Curved residue profiles

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This approach uses the real graph of the residue function. Curvature controls compatibility between congruences, without imposing coherence or restricting the prime factors of the moduli. It supplies a new endpoint theorem, not a solution of the full delayed-union problem.

Curved perturbations of every slope

Theorem. Fix $0<\alpha<1$ and $\theta\in[0,1)$, and set

$$a_q=\lfloor\theta q+q^\alpha\rfloor. \tag{1}$$

These residues satisfy $0\le a_q<q$ for all sufficiently large $q$. Let $D$ be a set of distinct moduli, and fix an integer $\ell\ge1$. Suppose the points $r_q=\ell q+a_q$ satisfy

$$r_q\not\equiv a_p\pmod p \qquad(p,q\in D,\ p<q). \tag{2}$$

Then $D$ and $\{r_q:q\in D\}$ have natural density zero.

Consequently, in every singleton system with residues (1), all first kills with bounded quotient $(n-a_q)/q$ form a naturally null set. The same assertion allows arbitrary uniformly bounded integer errors in (1). There is no Diophantine condition on $\theta$.

The earlier slope theorem required a Brjuno hypothesis for natural endpoint nullity and admitted a Liouville obstruction without that hypothesis. Here the sublinear curved term removes that restriction. Neither theorem controls unbounded quotients.

Curvature bounds the dyadic second moment

On Haar space let

$$Y_Q=\sum_{Q\le q<2Q}\mathbf1_{[a_q]_q}.$$

We prove, uniformly in $Q\ge2$ and $\theta$,

$$\mathbb E Y_Q^2\le M_\alpha<\infty. \tag{3}$$

The finitely many initially unnormalized residues cause no difficulty for this full-cylinder assertion.

Consider compatible distinct $q<r$ in the dyadic block and write

$$g=(q,r),\quad q=gu,\quad r=g(u+k),\quad T=Q/g, \quad 1\le k<T.$$

Compatibility and the error of less than one in a difference of floors give

$$\left\|\theta k+\Phi_{g,k}(u)\right\|<1/g, \qquad \Phi_{g,k}(u)=\frac{(g(u+k))^\alpha-(gu)^\alpha}{g}. \tag{4}$$

Put $w=Q^{\alpha-1}$. Since $F(x)=x^\alpha$ has second derivative of constant sign, with $|F''(x)|\asymp_\alpha w/Q$ on $[Q,2Q]$,

$$|\Phi'_{g,k}(u)|\asymp_\alpha kw/T. \tag{5}$$

The allowed interval of $u$ has length at most $T$, and its image has length $O_\alpha(kw)$. Thus (4) is the union of at most $O_\alpha(kw+1)$ intervals, each of length at most $O_\alpha(T/(gkw))$. Its number of integer points is at most

$$O_\alpha\left((kw+1)\left(\frac{T}{gkw}+1\right)\right). \tag{6}$$

This estimate is uniform in the phase $\theta k$.

Summing (6) over $1\le k<T$ bounds the number of candidate pairs for a fixed $g$ by

$$O_\alpha\left(\frac{T^2}{g} +\frac{T\log(2T)}{gw}+wT^2+T\right).$$

Dropping the coprimality condition on $u,u+k$ only increases this count. Every pair of exact gcd $g$ contributes $g/(qr)\le g/Q^2$ to the Haar second moment. Such distinct pairs have $g<Q$. After summing over $g$, including both orders, and adding the diagonal,

$$\mathbb E Y_Q^2 \ll_\alpha 1+w\log(2Q)+\frac{\log^2(2Q)}{Qw}. \tag{7}$$

For example the first term uses $\sum_g g^{-2}<\infty$, and the third uses $\sum_{g\le Q}\log(2Q/g)/g\ll\log^2(2Q)$. Since $Qw=Q^\alpha$ and $w=Q^{\alpha-1}$, the right side is bounded in $Q$. This proves (3).

The two residue coordinates become uniform

For every fixed integer $P\ge1$ and residues $t,b\pmod P$,

$$\#\{Q\le q<2Q:q\equiv t\pmod P,\ a_q\equiv b\pmod P\} =Q/P^2+o_{\alpha,\theta,P}(Q). \tag{8}$$

If $\theta$ is irrational, write $q=t+Pm$ and $y_m=(\theta q+q^\alpha)/P$. For every nonzero integer $h$,

$$e(hy_m)=e(h\theta t/P)e(h\theta m)e(hq^\alpha/P), \qquad e(x)=\exp(2\pi i x).$$

The partial sums of $e(h\theta m)$ are bounded. The last factor has total variation $O_{h,P}(Q^\alpha)$, so Abel summation bounds the exponential sum by $O_{h,\theta,P}(1+Q^\alpha)=o(Q)$. Trigonometric approximation therefore makes $y_m\bmod1$ uniform. The condition $a_q\equiv b\pmod P$ is precisely $\{y_m\}\in[b/P,(b+1)/P)$, proving (8).

If $\theta=A/d$ is rational, refine to $q\equiv r\pmod{dP}$. On this progression, $a_q\bmod P$ equals $\lfloor q^\alpha+Ar/d\rfloor\bmod P$. For fixed $L,M,\kappa$, the inverse-level intervals for $x^\alpha+\kappa$ on $[Q,2Q)$ number $O(Q^\alpha)$, each has length $O(Q^{1-\alpha})$, and the full interval lengths are increasing by convexity of $x^{1/\alpha}$. Abel summation against the mean-zero periodic sequence $\mathbf1_{j\equiv b\ (M)}-1/M$ consequently shows that the levels with index $j\equiv b\pmod M$ have total length $Q/M+O(Q^{1-\alpha})$. The two possibly clipped end intervals have the same error bound. Counting $q\equiv r\pmod L$ on each level incurs a total error $O(Q^\alpha)$. Hence

$$\#\{Q\le q<2Q:q\equiv r\pmod L, \lfloor q^\alpha+\kappa\rfloor\equiv b\pmod M\} =\frac{Q}{LM}+O_{\alpha,L,M,\kappa}(Q^\alpha+Q^{1-\alpha}).$$

Take $L=dP,M=P$ and sum over the $d$ possible $r$ to prove (8).

Transfer through an arbitrary old survivor set

Let $C_q=[a_q]_q$, $U=\bigcup_{q\in D}C_q$, and take a finite prefix $D_0\subset D$. Write

$$U_0=\bigcup_{q\in D_0}C_q,\qquad S=\widehat{\mathbb Z}\setminus U_0,\qquad \tau=\mu(U\setminus U_0).$$

Let $P$ be an old period and identify $S$ with its surviving residues modulo $P$. If $S$ is empty, (2) allows no later selected endpoints. Otherwise, for $q>\max D_0$, define

$$d=(q,P),\qquad f(q)=\frac{|S\cap[a_q]_d|}{P/d}.$$

Here $f(q)>0$ for selected $q$, by (2), and CRT gives $\mu(C_q\cap S)=f(q)/q$.

For fixed $t=q\bmod P$, put $d=(t,P)$ and define $f(t,b)$ in the same way. Translation by $\ell t$ preserves every fiber modulo $d$, so exact counting gives

$$\sum_{\substack{b\bmod P\\b+\ell t\in S}}\frac1{f(t,b)} =\frac Pd\,\#\{\text{nonempty }S\text{-fibers modulo }d\} \le P. \tag{9}$$

Equation (8), with this fixed period, therefore implies

$$\sum_{\substack{Q\le q<2Q\\a_q+\ell q\bmod P\in S}}\frac1{f(q)} \le Q+o(Q). \tag{10}$$

Put $m_Q=|D\cap[Q,2Q)|$ and $Z_Q=\sum_{q\in D\cap[Q,2Q)}\mathbf1_{C_q\cap S}$. Its support has measure at most $\tau$, and (3) gives $\mathbb E Z_Q^2\le M_\alpha$. Twice applying Cauchy--Schwarz yields

$$\begin{aligned} m_Q^2 &\le\left(\sum_{q\in D\cap[Q,2Q)}f(q)\right)(Q+o(Q))\\ &\le2Q(Q+o(Q))\,\mathbb E Z_Q\\ &\le2Q(Q+o(Q))\sqrt{M_\alpha\tau}. \end{aligned}$$

Thus

$$\limsup_{Q\to\infty}\frac{m_Q}{Q} \le\sqrt2\,M_\alpha^{1/4}\tau^{1/4}.$$

Exhausting $D$ by finite prefixes makes $\tau\to0$. Dyadic summation proves $d(D)=0$. Since $r_q\ge q$, the endpoint set is naturally null as well. The constant in this conclusion does not grow with the old period; the limit in $Q$ was taken while that period was fixed.

For actual bounded-quotient first kills, partition by the finitely many values of $\ell=(n-a_q)/q$. At each such point all earlier moduli are active, so (2) holds. A fixed integer shift of all the residues changes neither the compatibility proof nor uniformity in (8). Partitioning by the finitely many error values proves the stated bounded-perturbation extension.

A fixed flat profile can have large natural endpoint spikes

The curved theorem cannot be replaced by an assertion for every smooth increasing sublinear profile. In fact the fixed choice

$$a_q=\lfloor\log\log q\rfloor \tag{11}$$

admits a globally private endpoint family of positive upper natural density. Its full delayed union can fail to have natural density, although it has logarithmic density. All selected moduli below are large enough for (11) to be normalized.

Fix $0<\varepsilon<1/3$. At stage $j$, choose an integer $c_j$ exceeding every earlier endpoint, and put

$$Q_j=\left\lfloor\exp(\exp(c_j+1/2))\right\rfloor.$$

Writing $P_{\mathrm{old}}$ for the old period, choose $c_j$ so large that $P_{\mathrm{old}}/Q_j\le2^{-j}$ and

$$\mu(M([Q_j,2Q_j)\cap\mathbb N))\le\varepsilon2^{-j}. \tag{12}$$

This is possible by the uniform multiples-window limit. Throughout the block, $\lfloor\log\log q\rfloor=c_j$ for large $c_j$. Retain those $q\in[Q_j,2Q_j)$ for which $q+c_j$ avoids the earlier full union. Its Haar measure is at most $\varepsilon$, so periodic counting retains at least $(1-\varepsilon-o(1))Q_j$ moduli.

The new endpoints have pairwise differences less than $Q_j$, so are private within their block. They avoid earlier cylinders by selection. Every earlier endpoint is less than $c_j$, the least positive representative of each new cylinder, so the reverse privacy holds too. Also $q\le q+c_j<2q$. Consequently the full endpoint set $R$ satisfies

$$\overline d(R)\ge(1-\varepsilon)/2. \tag{13}$$

Let $D_j$ be the retained moduli and $B_j$ their delayed union. It is exactly $c_j+M(D_j)$. Positive translation and the exact multiples harmonic inequality give, for every $X$,

$$H_{B_j}(X)\le H_{M(D_j)}(X) \le\varepsilon2^{-j}H_{\lfloor X\rfloor}.$$

Thus finite-block approximation proves logarithmic density $\beta=\mu(\bigcup_j[c_j]_{D_j})\le\varepsilon$ for $B=\bigcup_jB_j$, where $[c_j]_{D_j}$ denotes the full block union. But $B$ contains $R$, so its upper natural density is greater than $\beta$ by (13). It has no natural density. Its lower natural density is $\beta$: finite prefixes give the lower bound, and logarithmic density gives the reverse inequality for the lower natural density.

The endpoint set $R$ itself has logarithmic density zero by the summable-footprint endpoint bound. The independent ordered-run criterion also gives logarithmic endpoint nullity for every system with (11). This example is not a counterexample to Problem 25.

The remaining gap

Even in the curved class, its full arithmetic cylinder trace cannot replace its delayed union. For $\theta=0$, $\alpha=1/2$, select $q=m^2$, $m\ge M\ge2$. The canonical residue is $m$, so the full cylinders cover every integer at least $M$ through an inactive root. Their delayed union, however, has natural density at most $\sum_{m\ge M}m^{-2}\le1/(M-1)$, by the summable-moduli theorem. Thus full-trace regularity cannot be assumed in a dynamical argument for these profiles.

The curved theorem proves natural nullity of every fixed quotient range, not tightness uniformly as the quotient bound tends to infinity. It therefore supplies no density theorem for the entire curved-profile union. The flat example shows why natural-density claims require care, while the run criterion gives a different logarithmic endpoint regime. No general endpoint theorem or control of unbounded-quotient first kills has resulted from this approach.