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Math-661-A5 · revision r1 · on Math-661

gpt pro 5.4 attempt on Math-661

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Contents
  1. Problem 66 --- Round 4 (continuation & spectral-rigidity mode)
  2. 1) ROUND-4 OBJECTIVE
  3. 2) ROUND-3 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-4)
  5. 4) ATTACK PLAN (ROUND-4)
  6. 5) WORK (ROUND-4)
  7. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL (EXACTLY ONE)
  9. 8) COMPLETION ESTIMATE (MANDATORY)
  10. 9) REFERENCES

Problem 66 --- Round 4 (continuation & spectral-rigidity mode)

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction direction. Round 3 proved that if \[ \frac{r(n)}{\log n}\to L\in(0,\infty),\qquad r(n):=(1_A\ast 1_A)(n), \] then $A$ is asymptotically equidistributed in every fixed residue class. The exact gap left after Round 3 is that this only controls rational Fourier modes (roots of unity). The most promising next step is therefore to pass from rational frequencies to all fixed frequencies on the unit circle.

In this round I prove a strictly stronger rigidity statement: the twisted Laplace transform \[ G_\theta(t):=\sum_{n\ge 1}1_A(n)e^{-tn}e^{in\theta} \] is lower-order for every fixed $\theta\notin 2\pi\mathbb Z$, and in fact this decay is uniform away from an $o(1)$-arc around $\theta=0$. As a consequence, the exponentially weighted measures supported on $\{n\alpha\}$ with $n\in A$ converge to Haar measure on the orbit closure of $\alpha$; in particular, for irrational $\alpha$, the set $A$ is Abel-weightedly equidistributed modulo $1$ along the rotation $n\mapsto n\alpha$.

This is a genuine extension of Round 3 from finite cyclic Fourier analysis to the full torus.

2) ROUND-3 FOUNDATION USED

I use the following previously established results as black boxes.
  • Round 2 Lemma 66.3: \[ \sum_{n\ge2}(\log n)e^{-tn}\sim \frac1t\log\frac1t. \]
  • Round 2 Lemma 66.4: with \[ F(t):=\sum_{n\ge 2} r(n)e^{-tn}, \] we have \[ F(t)\sim \frac{L}{t}\log\frac1t\qquad(t\to 0^+). \]
  • Round 2 Lemma 66.5: with \[ G(t):=\sum_{n\ge 1}1_A(n)e^{-tn}, \] we have \[ G(t)\sim \sqrt{L}\,t^{-1/2}\sqrt{\log\tfrac1t}. \]
  • Round 3 Theorem 66.10 and Corollary 66.11: every fixed modulus $q$ is equidistributed, equivalently every nontrivial rational Fourier mode is asymptotically negligible in natural counting. I use this as the structural benchmark that the present round must strictly strengthen.

3) NEW INSIGHT / TOOL (ROUND-4)

New tool: uniform summation-by-parts for twisted Laplace sums.

For \[ S_\theta(t):=\sum_{n\ge 2}(\log n)e^{-tn}e^{in\theta}, \] I prove the quantitative bound \[ S_\theta(t)\ll \frac{\log(1/t)}{\operatorname{dist}(\theta,2\pi\mathbb Z)} \] uniformly for $\theta\notin 2\pi\mathbb Z$. Since the main untwisted size is \[ \sum_{n\ge2}(\log n)e^{-tn}\asymp \frac1t\log\frac1t, \] this shows that twisting by any angle separated from $0$ by more than $\gg t$ forces a power-sized drop in the Laplace transform.

Inserted into the identity \[ \sum_{n\ge2} r(n)e^{-tn}e^{in\theta}=G_\theta(t)^2, \] this gives a uniform spectral exclusion principle: the only place where the generating function can have the principal singular size is inside a vanishing arc around $z=1$.

4) ATTACK PLAN (ROUND-4)

The remaining gap after Round 3 is:
Round 3 controls only rational phases $e^{2\pi i m/q}$; it does not rule out a persistent irrational or quasiperiodic bias of $1_A$.

I address exactly that gap by proving the following claims.

  1. For the model coefficients $(\log n)$, twisted Laplace sums at angle $\theta$ are only of size $O\big(\log(1/t)/\delta\big)$, where $\delta=\operatorname{dist}(\theta,2\pi\mathbb Z)$.
  2. Therefore the twisted Laplace transform of the representation function, \[ R_\theta(t):=\sum_{n\ge2}r(n)e^{-tn}e^{in\theta}, \] is $o\big((1/t)\log(1/t)\big)$ uniformly whenever $\delta\gg t$.
  3. Since $R_\theta(t)=G_\theta(t)^2$, the twisted transform $G_\theta(t)$ is $o\big(t^{-1/2}\sqrt{\log(1/t)}\big)$ uniformly away from a $t$-scale neighborhood of $0$.
  4. Normalizing by the untwisted mass $G(t)$ then yields convergence of the exponentially weighted orbital measures \[ \mu_t^{(\alpha)}:=\frac1{G(t)}\sum_{n\ge1}1_A(n)e^{-tn}\,\delta_{\{n\alpha\}} \] to Haar measure on the compact subgroup generated by $\alpha$.

This strictly strengthens Round 3: rational frequencies become a special case, while irrational frequencies are genuinely new.

5) WORK (ROUND-4)

Throughout, write \[ a_n:=1_A(n),\qquad r(n):=\sum_{k=1}^{n-1}a_ka_{n-k}, \] and assume \begin{equation} \label{wsref-0-65-71-3a-52-34-2d-61-73-73-75-6d-70-74-69-6f-6e} \frac{r(n)}{\log n}\to L\in(0,\infty). \end{equation} Also put \[ M(t):=\frac1t\log\frac1t\qquad(t\in(0,e^{-1}]). \]

Lemma 66.12 (uniform twisted model bound). Let \[ S_\theta(t):=\sum_{n\ge2}(\log n)e^{-tn}e^{in\theta} \qquad(t>0,\ \theta\in\mathbb R), \] and let \[ \delta(\theta):=\operatorname{dist}(\theta,2\pi\mathbb Z)\in[0,\pi]. \] Then for every $t\in(0,e^{-1}]$ and every $\theta\notin 2\pi\mathbb Z$, \[ S_\theta(t)\ll \frac{\log(1/t)}{\delta(\theta)}, \] with an absolute implied constant.

Proof: Fix $t$ and $\theta\notin 2\pi\mathbb Z$, and set \[ g_t(n):=(\log n)e^{-tn}. \] Let \[ B_\theta(N):=\sum_{n=2}^{N} e^{in\theta}. \] Since this is a finite geometric sum, \[ |B_\theta(N)|\le \frac{2}{|1-e^{i\theta}|} =\frac{1}{|\sin(\theta/2)|} \ll \frac{1}{\delta(\theta)}, \] because $|\sin u|\ge 2|u|/\pi$ for $|u|\le \pi/2$ and $\delta(\theta)\in[0,\pi]$.

Now summation by parts gives, for every $N\ge2$, \[ \sum_{n=2}^{N} e^{in\theta}g_t(n) =B_\theta(N)g_t(N)-\sum_{n=2}^{N-1} B_\theta(n)\big(g_t(n+1)-g_t(n)\big). \] Since $g_t(N)\to0$ as $N\to\infty$, letting $N\to\infty$ yields \[ |S_\theta(t)| \ll \frac{1}{\delta(\theta)}\sum_{n\ge2}|g_t(n+1)-g_t(n)|. \] By the fundamental theorem of calculus, \[ |g_t(n+1)-g_t(n)|\le \int_n^{n+1}|g_t'(x)|\,dx, \] so \[ \sum_{n\ge2}|g_t(n+1)-g_t(n)|\le \int_1^{\infty}|g_t'(x)|\,dx. \] As in Round 2 Lemma 66.3, \[ g_t'(x)=\Big(\frac1x-t\log x\Big)e^{-tx}, \] and therefore \[ \int_1^{\infty}|g_t'(x)|\,dx \le \int_1^{\infty}\Big(\frac1x+t\log x\Big)e^{-tx}\,dx \ll \log\frac1t. \] Combining the estimates gives \[ S_\theta(t)\ll \frac{\log(1/t)}{\delta(\theta)}, \] as required. ∎

Proposition 66.13 (uniform twisted cancellation for the representation transform). Define \[ R_\theta(t):=\sum_{n\ge2} r(n)e^{-tn}e^{in\theta} \qquad(t>0,\ \theta\in\mathbb R). \] Let $\eta:(0,e^{-1}]\to(0,\pi]$ satisfy \[ \frac{\eta(t)}{t}\to\infty\qquad(t\to0^+). \] Then \[ \sup_{\delta(\theta)\ge \eta(t)} \frac{|R_\theta(t)|}{M(t)}\to 0 \qquad(t\to0^+). \] In particular, for every fixed $\theta\notin2\pi\mathbb Z$, \[ R_\theta(t)=o\!\left(\frac1t\log\frac1t\right). \]

Proof: Fix $\varepsilon\in(0,1)$. By (eq:R4-assumption), there exists $N_\varepsilon$ such that for all $n\ge N_\varepsilon$, \[ |r(n)-L\log n|\le \varepsilon\log n. \] Split \[ R_\theta(t) =\sum_{2\le n< N_\varepsilon} r(n)e^{-tn}e^{in\theta} +L\sum_{n\ge N_\varepsilon}(\log n)e^{-tn}e^{in\theta} +\sum_{n\ge N_\varepsilon}\big(r(n)-L\log n\big)e^{-tn}e^{in\theta}. \] Hence, uniformly in $\theta$, \[ |R_\theta(t)| \le O_\varepsilon(1) +L\left|\sum_{n\ge N_\varepsilon}(\log n)e^{-tn}e^{in\theta}\right| +\varepsilon\sum_{n\ge N_\varepsilon}(\log n)e^{-tn}. \] By Lemma 66.12, deleting finitely many terms changes the twisted sum by $O_\varepsilon(1)$, so \[ \sup_{\delta(\theta)\ge \eta(t)} \left|\sum_{n\ge N_\varepsilon}(\log n)e^{-tn}e^{in\theta}\right| \ll \frac{\log(1/t)}{\eta(t)}. \] Also, by Round 2 Lemma 66.3, \[ \sum_{n\ge N_\varepsilon}(\log n)e^{-tn} \sim \frac1t\log\frac1t = M(t). \] Therefore \[ \sup_{\delta(\theta)\ge \eta(t)} |R_\theta(t)| \le O_\varepsilon(1)+O\!\left(\frac{\log(1/t)}{\eta(t)}\right)+\varepsilon(1+o(1))M(t). \] Divide by $M(t)=(1/t)\log(1/t)$: \[ \sup_{\delta(\theta)\ge \eta(t)} \frac{|R_\theta(t)|}{M(t)} \le o(1)+O\!\left(\frac{t}{\eta(t)}\right)+\varepsilon(1+o(1)). \] Since $\eta(t)/t\to\infty$, the middle term tends to $0$. Thus \[ \limsup_{t\to0^+} \sup_{\delta(\theta)\ge \eta(t)} \frac{|R_\theta(t)|}{M(t)} \le \varepsilon. \] Because $\varepsilon>0$ is arbitrary, the stated uniform $o(M(t))$ follows.

For fixed $\theta\notin 2\pi\mathbb Z$, choose for instance $\eta(t)=\tfrac12\delta(\theta)$ for all sufficiently small $t$; then $\eta(t)/t\to\infty$, so the pointwise statement follows. ∎

Theorem 66.14 (uniform spectral exclusion away from $1$). Let \[ G_\theta(t):=\sum_{n\ge1} a_n e^{-tn}e^{in\theta} \qquad(t>0,\ \theta\in\mathbb R). \] Then \[ R_\theta(t)=G_\theta(t)^2 \qquad\text{for all }t>0,\ \theta\in\mathbb R. \] Moreover, for every function $\eta(t)$ with $\eta(t)/t\to\infty$, \[ \sup_{\delta(\theta)\ge \eta(t)} \frac{|G_\theta(t)|}{t^{-1/2}\sqrt{\log(1/t)}}\to0 \qquad(t\to0^+). \] In particular, for every fixed $\theta\notin 2\pi\mathbb Z$, \[ G_\theta(t)=o\!\left(t^{-1/2}\sqrt{\log\tfrac1t}\right). \]

Proof: The convolution identity is immediate from absolute convergence: \[ G_\theta(t)^2 =\sum_{k,\ell\ge1} a_ka_\ell e^{-t(k+\ell)}e^{i(k+\ell)\theta} =\sum_{n\ge2}\Big(\sum_{k=1}^{n-1}a_ka_{n-k}\Big)e^{-tn}e^{in\theta} =R_\theta(t). \] Hence \[ |G_\theta(t)|^2=|R_\theta(t)|. \] By Proposition 66.13, \[ \sup_{\delta(\theta)\ge \eta(t)} |R_\theta(t)| = o(M(t)). \] Taking square roots gives \[ \sup_{\delta(\theta)\ge \eta(t)} |G_\theta(t)| = o\big(M(t)^{1/2}\big) =o\!\left(t^{-1/2}\sqrt{\log\tfrac1t}\right), \] which is exactly the claim. ∎

Corollary 66.15 (Abel-weighted orbit equidistribution on the circle). For $\alpha\in\mathbb R$, define a probability measure on $\mathbb T:=\mathbb R/\mathbb Z$ by \[ \mu_t^{(\alpha)}:=\frac{1}{G(t)}\sum_{n\ge1} a_n e^{-tn}\,\delta_{\{n\alpha\}} \qquad(t>0), \] where $\{x\}$ denotes the class of $x$ mod $1$. Let \[ H_\alpha:=\overline{\{n\alpha\pmod 1:n\in\mathbb Z\}}\le \mathbb T \] be the closed subgroup generated by $\alpha$, and let $m_{H_\alpha}$ denote Haar probability measure on $H_\alpha$.

Then \[ \mu_t^{(\alpha)}\Longrightarrow m_{H_\alpha} \qquad(t\to0^+) \] in the weak-* topology. Equivalently, for every continuous $1$-periodic function $f$, \[ \frac{\sum_{n\ge1} a_n e^{-tn}f(n\alpha)}{\sum_{n\ge1} a_n e^{-tn}} \to \int_{H_\alpha} f\,dm_{H_\alpha}. \] In particular:

  • if $\alpha\notin\mathbb Q$, then $H_\alpha=\mathbb T$ and \[ \frac{\sum_{n\ge1} a_n e^{-tn}f(n\alpha)}{\sum_{n\ge1} a_n e^{-tn}} \to \int_0^1 f(x)\,dx; \]
  • if $\alpha=p/q\in\mathbb Q$ in lowest terms, then $H_\alpha=\{0,1/q,\dots,(q-1)/q\}$ and the limit measure is uniform on this $q$-point orbit.

Proof: For each integer $m$, the $m$-th Fourier coefficient of $\mu_t^{(\alpha)}$ is \[ \widehat\mu_t^{(\alpha)}(m) :=\int_{\mathbb T} e^{2\pi i mx}\,d\mu_t^{(\alpha)}(x) =\frac{1}{G(t)}\sum_{n\ge1} a_n e^{-tn}e^{2\pi i mn\alpha} =\frac{G_{2\pi m\alpha}(t)}{G(t)}. \] If $m\alpha\in\mathbb Z$, then $G_{2\pi m\alpha}(t)=G(t)$, so \[ \widehat\mu_t^{(\alpha)}(m)=1. \] If $m\alpha\notin\mathbb Z$, then $2\pi m\alpha\notin2\pi\mathbb Z$, and Theorem 66.14 gives \[ \widehat\mu_t^{(\alpha)}(m)=\frac{G_{2\pi m\alpha}(t)}{G(t)}\to0, \] because Round 2 Lemma 66.5 gives $G(t)\sim \sqrt{L}\,t^{-1/2}\sqrt{\log(1/t)}$.

Thus, for every $m\in\mathbb Z$, \[ \widehat\mu_t^{(\alpha)}(m)\to \begin{cases} 1,& m\in H_\alpha^{\perp},\\ 0,& m\notin H_\alpha^{\perp}, \end{cases} \] where $H_\alpha^{\perp}=\{m\in\mathbb Z: e^{2\pi i mx}=1\ \forall x\in H_\alpha\}$ is the annihilator of $H_\alpha$. These are exactly the Fourier coefficients of Haar probability measure on $H_\alpha$. Since trigonometric polynomials are dense in $C(\mathbb T)$, the Fourier coefficients determine the weak-* limit, proving \[ \mu_t^{(\alpha)}\Longrightarrow m_{H_\alpha}. \] The stated formulation for continuous $f$ is exactly the definition of weak-* convergence. The irrational and rational cases are the standard descriptions of $H_\alpha$. ∎

6) ADVERSARIAL VERIFICATION

  • Interaction with Round 3. Round 3 handled rational frequencies by discrete Fourier analysis modulo $q$. The present round does not re-prove that theorem; it extends the cancellation mechanism to all real frequencies via a different summation-by-parts argument. Rational phases become a special case of Theorem 66.14.
  • Uniformity near $\theta=0$. The bound in Lemma 66.12 necessarily degenerates like $1/\delta(\theta)$. Accordingly, Theorem 66.14 only claims uniform decay when $\delta(\theta)\gg t$; no false uniformity across a $t$-scale arc around $0$ is asserted.
  • Ordered convolution. The identity $R_\theta(t)=G_\theta(t)^2$ depends on the ordered representation function. For unordered representations the transform would differ by a diagonal correction, so the present constants and formulas are specific to the ordered problem, consistently with earlier rounds.
  • Complex square-root issue. No branch choice is used. We only infer \[ |G_\theta(t)|^2=|R_\theta(t)|, \] so the uniform $o$-estimate follows from absolute values alone.
  • Haar-limit corollary. The proof uses only Fourier coefficients of probability measures on the compact group $\mathbb T$ and density of trigonometric polynomials in $C(\mathbb T)$. No unjustified de-Abelization is claimed.
  • What is and is not proved. Corollary 66.15 is an Abel-weighted equidistribution theorem. It does not imply that \[ \frac{1}{A(x)}\sum_{\substack{n\le x\\ n\in A}} e^{2\pi i mn\alpha}\to0 \] for irrational $\alpha$; such a Tauberian de-Abelization step remains open here.
  • Consistency with earlier rounds. If $\alpha=p/q$, Corollary 66.15 gives only exponential-weight equidistribution on the $q$-point orbit. Round 3 Theorem 66.10 is stronger in that special case, because it upgrades the conclusion to ordinary counting in each residue class mod $q$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 4 proves a new rigidity theorem beyond Round 3: \[ \sum_{n\ge1}1_A(n)e^{-tn}e^{in\theta} =o\!\left(t^{-1/2}\sqrt{\log\tfrac1t}\right) \qquad(\theta\notin2\pi\mathbb Z), \] and in fact the decay is uniform outside every arc of width $\eta(t)$ with $\eta(t)\gg t$. Equivalently, the principal singular behavior of the generating function is spectrally localized at $z=1$; every other fixed boundary phase is negligible.

As a consequence, for every fixed $\alpha$, the exponentially weighted orbital measures supported on $\{n\alpha:n\in A\}$ converge to Haar measure on the subgroup generated by $\alpha$. In particular, every irrational rotation is uniformly distributed along $A$ in Abel means. This rules out an entire class of constructions with persistent fixed-frequency or quasiperiodic bias.

The original existence/nonexistence problem remains open: I have not produced either a contradiction or a construction of such a set $A$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 63%

9) REFERENCES

  • N. H. Bingham, C. M. Goldie, J. L. Teugels, Regular Variation, Cambridge Univ. Press.
  • W. Rudin, Fourier Analysis on Groups, Wiley.