most $0.9k$. A first product after time $1.1k$ requires fewer than $k$ successes in a prefix of mean $1.1k+O(k^{-3})$ (or one with mean just below that threshold). Both probabilities are $O(1/k)$ by Chebyshev; conditioning costs a factor $1/\beta=1+O(1/k)$.

Size bias and concentration

The size-biased probability $d\nu=Y\,d\mu/\mathbb EY$ can be sampled by choosing $q$ with probability $e_q/\beta$ and then choosing a Haar integer conditional only on $q\mid z$, not on $F_q$. Its density is exactly $\sum_q(e_q/\beta)q\mathbf1_{q\mid z}=2Y/\beta$.

For the chosen $q$, let $N_q$ count dividing pool primes at most $u(q)$. Under this conditional Haar law,

$$N_q=k+\sum_{\substack{p\in P,\ p\le u(q)\\p\nmid q}} \mathbf1_{p\mid z},$$

with independent Bernoulli terms. When (5) holds,

$$\mathbb E(N_q\mid q)\ge1.9k-k^{-2}, \qquad \operatorname{Var}(N_q\mid q)\le1.1k.$$

Another application of Chebyshev gives $\Pr(N_q\ge1.8k\mid q)=1-O(1/k)$, uniformly over these $q$.

Every $k$-product $r$ of these dividing primes is among our moduli, and (4) gives

$$re_r\ge\prod_{\substack{p\in P\\p\le u(q)}}(1-1/p) \ge e^{-T(q)-O(k^{-2})}.$$

It follows, with $\nu$-probability $1-O(1/k)$, that

$$Y\ge\frac12e^{-1.1k-O(k^{-2})} {\lfloor1.8k\rfloor\choose k}.$$

Stirling's formula gives

$$\log {\lfloor1.8k\rfloor\choose k} =(1.8\log1.8-0.8\log0.8)k+O(\log k),$$

whose linear coefficient exceeds $1.1$. Hence, with $\gamma=1/20$, for all sufficiently large $k$,

$$\nu(Y\ge e^{\gamma k})=1-O(1/k), \qquad \int_{\{Y\ge e^{\gamma k}\}}Y\,d\mu\longrightarrow1/2.$$

Since $Y\le Z_X$, this proves (1).

If $\Phi$ is convex and increasing, $\Phi(0)=0$, and $\Phi(t)/t\to\infty$, then $\Phi(t)/t$ is nondecreasing and (1) implies $\int\Phi(Z_{X_k})\,d\mu\to\infty$ as well.

The distinction from the disjoint potential is essential. In these zero-residue examples, the geometric valuation model gives $c_q(X)\le (Z_X^{\rm Euler}/H_X)e_q\ll e_q$, where $Z_X^{\rm Euler}=\prod_{p\le X}(1-1/p)^{-1}$. Thus $\sum_q(c_q/e_q)\mathbf1_{F_q}$ is uniformly bounded. Replacing these disjoint cells by the cylinders loses precisely the information needed by the still-unproved entropy argument.