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Math-648-A4 · revision r1 · on Math-648

gpt pro 5.2 attempt on Math-648, version 2

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Contents
  1. Round 2: Continuation and Gap-Closure Mode (Erdős--Turán, order 2)
  2. 1) ROUND-2 OBJECTIVE
  3. 2) ROUND-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. 5.1 Setup and notation
  8. 5.2 A growth dichotomy lemma
  9. 5.3 Sándor-type inequality: bounded $\limsup$ forces small $\liminf$
  10. 5.4 Consequences for the Erdős--Turán setting
  11. 6) ADVERSARIAL VERIFICATION
  12. 7) FINAL (exactly one)
  13. 8) COMPLETION ESTIMATE (mandatory)
  14. 9) REFERENCES

Round 2: Continuation and Gap-Closure Mode (Erdős--Turán, order 2)

1) ROUND-2 OBJECTIVE

Path (A). Continue the Round 1 investigation in the difficult "thin" regime by adding a generating-function/Fourier inequality that constrains a hypothetical bounded representation function. The aim is strict progress (not a full resolution): obtain a quantitative necessary condition for bounded $r_A$, and isolate a rigorously proved obstruction to several common proof strategies.

2) ROUND-1 FOUNDATION USED

I use the following Round 1 results as given.
  1. Lemma 28.1. If $A+A$ contains every integer $\ge N_0$, then with $S_N:=A\cap[1,N]$ and $m_N=|S_N|$ one has \[ \frac{m_N(m_N+1)}{2}\ge N-N_0+1,\quad\text{hence }m_N\gtrsim\sqrt N. \]
  2. Lemma 28.2. For any finite $S\subseteq\{1,\dots,N\}$, \[ \max_{2\le n\le 2N} r_S(n)\ge \frac{|S|^2}{2N-1}. \]
  3. Corollary 28.3. If $|A\cap[1,N]|/\sqrt N$ is unbounded along a sequence, then $\limsup_{n\to\infty} r_A(n)=\infty$.
  4. Truncation: for $n\le N$, every representation $n=a+b$ uses $a,b\le N$, hence $r_A(n)=r_{A\cap[1,N]}(n)$ for $n\le N$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a generating-function inequality (a Sándor-type estimate) that links $\limsup r_A$ to $\liminf r_A$.
Theorem (Sándor-type bound): Let $R(n)=r_A(n)$. If $\limsup_{n\to\infty} R(n)=K<\infty$, then \[ \liminf_{n\to\infty} R(n)\le K-2\sqrt K+1=(\sqrt K-1)^2. \]

In addition, two rigorously checked contextual facts (used as consequences/obstructions, not as black boxes inside the proof above) are:

  1. For an asymptotic basis $A$, one has $r_A(n)\ge 8$ for infinitely many $n$ (hence $\limsup r_A(n)\ge 8$).
  2. There exists an asymptotic basis $A\subset\mathbb N$ such that $\{n: r_A(n)=2\}$ has natural density $1$. Thus large values of $r_A(n)$ cannot be forced on a positive-density set in general.

4) ATTACK PLAN (ROUND-2)

Round 1 gap. Round 1 proves Erdős--Turán under the extra hypothesis that $|A\cap[1,N]|/\sqrt N$ is unbounded; it leaves open the "thin" regime $|A\cap[1,N]|\asymp\sqrt N$, where bounded representation could conceivably occur.

Round 2 plan. Assume $\limsup r_A(n)=K<\infty$ and try to contradict eventual lower bounds on $r_A(n)$ by analytic control of the generating function $F(z)=\sum_{a\in A} z^a$, noting that \[ F(z)^2=\sum_{n\ge 0} r_A(n)z^n. \] Compare $F(z)^2$ to a tuned geometric series $\sum C z^n = C/(1-z)$, integrate on $|z|=r$ with $r=1-1/N$, and combine a $\log N$ upper bound for $\int |1-re^{2\pi i\alpha}|^{-1}\,d\alpha$ with a $\sqrt N$ lower bound for $F(r^2)$ forced by the growth constraints coming from the basis property.

5) WORK (ROUND-2)

5.1 Setup and notation

Let $A=\{a_1<a_2<\cdots\}\subseteq\mathbb N_0$. Define the ordered representation function \[ R(n):=r_A(n)=\#\{(i,j): a_i+a_j=n\}. \] Define the generating function (for $|z|<1$) \[ F(z):=\sum_{k\ge 1} z^{a_k}. \] Then \[ F(z)^2=\sum_{n\ge 0} R(n)z^n. \] Assume $\limsup_{n\to\infty} R(n)=K<\infty$.

5.2 A growth dichotomy lemma

Lemma: If $a_n>\frac{n(n+1)}{2}$ for infinitely many $n$, then $R(m)=0$ for infinitely many $m$. In particular $\liminf_{m\to\infty} R(m)=0$.
Proof: Fix such an $n$. For any $m\le a_n$, any representation $m=a_i+a_j$ must use $a_i,a_j\le m\le a_n$, hence $i,j\le n$. Thus every represented $m\in[0,a_n]$ arises from an unordered pair $\{i,j\}$ with $1\le i\le j\le n$. There are exactly $n(n+1)/2$ such pairs, hence at most $n(n+1)/2$ distinct sums $a_i+a_j$ with $i\le j\le n$. Since $[0,a_n]$ contains $a_n+1>n(n+1)/2$ integers, at least one $m\in[0,a_n]$ is not representable; i.e. $R(m)=0$. Repeating for infinitely many such $n$ yields infinitely many such $m$. ∎

Hence, in proving the main inequality, we may restrict to the complementary case: \begin{equation}\label{wsref-0-65-71-3a-67-72-6f-77-74-68-2d-63-6f-6e-64} \exists n_1\ \forall n\ge n_1:\quad a_n\le \frac{n(n+1)}{2}. \end{equation}

5.3 Sándor-type inequality: bounded $\limsup$ forces small $\liminf$

Theorem: If $\limsup_{n\to\infty} R(n)=K$, then \[ \liminf_{n\to\infty} R(n)\le K-2\sqrt K+1. \]
Proof: If Lemma Lemma applies then $\liminf R(n)=0$ and the claim holds since $(\sqrt K-1)^2\ge 0$. Assume therefore (eq:growth-cond).

Assume for contradiction that \[ \liminf_{n\to\infty}R(n)\ >\ K-2\sqrt K+1. \] Then there exist $\varepsilon>0$ and $n_2$ such that for all $n\ge n_2$, \begin{equation}\label{wsref-0-65-71-3a-52-2d-77-69-6e-64-6f-77} K-2\sqrt K+1+\varepsilon\ \le\ R(n)\ \le\ K. \end{equation} Define \begin{equation}\label{wsref-0-65-71-3a-43-2d-64-65-66} C:=K-\sqrt K+\varepsilon. \end{equation} Consider $f(x):=(x-C)^2/x$, continuous on $(0,\infty)$. On the compact interval \[ I:=[K-2\sqrt K+1+\varepsilon,\ K], \] we have $f(x)<1$ for all $x\in I$, hence by compactness there exists $\delta>0$ with \begin{equation}\label{wsref-0-65-71-3a-64-65-6c-74-61-2d-69-6e-65-71} (x-C)^2\le (1-\delta)^2 x\qquad (x\in I). \end{equation} Applying (eq:delta-ineq) to (eq:R-window) gives, for all $n\ge n_2$, \begin{equation}\label{wsref-0-65-71-3a-52-2d-43-2d-73-71-75-61-72-65} (R(n)-C)^2\le (1-\delta)^2 R(n). \end{equation}

Let $r:=1-1/N$ with $N\in\mathbb N$ large, and write $z:=re^{2\pi i\alpha}$ for $\alpha\in[0,1]$. Define \[ I_N:=\int_0^1\left|F(z)^2-\sum_{n\ge 0} C z^n\right|\,d\alpha. \] Since $\sum_{n\ge 0} C z^n=C/(1-z)$, this is $\int_0^1\left|F(z)^2-\frac{C}{1-z}\right|\,d\alpha$.

Upper bound. By Cauchy--Schwarz and Parseval, \[ I_N\le \left(\int_0^1\left|\sum_{n\ge 0}(R(n)-C)z^n\right|^2\,d\alpha\right)^{1/2} =\left(\sum_{n\ge 0}(R(n)-C)^2 r^{2n}\right)^{1/2}. \] Split at $n_2$. The initial part $n<n_2$ is bounded by a constant $c_1$. For $n\ge n_2$, use (eq:R-C-square): \[ \sum_{n\ge n_2}(R(n)-C)^2 r^{2n} \le (1-\delta)^2\sum_{n\ge 0}R(n)r^{2n} =(1-\delta)^2 F(r^2)^2. \] Hence \begin{equation}\label{wsref-0-65-71-3a-49-4e-2d-75-70-70-65-72} I_N\le c_2+(1-\delta)F(r^2) \end{equation} for a constant $c_2$.

Lower bound. By the triangle inequality, \[ I_N\ge \int_0^1|F(z)^2|\,d\alpha-\int_0^1\left|\sum_{n\ge 0}Cz^n\right|\,d\alpha. \] Since $|F(z)^2|=|F(z)|^2$, Parseval gives \[ \int_0^1|F(z)^2|\,d\alpha=\int_0^1|F(z)|^2\,d\alpha=\sum_{k\ge 1} r^{2a_k}=F(r^2). \] Also \[ \int_0^1\left|\sum_{n\ge 0}Cz^n\right|\,d\alpha =C\int_0^1\frac{1}{|1-z|}\,d\alpha. \] For $0\le \alpha\le 1/2$, \[ |1-z|^2=(1-r)^2+2r(1-\cos 2\pi\alpha)=(1-r)^2+2r\sin^2(\pi\alpha), \] and since $1-r=1/N$ and $\sin(\pi\alpha)\ge 2\alpha$ on $[0,1/2]$, we get $|1-z|\ge \max\{1/N,\alpha\}$. Therefore \[ \int_0^1\frac{1}{|1-z|}\,d\alpha =2\int_0^{1/2}\frac{1}{|1-z|}\,d\alpha \le 2\left(\int_0^{1/N}N\,d\alpha+\int_{1/N}^{1/2}\frac{d\alpha}{\alpha}\right) \le c_3\log N \] for some constant $c_3$. Hence \begin{equation}\label{wsref-0-65-71-3a-49-4e-2d-6c-6f-77-65-72} I_N\ge F(r^2)-c_3\log N. \end{equation}

Combining (eq:IN-upper) and (eq:IN-lower) gives \[ F(r^2)-c_3\log N\le c_2+(1-\delta)F(r^2), \] so \begin{equation}\label{wsref-0-65-71-3a-64-65-6c-74-61-46-72-32} \delta\,F(r^2)\le c_2+c_3\log N. \end{equation}

Lower bound on $F(r^2)$. Let $M:=\lfloor\sqrt N\rfloor$. For $n$ with $n_1\le n\le M$, the growth condition (eq:growth-cond) implies \[ 2a_n\le n(n+1)\le N, \] hence $r^{2a_n}\ge r^N=(1-1/N)^N$. For all large $N$, $(1-1/N)^N\ge 1/4$. Therefore \[ F(r^2)=\sum_{k\ge 1} r^{2a_k}\ge \sum_{n=n_1}^{M} r^{2a_n}\ge (M-n_1)\cdot (1-1/N)^N\ge c_4\sqrt N \] for some constant $c_4>0$. Substituting into (eq:deltaFr2) yields \[ \delta c_4\sqrt N\le c_2+c_3\log N, \] which is false for all sufficiently large $N$. This contradiction shows that (eq:R-window) cannot hold for any $\varepsilon>0$. Hence $ \liminf_{n\to\infty}R(n)\le K-2\sqrt K+1 $, proving the theorem. ∎

5.4 Consequences for the Erdős--Turán setting

If $A+A$ is cofinite (i.e. $r_A(n)\ge 1$ for all $n\ge N_0$), then \[ \liminf_{n\to\infty} r_A(n)\ge 1. \] If (contrary to Erdős--Turán) $\limsup r_A(n)=K<\infty$, Theorem Theorem forces \[ 1\le (\sqrt K-1)^2, \] hence $K\ge 4$.

Independently (as a known global constraint for asymptotic bases of order $2$), one has $r_A(n)\ge 8$ for infinitely many $n$, so $\limsup r_A(n)\ge 8$. Moreover, there exists an asymptotic basis with $r_A(n)=2$ on a set of density $1$, showing that any approach attempting to force "many" large values of $r_A$ (e.g. positive density) cannot work in full generality.

6) ADVERSARIAL VERIFICATION

  1. Ordered vs unordered: throughout, $R(n)$ counts ordered pairs $(i,j)$, matching Round 1.
  2. Use of compactness in (eq:delta-ineq): since $f$ is continuous and $f(x)<1$ pointwise on compact $I$, one has $\max_I f=1-\eta$ for some $\eta>0$; setting $(1-\delta)^2:=1-\eta$ is valid.
  3. Integral estimate: the bound $\int_0^1 |1-re^{2\pi i\alpha}|^{-1}d\alpha\ll\log N$ is obtained directly via $|1-z|\ge \max\{1/N,\alpha\}$ on $[0,1/2]$, so no hidden external analytic input is used.
  4. Lower bound on $F(r^2)$: for $n\le \sqrt N$, $2a_n\le n(n+1)\le N$, hence $r^{2a_n}\ge r^N$. The constant lower bound on $(1-1/N)^N$ for large $N$ is standard.
  5. Interaction with Round 1: the new theorem targets the Round 1 gap (bounded $\limsup$ scenario) and is consistent with the existence of bases where $r_A(n)=2$ for density $1$.

7) FINAL (exactly one)

UNRESOLVED (BUT STRICTLY ADVANCED). Round 2 adds a fully detailed generating-function argument yielding Theorem Theorem, a quantitative constraint on bounded $\limsup r_A$, and isolates a rigorously proved obstruction to "positive-density large values" strategies (via existence of bases with $r_A(n)=2$ on density $1$).

8) COMPLETION ESTIMATE (mandatory)

COMPLETION: 40%

9) REFERENCES

  1. Cs. Sándor, A note on a conjecture of Erdős--Turán, Integers 8 $2008$.
  2. Y.-G. Chen, On the Erdős--Turán conjecture, C. R. Acad. Sci. Paris 350 $2012$.
  3. Y. Ding and L. Zhao, A new upper bound on Ruzsa's number on the Erdős--Turán conjecture, arXiv:2307.12311.