Local energy for the full endpoint class

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This examines the comparison-correlation part of Section 5 and its Section 2 inputs in the preserved source. The source proof is OpenAI's; the independent examination is prepared for Plasma AI. No restriction to the special pair in the signed application is made here. The first endpoint may have arbitrary allowed $\alpha,c_r$, and the second endpoint may have completely different allowed data.

The residual coefficient stays arbitrary

In a comparison pattern all freed labels are big. The number of small marks in every small group therefore remains $\ell\ge1$. Select one designated small mark. If its group has no prime square dividing $w$, removing that prime lowers both $\omega_g$ and the mark count by one. It gives

$$W_{\mathbf t}^{\mathbf b}(w)= \sum_{p_s\in\mathcal P_{g_s},\ p_s\mid w} V_{g_s}^{-1}W_{\mathbf t-\mathbf e_{g_s}}^{\mathbf b}(w/p_s). \tag{F1}$$

There is no factorial multiplier: fixing the designated slot and ordering the other slots gives a bijection. The mark function uses only big labels, so it is unaffected. The discrepancy-bearing endpoint is invariant under removing group primes.

For the excluded square multiples, both sides have fixed logarithmic bounds. Their endpoint-weighted squared error is bounded by

$$Y L^{O(1)}\sum_{p_s\in\mathcal P_{g_s}}p_s^{-2} \ll Y L^{O(1)}e^{-L^a}. \tag{F2}$$

Indeed $w_*\mid w/p_s^2$, so a fixed fourth divisor moment bounds the sum after writing $w=p_s^2v$; every $Y/p_s^2$ is $x^{1+o(1)}$. Parseval disposes of this error in local Fourier energy.

Because $m$ and the long integer $p$ are rough above $W$, neither has a group prime factor. The remaining factorization is exactly $w=mp_s p r$ with residual coefficient

$$c_{r_*}W_{\mathbf t-\mathbf e_{g_s}}^{\mathbf b}(r). \tag{F3}$$

It is bounded by a fixed power of $L$ and has no roughness requirement. This is valid for every original residual sequence; no multiplicativity or regularity was introduced.

At a major-arc denominator $q\le L^{C_{\rm arc}}$, the variables $m,p_s,p$ are units for large $x$. Split $r$ by $d=(r,q)$ and expand the rational phase on the units modulo $q/d$. The argument $mp_s p(r/d)$ is a unit there. Character orthogonality includes all characters of this quotient, including principal and imprimitive ones. The condition $(r,q)=d$ and the factor $\chi(r/d)$ go into (F3). The $m$ factor is still the original restricted $\alpha_m$, multiplied only by its original character, an additional character of bounded logarithmic modulus, and a bounded logarithmic power twist.

After extracting the polynomial logarithmic costs, each sequence has form

$$a_w=\Psi_1(w/Y)w^{iv} \sum_{mp_s p r=w} \alpha_m\chi_m(m)m^{i\sigma_0} b_s(p_s)\mathbf1_{p\in I_p,\ P^-(p)>W} \chi_p(p)p^{i\sigma_1}d_r, \tag{F4}$$

with $|b_s|=O(1)$ and $|d_r|\le L^{O(1)}$. The first coefficient sequence has not been replaced by a convolution or a residue-class restriction. Products of the two $m$ characters have modulus bounded by the product of their moduli, a fixed power of $L$ chosen before $B$.

Fourier-to-Mellin normalization

Let $H=H'/L^{C_{\rm arc}}$, where the product of free-cell sizes satisfies $H'\ge\exp(L^c)$, and put $Y=ZH'=x^{1+o(1)}$. Choose a fixed smooth kernel $K$ of integral one, with small compact support, whose Fourier transform has modulus at least $1/2$ on $[-1,1]$. Plancherel for the smoothed discrete measure gives

$$\int_{|\beta|\le1/H}|\widehat a(\beta)|^2\,d\beta \ll H^{-2}\int \left|\sum_w a_wK((w-v)/H)\right|^2dv. \tag{F5}$$

Only $v\asymp Y$ contributes. On the union of the supports of this kernel and $K(v\log(w/v)/H)$, one has $|w-v|=O(H)$. Taylor's formula therefore bounds the difference of the two arguments by $O(H/Y)$. For each $v$ there are $O(H)$ possible integers $w$; Cauchy–Schwarz and integration over the $O(H)$ possible centers for each $w$ show that the normalized squared error is

$$O\left((H/Y)^2\sum_w|a_w|^2\right). \tag{F6}$$

The logarithmic kernel is used only for $v\asymp Y$, not near zero.

Set $h=H/Y$, $v=Ye^u$, and $P(t)=\sum_w a_ww^{it}$. Logarithmic Fourier inversion expresses the kernel sum as

$$\frac h{2\pi}\int e^{-u}\widehat K(he^{-u}t/(2\pi))v^{-it}P(t)\,dt.$$

Insert a compact cutoff in $u$ that is one on the required range. The Fourier transform in $u$ of its amplitude is bounded, for each fixed $j$, by $C_j(1+|s|)^{-2}(1+|ht|)^{-j}$. Two integrations by parts give the first factor; the Schwartz bounds for $\widehat K$ absorb the factors $ht$ produced by differentiation. Minkowski in $s$ and Plancherel in $u$ yield

$$\int_{|\beta|\le1/H}|\widehat a(\beta)|^2d\beta \ll_j Y^{-1}\int(1+|ht|)^{-j}|P(t)|^2dt +(H/Y)^2\sum_w|a_w|^2. \tag{F7}$$

The normalization is $H^{-2}Yh^2=Y^{-1}$. The ordinary coefficient mean square bounds a dyadic time interval by $Y^{-2}\int_{T'\le|t|\le2T'}|P(t)|^2dt \ll(1+T'/Y)L^{O(1)}$. Thus the tail $|t|>L/h$ contributes $O(YL^{-A})$ to (F7) with large fixed $j$. The error (F6) has that size for every $A$ as well, since $H=x^{o(1)}$.

Split the four factors in (F4) into dyads and separate the smooth product cutoff in logarithmic coordinates. The number of boxes and the integral coefficient mass are fixed powers of $L$. Integration by parts fixes the frequency cutoff exponent before the desired tail saving: with a derivative bound $L^{C(j+1)}$ and truncation $L^{C'}$, $C'>C$, the tail exponent is $C(j+1)-C'(j-b)$ in fixed dimension $b$ and decreases without bound as $j$ increases. This does not change the earlier ideal frequency range.

Passing from $P(t)/Y$ to reciprocal-index coefficients is exact after absorbing $w/Y$ into the smooth product cutoff: $1/Y=(w/Y)/w$. The boundedness of $w/Y$ alone would not justify a termwise replacement inside an oscillatory sum; it is this additional smooth separation that does.

The retained frequency shifts enlarge the time interval only to $|t|\le2L/h$. The task is reduced to bounding

$$\int_{|t|\le2L/h}|P_s(t)P_l(t)R(t)|^2dt, \tag{F8}$$

where $P_s$ is the reciprocal small-prime polynomial at scale $P$, $P_l$ the reciprocal rough-integer polynomial at scale $P'$, and

$$R(t)=\left(\sum_{m\asymp M_0} \alpha_m\chi_m(m)m^{-1+i\sigma_0+it}\right) \left(\sum_{r\asymp R_0}d_rr^{-1+it}\right).$$

Their scales satisfy

$$e^{L^a}/2\le P\le e^{L^b},\quad x^\tau/2\le P'\le x^\eta,\quad PM_0P'R_0\asymp Y. \tag{F9}$$

Every dyad is intersected with its original restrictions. A subproduct at scale $U$ has coefficient-square sum $L^{O(1)}/U$: a convolution of a fixed number of bounded coefficients has at most a fixed divisor power of representations. The number of marked groups was already paid for in the bound on (F3). It is not a growing divisor-moment order.

Exceptional intervals do not cost their cardinality

Outside $E=\{t:|P_s(t)|>L^{-A_s}\}$, the mean square of $P_lR$ gives $L^{-2A_s+O(1)}$, because its scale $Y/P$ exceeds the time interval:

$$\frac{Y/P}{2L/h}=\frac H{2LP}\longrightarrow\infty$$

by $b<c$.

For $j_0=\lfloor\log Y/\log(2P)\rfloor$, $P_s^{j_0}$ has indices at most $Y$. Unique factorization gives at most $j_0!$ ordered prime representations of an index, including repeated primes. Its squared coefficient sum is at most $j_0!(C/P)^{j_0}$. Choose an exceptional point in each occupied unit interval and split interval indices into three classes to obtain unit separation. The coefficient mean square and the elementary unit-interval Sobolev inequality give

$$N_E\ll YL^{O(1)}j_0! (CL^{2A_s}/P)^{j_0}.$$

Here differentiation of the powered polynomial costs only $\log Y$, not an uncontrolled constant depending on $j_0$. Since $j_0=O(L^{1-a})$ and $\log Y-j_0\log P=O(L^b+j_0)$,

$$N_E\le\exp(O(L^b+L^{1-a}\log L))=Y^{o(1)}. \tag{F10}$$

Multiplying a logarithmic saving by (F10) would not suffice. Instead collapse $R$ to coefficients on an interval at scale

$$U\asymp Y/(PP')\ge Y^{1-\eta-o(1)}>Y^{0.6}. \tag{F11}$$

This holds for the whole claimed $\eta<1/4$ class. For $|z|\ll Y$, the Gram kernel of the evaluation map satisfies

$$\left|\sum_{n\asymp U}n^{iz}\right| \ll U/(1+|z|)+Y^{1/2}. \tag{F12}$$

For $1\le|z|\le cU$, the first derivative test applied to $z\log n/(2\pi)$ gives $O(U/|z|)$; negative derivatives use the interval between the integers $-1$ and $0$. For $cU<|z|\ll Y$, the second derivative test gives $O(\sqrt{|z|}+U/\sqrt{|z|})=O(Y^{1/2})$. The range $|z|<1$ is trivial.

Take a maximum of $|R|$ on each closed exceptional unit interval, and again split indices into three separated classes. For each class the absolute Gram row sums are $O(U\log(2Y)+N_EY^{1/2})=O(U\log(2Y))$, by (F10)–(F11). The coefficient norm $L^{O(1)}/U$ then gives

$$\sum_{I\ {\rm occupied}}\sup_{t\in I}|R(t)|^2\ll L^{O(1)}. \tag{F13}$$

This is the estimate that defeats arbitrary concentration of the residual coefficients on the exceptional times.

The long integer factor on every subinterval

The long factor has its stated special form: a character and a power twist on rough integers in an interval. It is not an arbitrary coefficient sequence. Let that interval be any $J'\subset[P',2P']$, even very short, and put $u=t+\sigma$.

For completeness, the needed sieve polynomial can be reconstructed elementarily. In the block $z^{2^{-j-1}}<p\le z^{2^{-j}}$ truncate inclusion–exclusion at the even degree $h_j=(j+1)h_s$. If $t$ bad conditions hold, its value for $t>0$ is $\binom{t-1}{h_j}$; for $t=0$ it is one. It is an upper bound and its overcount is at most $\binom t{h_j+1}$. A telescoping product bounds the full overcount by the sum of one-block corrections. The correction supports are mutually disjoint and disjoint from the upper polynomial's support, so their coefficients have modulus at most one. Their level is at most $z^{4h_s+2}$, because $\sum_{j\ge0}(j+1)2^{-j}=4$.

For independent bad probabilities $g(p)$ satisfying the source's block mass and upper-bound hypotheses, each block avoidance probability is bounded below, while its correction expectation is at most $C^{h_j+1}/(h_j+1)!$. The sum of relative corrections is $O(e^{-h_s})$ for large even $h_s$, uniformly as it grows. Apply this to $g(p)=1/p$; Mertens supplies the bounded block masses. Counting multiples on any interval gives the claimed overcount $O(|I|e^{-h_s}+D)$ with $D=z^{4h_s+2}$.

Now take $z=W$, $h_s=2\lceil T^2\rceil$, and $D=W^{4h_s+2}=x^{o(1)}$. Replacing roughness by the upper polynomial $\sum_{d\mid n}\lambda_d$ costs $O(e^{-h_s}+D/P')$ after reciprocal normalization, also on a subinterval because the overcount is nonnegative. Further,

$$\sum_d|\lambda_d|/d \le\prod_{p\le W}(1+1/p)\ll\sqrt L. \tag{F14}$$

Let $q'$ be the character modulus. Terms with $(d,q')>1$ vanish. For the others write $n=d(q'y+a)$ in each residue class. The new scale is $N=P'/(dq')\ge x^{\tau/2}$ eventually. A sufficiently large fixed $B_0>C+1$ ensures $|u|\ge L^{B_0}/2$ when $|t|\ge L^{B_0}$.

The phase $u\log(q'y+a)/(2\pi)$ has $k$th derivative of constant sign and magnitude comparable to $|u|N^{-k}$. If $|u|\le cN$, the monotone first derivative test gives $O(N/|u|)$ uniformly on every subinterval. Otherwise the overlapping ranges

$$N^{k-2+\varepsilon}\le|u|\le N^{k-\varepsilon},\qquad \varepsilon=1/10,\quad 2\le k\le K=\lceil4/\tau\rceil+3, \tag{F15}$$

cover $cN<|u|\le x^2$. For $k=2$ use the second derivative test. For $k>2$, difference $k-2$ times with positive shifts at most $H_0=\lfloor N^{\varepsilon/(2k)}\rfloor$. The resulting second derivative is an iterated integral of the $k$th derivative and lies, up to fixed factors, between $N^{-2+\varepsilon}$ and $N^{-\varepsilon/2}$. The second derivative test on any intersection of the shifted supports gives $O(N^{1-\varepsilon/4})$.

To see the short-interval uniformity, extend the original phase sequence by zero within a containing interval of length $O(N)$. Averaging $H_0$ translates and applying Cauchy–Schwarz gives a normalized recurrence $B_j^2\ll H_0^{-1}+B_{j+1}$. Every correlation support is again an interval, and all shifts are $o(N)$. Iteration yields $B_0\ll N^{-\delta_k}$ with $\delta_k=\varepsilon/(k2^{k-1})$. Thus, with a fixed positive $\delta$ depending only on $\tau$,

$$\left|\sum_y \exp(iu\log(q'y+a))\right| \ll N(|u|^{-1}+N^{-\delta}) \tag{F16}$$

on every subinterval. The same proof treats negative $u$ by absolute derivative sizes. Partial summation of the reciprocal weight and summing the $q'$ classes give the bound $d^{-1}(|u|^{-1}+x^{-\tau\delta/2})$. By (F14), the long factor is at most

$$O(e^{-h_s}+D/P')+ O\bigl(\sqrt L(L^{-B_0}+x^{-\tau\delta/2})\bigr). \tag{F17}$$

Taking also $B_0>A_l+2$ proves any prescribed $L^{-A_l}$ saving. No lower bound on the length of $J'$ was imposed. A one-integer interval, for example, is already smaller than every fixed negative logarithmic power after its reciprocal normalization.

Closing the time split and choosing the discrepancy exponent

On the exceptional set with $|t|\ge L^{B_0}$, the actual retained range obeys $|t+\sigma_1|<x^2$. The small-prime harmonic sum is bounded. Combining (F13) and (F17), rather than multiplying by $N_E$, bounds this part of (F8) by $L^{-2A_l+O(1)}$.

For $|t|\le L^{B_0}$, use the original discrepancy on the $m$ factor in $R$. Its support is an interval inside $[1,x^2]$, since $m\ll Y=x^{1+o(1)}$. Its character modulus and power twist have fixed logarithmic exponents determined earlier. With $B$ above those exponents, the factor is at most $L^{-B}$. All other harmonic sums have fixed logarithmic bounds. This part is therefore $L^{-2B+B_0+O(1)}$.

For constants $K_i$ fixed independently of $B$ and proxy precision, the three ranges give

$$\int_{|t|\le2L/h}|P_sP_lR|^2dt \ll L^{-2A_s+K_1}+L^{-2A_l+K_2} +L^{-2B+B_0+K_3}+o(L^{-Q}). \tag{F18}$$

For a desired $Q$, first fix $A_s,A_l$, then $B_0$ using the long factor estimate, and finally $B$ for the low-time saving and the modulus and frequency requirements. This gives (F8) with arbitrary logarithmic saving, and (F7) gives local Fourier energy $O(YL^{-A})$ after paying the earlier separation costs.

This derivation uses no relation between the two endpoints, no roughness of $c_r$, no regularity of $\alpha$, no minimum interval length for $I_p$, and no information about proxy cells. Restricting $\alpha$ to another interval preserves its discrepancy exactly because intersections of intervals are intervals. Conjugation preserves it by $(\chi,u)\mapsto(\overline\chi,-u)$. Additional fixed twists are handled inside the endpoint representation by the final choice of $B$; no unchanged-$B$ stability under arbitrary extra twists is asserted.

No computational experiment is used in this derivation.