Harmonic affine completion

---

The following uniform coupling theorem is proved. Its application to Problem 25 includes a return-probability weight that has not been removed. No singleton-distinctness assertion is implicit in the theorem.

Total-variation convergence

Write $H_N=\sum_{n\le N}1/n$, with $H_0=0$. For integers $1\le L\le X$, define a subprobability measure $\nu_{L,X}$ on $G=\widehat{\mathbb Z}$ as follows. Give each $n\in[L,X]$ mass $1/(nH_X)$, choose

$$1\le r\le\lfloor n/L\rfloor \quad\hbox{with probability}\quad \frac1{rH_{\lfloor n/L\rfloor}},$$

and then choose a Haar-uniform point in the cylinder $[n]_{Lr}$. This preserves $n\pmod L$. Its total mass is

$$b_{L,X}=\frac{H_X-H_{L-1}}{H_X}.$$

Uniformly in $L,X$,

$$\sup_{E\subseteq G\ { m measurable}} |\nu_{L,X}(E)-b_{L,X}\mu(E)| \ll\frac{1+\log\log(3X/L)}{H_X}. \tag{1}$$

In particular $\nu_{1,X}$ converges to Haar in total variation, not merely against fixed continuous test functions.

Its density is

$$\rho_{L,X}(z)=\frac L{H_X} \sum_{r\le X/L} \sum_{\substack{Lr\le n\le X\\n\equiv z\pmod{Lr}}} \frac1{nH_{\lfloor n/L\rfloor}}. \tag{2}$$

At fixed $r$, write the canonical residue of $z$ modulo $Lr$ as $a\in[0,Lr)$. Every term has $n=kLr+a$, $k\ge1$, so $n\ge kLr$ and $H_{\lfloor n/L\rfloor}\ge H_{kr}$. Consequently, with $T=\lfloor X/L\rfloor$,

$$\rho_{L,X}(z)\le\frac{D_T}{H_X},\qquad D_T:=\sum_{j\le T}\frac{d(j)}{jH_j}. \tag{3}$$

Here $d(j)$ is the divisor-counting function. The elementary identity

$$\sum_{j\le t}d(j)=\sum_{a\le t}\lfloor t/a\rfloor =t\log t+O(t)$$

and partial summation give

$$D_T=\log T+O(1+\log\log(3T)). \tag{4}$$

For detail, for $j\ge2$ replace $1/H_j$ by $1/\log j+O(1/\log^2j)$. Partial summation with the displayed divisor estimate makes the main sum $\log T+O(1+\log\log(3T))$, and the error sum $O(1+\log\log(3T))$.

Uniformly, $H_X-H_{L-1}=\log T+O(1)$. Thus (3) bounds $\rho_{L,X}$ above by $b_{L,X}+\eta$, where

$$0\le\eta=\frac{D_T-H_X+H_{L-1}}{H_X} \ll\frac{1+\log\log(3T)}{H_X}.$$

Nonnegativity also follows by integrating (3). The density has mean $b_{L,X}$, so its positive and negative deviations from this mean each have integral at most $\eta$. This proves (1). For $L=1$, the same bound implies the relative entropy estimate

$$D(\nu_{1,X}\Vert\mu) \le\log(D_X/H_X)=O(\log\log(3X)/H_X).$$

An averaged conditioned return inequality

Consider any finite or countable congruence family and put

$$U=\bigcup_q[a_q]_q,\qquad D_n=\{q\le n:n\equiv a_q\pmod q\}.$$

For a set $D$, let $M(D)$ be its set of positive multiples. If the random completion period is divisible by a member of $D_n$, the completed point belongs to $U$. For $L=1$, (1) therefore gives

$$\frac1{H_X}\sum_{n\le X} \frac{H_{M(D_n)}(n)}{nH_n} \le\mu(U)+O\!\left(\frac{\log\log(3X)}{H_X}\right). \tag{5}$$

More generally, let $S$ be $L$-periodic and put

$$D'_{n,L}=\{q/(q,L):q\in D_n\},\qquad N_n=\lfloor n/L\rfloor.$$

The completion preserves $S$, and $q\mid Lr$ if and only if $q/(q,L)\mid r$. Hence

$$\frac1{H_X}\sum_{\substack{L\le n\le X\\n\in S}} \frac{H_{M(D'_{n,L})}(N_n)}{nH_{N_n}} \le b_{L,X}\mu(S\cap U) +O\!\left(\frac{1+\log\log(3X/L)}{H_X}\right). \tag{6}$$

Indeed, completion restricted to starting points in $S$ is exactly the restriction of $\nu_{L,X}$ to $S$, so its mass in $U$ is $\nu_{L,X}(S\cap U)$. This also explains the coefficient $b_{L,X}$. The theorem allows $S,U,L$ to depend on $X$.

One may take $S$ to be the full periodic survivor set of a finite prefix and $U$ its tail union; then the right side has the desired incremental Haar footprint. Starting with $n\ge L$, where $L$ is the prefix period, makes every prefix modulus active.

The remaining loss

The weights on the left of (5)--(6) are return probabilities, not indicators of deletion. For a sole active witness $q$, the return probability at $L=1$ is only

$$\frac{H_{\lfloor n/q\rfloor}}{qH_n}.$$

In (6), a witness with $q/(q,L)>n/L$ contributes zero. Thus comparable-modulus first kills can be lost altogether, and the coupling does not establish the required unweighted logarithmic-tail estimate. Neither the coupling proof nor (5)--(6) uses singleton distinctness. The counterexample below rules out a universal lower bound for these return weights even after imposing distinctness and primitivity. A solution by this route would need a different completion law or additional footprint-sensitive information. No such adaptive kernel or uniform first-kill relative-entropy bound was obtained.

The averaged return can be arbitrarily small despite full deletion

There is no universal positive lower bound for the return weight in (5), even for a primitive zero-residue system with canonical private integer witnesses. Fix an integer $Q\ge2$, and put

$$P_Q=\{q\ge Q:\text{every proper divisor of }q\text{ is below }Q\}.$$

This is primitive. Every integer $n\ge Q$ has a smallest divisor at least $Q$, which belongs to $P_Q$. Thus the zero classes at these distinct moduli, activated at $n\ge q$, delete exactly $[Q,\infty)\cap\mathbb N$. Their canonical witnesses $r_q=q$ are globally private, since no other member of $P_Q$ divides $q$. In particular all their full first-kill cells are nonempty.

Nevertheless the left side of (5) is at most $1/(Q-1)$, uniformly in $X$. To see this, sample $n$ harmonically up to $X$, and then $m$ harmonically up to $n$, as in the completion. Its guaranteed-return event is exactly $(n,m)\ge Q$. For every $d\ge1$,

$$\Pr(d\mid m\,\vert\,n) =\frac{H_{\lfloor n/d\rfloor}}{dH_n}\le\frac1d, \qquad \Pr(d\mid n)=\frac{H_{\lfloor X/d\rfloor}}{dH_X}\le\frac1d.$$

Here the first probability is conditional on the sampled integer $n$. Consequently

$$\Pr((n,m)\ge Q) \le\sum_{d=Q}^{\infty}\Pr(d\mid n,\ d\mid m) \le\sum_{d=Q}^{\infty}\frac1{d^2}\le\frac1{Q-1}.$$

The deletion proportion tends to one, whereas its averaged return can therefore be as small as desired. This example does not have a small full Haar footprint, and it is not a counterexample to Problem 25. It rules out deducing the desired tail bound merely by placing a universal lower bound on this return probability, even after passing to primitive moduli or requiring private witnesses.